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• Insane CPU usage in QT 5.0

  - by GravityScore

  I'm having trouble using the QT framework, particularly with the paintEvent of QWidget. I have a QWidget set up, and am overriding the paintEvent of it. I need to render a bunch of rectangles (grid system), 51 by 19, leading to 969 rectangles being drawn. This is done in a for loop. Then I also need to draw an image on each on of these grids. The QWidget is added to a QMainWindow, which is shown. This works nicely, but it's using up 47% of CPU per window open! And I want to allow the user to open multiple windows like this, likey having 3-4 open at a time, which puts the CPU close to 150%. Why does this happen? Here is the paintEvent contents. The JNI calls don't cause the CPU usage, commenting them out doesn't lower it, but commenting out the p.fillRect and Renderer::renderString (which draws the image) lowers the CPU to about 5%. // Background QPainter p(this); p.fillRect(0, 0, this->width(), this->height(), QBrush(QColor(0, 0, 0))); // Lines for (int y = 0; y < Global::terminalHeight; y++) { // Line and color method ID jmethodID lineid = Manager::jenv->GetMethodID(this->javaClass, "getLine", "(I)Ljava/lang/String;"); error(); jmethodID colorid = Manager::jenv->GetMethodID(this->javaClass, "getColorLine", "(I)Ljava/lang/String;"); error(); // Values jstring jl = (jstring) Manager::jenv->CallObjectMethod(this->javaObject, lineid, jint(y)); error(); jstring cjl = (jstring) Manager::jenv->CallObjectMethod(this->javaObject, colorid, jint(y)); error(); // Convert to C values const char *l = Manager::jenv->GetStringUTFChars(jl, 0); const char *cl = Manager::jenv->GetStringUTFChars(cjl, 0); QString line = QString(l); QString color = QString(cl); // Render line for (int x = 0; x < Global::terminalWidth; x++) { QColor bg = Renderer::colorForHex(color.mid(x + color.length() / 2, 1)); // Cell location on widget int cellx = x * Global::cellWidth + Global::xoffset; int celly = y * Global::cellHeight + Global::yoffset; // Background p.fillRect(cellx, celly, Global::cellWidth, Global::cellHeight, QBrush(bg)); // String // Renders the image to the grid Renderer::renderString(p, tc, text, cellx, celly); } // Release Manager::jenv->ReleaseStringUTFChars(jl, l); Manager::jenv->ReleaseStringUTFChars(cjl, cl); }

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• Why wont my while loop take new input (c++)

  - by Van

  I've written a program to get a string input from a user and parse it into tokens and move a robot according to the input. My problem is trying to issue more than one command. The code looks like: void Navigator::manualDrive() { const int bufSize = 42; char uinput[bufSize]; char delim[] = " "; char *token; while(true) { Navigator::parseInstruction(uinput); } } /* parseInstruction(char *c) -- parses cstring instructions received * and moves robot accordingly */ void Navigator::parseInstruction(char * c) { const int bufSize = 42; char uinput[bufSize]; char delim[] = " "; char *token; cout << "Enter your directions below: \n"; cin.ignore(); cin.getline (uinput, bufSize); token=strtok(uinput, delim); if(strcmp("forward", token) == 0) { int inches; token = strtok(NULL, delim); inches = atoi (token); Navigator::travel(inches); } if(strcmp("back",token) == 0) { int inches; token = strtok(NULL, delim); inches = atoi (token); double value = fabs(0.0735 * fabs(inches) - 0.0550); myRobot.backward(1/*speed*/, value/*time*/); } if(strcmp("turn",token) == 0) { int degrees; token = strtok(NULL, delim); if(strcmp("left",token) == 0) { token = strtok(uinput, delim); degrees = atoi (token); double value = fabs(0.0041 * degrees - 0.0523); myRobot.turnLeft(1/*speed*/, value/*time*/); } } if(strcmp("turn",token) == 0) { int degrees; token = strtok(NULL, delim); if(strcmp("right",token) == 0) { token = strtok(uinput, delim); degrees = atoi (token); double value = fabs(0.0041 * degrees - 0.0523); myRobot.turnRight(1/*speed*/, value/*time*/); } } if(strcmp("stop",token) == 0) { myRobot.motors(0,0); } } In the function manualDrive I have a while loop calling the function parseInstruction infinitely. The program outputs "Enter your directions below: " When I give the program instructions it executes them, and then it outputs "enter your directions below: " again and when I input my directions again it does not execute them and outputs "Enter your directions below: " instead. I'm sure this is a very simple fix I'm just very new to c++. So if you could please help me out and tell me why the program only takes the first set of directions. thanks

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• assistance required, hangman game.

  - by Phillip Gibson

  I am making a hangman game and am having trouble with part of it. I have selected a random word from a file, but I want to display the word as a series of undersocres __ and then match the letter chosen to a position in the undersocres. Can anyone help me? cout <<"1. Select to play the game\n"; cout <<"2. Ask for help\n"; cout <<"3. Select to quit the game\n"; cout << "Enter a selection: "; int number; cin >> number; while(number < 1 || number > 3 || cin.fail()) { if(cin.fail()) { cin.sync(); cin.clear(); cout << "You have not entered a number, please enter a menu selection between 1 and 3\n"; cin >> number; } else { cout << "Your selection must be between 1 and 3!\n"; cin >> number; } } switch (number) { case 1: { string word; string name; cout << " Whats your name? "; cin >> name; Player player(); ifstream FileReader; FileReader.open("words.txt"); if(!FileReader.is_open()) cout << "Error"; //this is for the random selection of words srand(time(0)); int randnum = rand()%10+1; for(int counter = 0; counter < randnum; counter++) { getline(FileReader, word, '\n'); } cout << "my word: " << word << "\n"; // get length of word int length; //create for loop for(int i = 0; i < length; i++) cout << "_"; //_ _ _ _ _ SetCursorPos(2,10); FileReader.close(); break;

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• zlib gzgets extremely slow?

  - by monkeyking

  I'm doing stuff related to parsing huge globs of textfiles, and was testing what input method to use. There is not much of a difference using c++ std::ifstreams vs c FILE, According to the documentation of zlib, it supports uncompressed files, and will read the file without decompression. I'm seeing a difference from 12 seconds using non zlib to more than 4 minutes using zlib.h This I've tested doing multiple runs, so its not a disk cache issue. Am I using zlib in some wrong way? thanks #include <zlib.h> #include <cstdio> #include <cstdlib> #include <fstream> #define LENS 1000000 size_t fg(const char *fname){ fprintf(stderr,"\t-> using fgets\n"); FILE *fp =fopen(fname,"r"); size_t nLines =0; char *buffer = new char[LENS]; while(NULL!=fgets(buffer,LENS,fp)) nLines++; fprintf(stderr,"%lu\n",nLines); return nLines; } size_t is(const char *fname){ fprintf(stderr,"\t-> using ifstream\n"); std::ifstream is(fname,std::ios::in); size_t nLines =0; char *buffer = new char[LENS]; while(is. getline(buffer,LENS)) nLines++; fprintf(stderr,"%lu\n",nLines); return nLines; } size_t iz(const char *fname){ fprintf(stderr,"\t-> using zlib\n"); gzFile fp =gzopen(fname,"r"); size_t nLines =0; char *buffer = new char[LENS]; while(0!=gzgets(fp,buffer,LENS)) nLines++; fprintf(stderr,"%lu\n",nLines); return nLines; } int main(int argc,char**argv){ if(atoi(argv[2])==0) fg(argv[1]); if(atoi(argv[2])==1) is(argv[1]); if(atoi(argv[2])==2) iz(argv[1]); }

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• Problems with making a simple UNIX shell

  - by Kodemax

  Hai, I am trying to create a simple shell in UNIX. I read a lot and found that everybody uses the strtok a lot. But i want to do without any special functions. So i wrote the code but i cant seem to get it to work. Can anybody point out what i am doing wrong here? void process(char**); int arg_count; char **splitcommand(char* input) { char temp[81][81] ,*cmdptr[40]; int k,done=0,no=0,arg_count=0; for(int i=0 ; input[i] != '\0' ; i++) { k=0; while(1) { if(input[i] == ' ') { arg_count++; break; } if(input[i] == '\0') { arg_count++; done = 1; break; } temp[arg_count][k++] = input[i++]; } temp[arg_count][k++] = '\0'; if(done == 1) { break; } } for(int i=0 ; i<arg_count ; i++) { cmdptr[i] = temp[i]; cout<<endl; } cout<<endl; } void process(char* cmd[]) { int pid = fork(); if(pid < 0) { cout << "Fork Failed" << endl; exit(-1); } else if( pid == 0) { cout<<endl<<"in pid"; execvp(cmd[0], cmd); } else { wait(NULL); cout << "Job's Done" << endl; } } int main() { cout<<"Welcome to shell !!!!!!!!!!!"<<endl; char input[81]; cin.getline(input,81); splitcommand(input); }

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• Why wont my while loop wont take new input (c++)

  - by Van

  I've written a program to get a string input from a user and parse it into tokens and move a robot according to the input. My problem is trying to issue more than one command. The code looks like: void Navigator::manualDrive() { const int bufSize = 42; char uinput[bufSize]; char delim[] = " "; char *token; while(true) { Navigator::parseInstruction(uinput); } } /* parseInstruction(char *c) -- parses cstring instructions received * and moves robot accordingly */ void Navigator::parseInstruction(char * c) { const int bufSize = 42; char uinput[bufSize]; char delim[] = " "; char *token; cout << "Enter your directions below: \n"; cin.ignore(); cin.getline (uinput, bufSize); token=strtok(uinput, delim); if(strcmp("forward", token) == 0) { int inches; token = strtok(NULL, delim); inches = atoi (token); Navigator::travel(inches); } if(strcmp("back",token) == 0) { int inches; token = strtok(NULL, delim); inches = atoi (token); double value = fabs(0.0735 * fabs(inches) - 0.0550); myRobot.backward(1/*speed*/, value/*time*/); } if(strcmp("turn",token) == 0) { int degrees; token = strtok(NULL, delim); if(strcmp("left",token) == 0) { token = strtok(uinput, delim); degrees = atoi (token); double value = fabs(0.0041 * degrees - 0.0523); myRobot.turnLeft(1/*speed*/, value/*time*/); } } if(strcmp("turn",token) == 0) { int degrees; token = strtok(NULL, delim); if(strcmp("right",token) == 0) { token = strtok(uinput, delim); degrees = atoi (token); double value = fabs(0.0041 * degrees - 0.0523); myRobot.turnRight(1/*speed*/, value/*time*/); } } if(strcmp("stop",token) == 0) { myRobot.motors(0,0); } } In the function manualDrive I have a while loop calling the function parseInstruction infinitely. The program outputs "Enter your directions below: " When I give the program instructions it executes them, and then it outputs "enter your directions below: " again and when I input my directions again it does not execute them and outputs "Enter your directions below: " instead. I'm sure this is a very simple fix I'm just very new to c++. So if you could please help me out and tell me why the program only takes the first set of directions. thanks

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• Maps with a nested vector

  - by wawiti

  For some reason the compiler won't let me retrieve the vector of integers from the map that I've created, I want to be able to overwrite this vector with a new vector. The error the compiler gives me is ridiculous. Thanks for your help!! The compiler didn't like this part of my code: line_num = miss_words[word_1]; Error: [Wawiti@localhost Lab2]$ g++ -g -Wall *.cpp -o lab2 main.cpp: In function ‘int main(int, char**)’: main.cpp:156:49: error: no match for ‘operator=’ in ‘miss_words.std::map<_Key, _Tp, _Compare, _Alloc>::operator[]<std::basic_string<char>, std::vector<int>, std::less<std::basic_string<char> >, std::allocator<std::pair<const std::basic_string<char>, std::vector<int> > > >((*(const key_type*)(& word_1))) = line_num.std::vector<_Tp, _Alloc>::push_back<int, std::allocator<int> >((*(const value_type*)(& line)))’ main.cpp:156:49: note: candidate is: In file included from /usr/lib/gcc/x86_64-redhat->linux/4.7.2/../../../../include/c++/4.7.2vector:70:0, from header.h:19, from main.cpp:15: /usr/lib/gcc/x86_64-redhat-linux/4.7.2/../../../../include/c++/4.7.2/bits/vector.tcc:161:5: note: std::vector<_Tp, _Alloc>& std::vector<_Tp, _Alloc>::operator=(const std::vector<_Tp, _Alloc>&) [with _Tp = int; _Alloc = std::allocator<int>] /usr/lib/gcc/x86_64-redhat-linux/4.7.2/../../../../include/c++/4.7.2/bits/vector.tcc:161:5: note: no known conversion for argument 1 from ‘void’ to ‘const std::vector<int>&’ CODE: map<string, vector<int> > miss_words; // Creates a map for misspelled words string word_1; // String for word; string sentence; // To store each line; vector<int> line_num; // To store line numbers ifstream file; // Opens file to be spell checked file.open(argv[2]); int line = 1; while(getline(file, sentence)) // Reads in file sentence by sentence { sentence=remove_punct(sentence); // Removes punctuation from sentence stringstream pars_sentence; // Creates stringstream pars_sentence << sentence; // Places sentence in a stringstream while(pars_sentence >> word_1) // Picks apart sentence word by word { if(dictionary.find(word_1)==dictionary.end()) { line_num = miss_words[word_1]; //Compiler doesn't like this miss_words[word_1] = line_num.push_back(line); } } line++; // Increments line marker }

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• No Matching Function Error for inserting into a list in c++

  - by Josh Curren

  I am getting an error when I try to insert an item into a list (in C++). The error is that there is no matching function for call to the insert(). I also tried push_front() but got the same error. Here is the error message: main.cpp:38: error: no matching function for call to ‘std::list<Salesperson, std::allocator<Salesperson> >::insert(Salesperson&)’ /usr/lib/gcc/i686-pc-cygwin/4.3.4/include/c++/bits/list.tcc:99: note: candidates are: std::_List_iterator<_Tp> std::list<_Tp, _Alloc>::insert(std::_List_iterator<_Tp>, const _Tp&) [with _Tp = Salesperson, _Alloc = std::allocator<Salesperson>] /usr/lib/gcc/i686-pc-cygwin/4.3.4/include/c++/bits/stl_list.h:961: note: void std::list<_Tp, _Alloc>::insert(std::_List_iterator<_Tp>, size_t, const _Tp&) [with _Tp = Salesperson, _Alloc = std::allocator<Salesperson>] Here is the code: #include <stdlib.h> #include <iostream> #include <fstream> #include <string> #include <list> #include "Salesperson.h" #include "Salesperson.cpp" #include "OrderedList.h" #include "OrderedList.cpp" using namespace std; int main(int argc, char** argv) { cout << "\n------------ Asn 8 - Sales Report ------------" << endl; list<Salesperson> s; int id; string fName, lName; int numOfSales; string year; std::ifstream input("Sales.txt"); while( !std::getline(input, year, ',').eof() ) { input >> id; input >> lName; input >> fName; input >> numOfSales; Salesperson sp = Salesperson( id, fName, lName ); s.insert( sp ); //THIS IS LINE 38 ************************** for( int i = 0; i < numOfSales; i++ ) { double sale; input >> sale; sp.sales.insert( sale ); } } cout << endl; return (EXIT_SUCCESS); }

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• C++ vector of strings, pointers to functions, and the resulting frustration.

  - by Kyle

  So I am a first year computer science student, for on of my final projects, I need to write a program that takes a vector of strings, and applies various functions to these. Unfortunately, I am really confused on how to use pointer to pass the vector from function to function. Below is some sample code to give an idea of what I am talking about. I also get an error message when I try to deference any pointer. thanks. #include <iostream> #include <cstdlib> #include <vector> #include <string> using namespace std; vector<string>::pointer function_1(vector<string>::pointer ptr); void function_2(vector<string>::pointer ptr); int main() { vector<string>::pointer ptr; vector<string> svector; ptr = &svector[0]; function_1(ptr); function_2(ptr); } vector<string>::pointer function_1(vector<string>::pointer ptr) { string line; for(int i = 0; i < 10; i++) { cout << "enter some input ! \n"; // i need to be able to pass a reference of the vector getline(cin, line); // through various functions, and have the results *ptr.pushback(line); // reflectedin main(). But I cannot use member functions } // of vector with a deferenced pointer. return(ptr); } void function_2(vector<string>::pointer ptr) { for(int i = 0; i < 10; i++) { cout << *ptr[i] << endl; } }

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• Passing an array of an array of char to a function

  - by L.A. Rabida

  In my program, I may need to load a large file, but not always. So I have defined: char** largefilecontents; string fileName="large.txt"; When I need to load the file, the program calles this function: bool isitok=LoadLargeFile(fileName,largefilecontents); And the function is: bool LoadLargeFile(string &filename, char ** &lines) { if (lines) delete [] lines; ifstream largeFile; #ifdef LINUX largeFile.open(filename.c_str()); #endif #ifdef WINDOWS largeFile.open(filename.c_str(),ios::binary); #endif if (!largeFile.is_open()) return false; lines=new char *[10000]; if (!lines) return false; largeFile.clear(); largeFile.seekg(ios::beg); for (int i=0; i>-1; i++) { string line=""; getline(largeFile,line); if (largeFile.tellg()==-1) break; //when end of file is reached, tellg returns -1 lines[i]=new char[line.length()]; lines[i]=const_cast<char*>(line.c_str()); cout << lines[i] << endl; //debug output } return true; } When I view the debug output of this function, "cout << lines[i] << endl;", it is fine. But when I then check this in the main program like this, it is all messed up: for (i=0; i<10000; i++) cout << largefilecontents[i] << endl; So within the function LoadLargeFile(), the results are fine, but without LoadLargeFile(), the results are all messed up. My guess is that the char ** &lines part of the function isn't right, but I do not know what this should be. Could someone help me? Thank you in advance!

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• How do I convert Data::Dumper output back into a Perl data structure?

  - by newbee_me

  Hi all! I was wondering if you could shed some lights regarding the code I've been doing for a couple of days. I've been trying to convert a Perl-parsed hash back to XML using the XMLout() and XMLin() method and it has been quite successful with this format. #!/usr/bin/perl -w use strict; # use module use IO::File; use XML::Simple; use XML::Dumper; use Data::Dumper; my $dump = new XML::Dumper; my ( $data, $VAR1 ); Topology:$VAR1 = { 'device' => { 'FOC1047Z2SZ' => { 'ChassisID' => '2009-09', 'Error' => undef, 'Group' => { 'ID' => 'A1', 'Type' => 'Base' }, 'Model' => 'CATALYST', 'Name' => 'CISCO-SW1', 'Neighbor' => {}, 'ProbedIP' => 'TEST', 'isDerived' => 0 } }, 'issues' => [ 'TEST' ] }; # create object my $xml = new XML::Simple (NoAttr=>1, RootName=>'data', SuppressEmpty => 'true'); # convert Perl array ref into XML document $data = $xml->XMLout($VAR1); #reads an XML file my $X_out = $xml->XMLin($data); # access XML data print Dumper($data); print "STATUS: $X_out->{issues}\n"; print "CHASSIS ID: $X_out->{device}{ChassisID}\n"; print "GROUP ID: $X_out->{device}{Group}{ID}\n"; print "DEVICE NAME: $X_out->{device}{Name}\n"; print "DEVICE NAME: $X_out->{device}{name}\n"; print "ERROR: $X_out->{device}{error}\n"; I can access all the element in the XML with no problem. But when I try to create a file that will house the parsed hash, problem arises because I can't seem to access all the XML elements. I guess, I wasn't able to unparse the file with the following code. #!/usr/bin/perl -w use strict; #!/usr/bin/perl # use module use IO::File; use XML::Simple; use XML::Dumper; use Data::Dumper; my $dump = new XML::Dumper; my ( $data, $VAR1, $line_Holder ); #this is the file that contains the parsed hash my $saveOut = "C:/parsed_hash.txt"; my $result_Holder = IO::File->new($saveOut, 'r'); while ($line_Holder = $result_Holder->getline){ print $line_Holder; } # create object my $xml = new XML::Simple (NoAttr=>1, RootName=>'data', SuppressEmpty => 'true'); # convert Perl array ref into XML document $data = $xml->XMLout($line_Holder); #reads an XML file my $X_out = $xml->XMLin($data); # access XML data print Dumper($data); print "STATUS: $X_out->{issues}\n"; print "CHASSIS ID: $X_out->{device}{ChassisID}\n"; print "GROUP ID: $X_out->{device}{Group}{ID}\n"; print "DEVICE NAME: $X_out->{device}{Name}\n"; print "DEVICE NAME: $X_out->{device}{name}\n"; print "ERROR: $X_out->{device}{error}\n"; Do you have any idea how I could access the $VAR1 inside the text file? Regards, newbee_me

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• C++ match string in file and get line number

  - by Corey

  I have a file with the top 1000 baby names. I want to ask the user for a name...search the file...and tell the user what rank that name is for boy names and what rank for girl names. If it isn't in boy names or girl names, it tells the user it's not among the popular names for that gender. The file is laid out like this: Rank Boy-Names Girl-Names 1 Jacob Emily 2 Michael Emma . . . Desired output for input Michael would be: Michael is 2nd most popular among boy names. If Michael is not in girl names it should say: Michael is not among the most popular girl names Though if it was, it would say: Micheal is (rank) among girl names The code I have so far is below.. I can't seem to figure it out. Thanks for any help. #include <iostream> #include <fstream> #include <string> #include <cctype> using namespace std; void find_name(string name); int main(int argc, char **argv) { string name; cout << "Please enter a baby name to search for:\n"; cin >> name; /*while(!(cin>>name)) { cout << "Please enter a baby name to search for:\n"; cin >> name; }*/ find_name(name); cin.get(); cin.get(); return 0; } void find_name(string name) { ifstream input; int line = 0; string line1 = " "; int rank; string boy_name = ""; string girl_name = ""; input.open("/<path>/babynames2004.rtf"); if (!input) { cout << "Unable to open file\n"; exit(1); } while(input.good()) { while(getline(input,line1)) { input >> rank >> boy_name >> girl_name; if (boy_name == name) { cout << name << " is ranked " << rank << " among boy names\n"; } else { cout << name << " is not among the popular boy names\n"; } if (girl_name == name) { cout << name << " is ranked " << rank << " among girl names\n"; } else { cout << name << " is not among the popular girl names\n"; } } } input.close(); }

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• Input not cleared.

  - by SoulBeaver

  As the question says, for some reason my program is not flushing the input or using my variables in ways that I cannot identify at the moment. This is for a homework project that I've gone beyond what I had to do for it, now I just want the program to actually work :P Details to make the finding easier: The program executes flawlessly on the first run through. All throws work, only the proper values( n 0 ) are accepted and turned into binary. As soon as I enter my terminate input, the program goes into a loop and only asks for the termiante again like so: When I run this program on Netbeans on my Linux Laptop, the program crashes after I input the terminate value. On Visual C++ on Windows it goes into the loop like just described. In the code I have tried to clear every stream and initialze every variable new as the program restarts, but to no avail. I just can't see my mistake. I believe the error to lie in either the main function: int main( void ) { vector<int> store; int terminate = 1; do { int num = 0; string input = ""; if( cin.fail() ) { cin.clear(); cin.ignore( numeric_limits<streamsize>::max(), '\n' ); } cout << "Please enter a natural number." << endl; readLine( input, num ); cout << "\nThank you. Number is being processed..." << endl; workNum( num, store ); line; cout << "Go again? 0 to terminate." << endl; cin >> terminate // No checking yet, just want it to work! cin.clear(); }while( terminate ); cin.get(); return 0; } or in the function that reads the number: void readLine( string &input, int &num ) { int buf = 1; stringstream ss; vec_sz size; if( ss.fail() ) { ss.clear(); ss.ignore( numeric_limits<streamsize>::max(), '\n' ); } if( getline( cin, input ) ) { size = input.size(); for( int loop = 0; loop < size; ++loop ) if( isalpha( input[loop] ) ) throw domain_error( "Invalid Input." ); ss << input; ss >> buf; if( buf <= 0 ) throw domain_error( "Invalid Input." ); num = buf; ss.clear(); } }

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• C++, using stack.h read a string, then display it in reverse

  - by user1675108

  For my current assignment, I have to use the following header file, #ifndef STACK_H #define STACK_H template <class T, int n> class STACK { private: T a[n]; int counter; public: void MakeStack() { counter = 0; } bool FullStack() { return (counter == n) ? true : false ; } bool EmptyStack() { return (counter == 0) ? true : false ; } void PushStack(T x) { a[counter] = x; counter++; } T PopStack() { counter--; return a[counter]; } }; #endif To write a program that will take a sentence, store it into the "stack", and then display it in reverse, and I have to allow the user to repeat this process as much as they want. The thing is, I am NOT allowed to use arrays (otherwise I wouldn't need help with this), and am finding myself stumped. To give an idea of what I am attempting, here is my code as of posting, which obviously does not work fully but is simply meant to give an idea of the assignment. #include <iostream> #include <cstring> #include <ctime> #include "STACK.h" using namespace std; int main(void) { auto time_t a; auto STACK<char, 256> s; auto string curStr; auto int i; // Displays the current time and date time(&a); cout << "Today is " << ctime(&a) << endl; s.MakeStack(); cin >> curStr; i = 0; do { s.PushStack(curStr[i]); i++; } while (s.FullStack() == false); do { cout << s.PopStack(); } while (s.EmptyStack() == false); return 0; } // end of "main" **UPDATE This is my code currently #include <iostream> #include <string> #include <ctime> #include "STACK.h" using namespace std; time_t a; STACK<char, 256> s; string curStr; int i; int n; // Displays the current time and date time(&a); cout << "Today is " << ctime(&a) << endl; s.MakeStack(); getline(cin, curStr); i = 0; n = curStr.size(); do { s.PushStack(curStr[i++]); i++; }while(i < n); do { cout << s.PopStack(); }while( !(s.EmptyStack()) ); return 0;

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• map<string, vector<string>> reassignment of vector value

  - by user2950936

  I am trying to write a program that takes lines from an input file, sorts the lines into 'signatures' for the purpose of combining all words that are anagrams of each other. I have to use a map, storing the 'signatures' as the keys and storing all words that match those signatures into a vector of strings. Afterwards I must print all words that are anagrams of each other on the same line. Here is what I have so far: #include <iostream> #include <string> #include <algorithm> #include <map> #include <fstream> using namespace std; string signature(const string&); void printMap(const map<string, vector<string>>&); int main(){ string w1,sig1; vector<string> data; map<string, vector<string>> anagrams; map<string, vector<string>>::iterator it; ifstream myfile; myfile.open("words.txt"); while(getline(myfile, w1)) { sig1=signature(w1); anagrams[sig1]=data.push_back(w1); //to my understanding this should always work, } //either by inserting a new element/key or //by pushing back the new word into the vector<string> data //variable at index sig1, being told that the assignment operator //cannot be used in this way with these data types myfile.close(); printMap(anagrams); return 0; } string signature(const string& w) { string sig; sig=sort(w.begin(), w.end()); return sig; } void printMap(const map& m) { for(string s : m) { for(int i=0;i<m->second.size();i++) cout << m->second.at(); cout << endl; } } The first explanation is working, didn't know it was that simple! However now my print function is giving me: prob2.cc: In function âvoid printMap(const std::map<std::basic_string<char>, std::vector<std::basic_string<char> > >&)â: prob2.cc:43:36: error: cannot bind âstd::basic_ostream<char>::__ostream_type {aka std::basic_ostream<char>}â lvalue to âstd::basic_ostream<char>&&â In file included from /opt/centos/devtoolset-1.1/root/usr/lib/gcc/x86_64-redhat-linux/4.7.2/../../../../include/c++/4.7.2/iostream:40:0, Tried many variations and they always complain about binding void printMap(const map<string, vector<string>> &mymap) { for(auto &c : mymap) cout << c.first << endl << c.second << endl; }

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• How to play audio in Java Application

  - by user577829

  I'm making a java application and I need to play audio. I'm playing mainly small sound files of my cannon firing (its a cannon shooting game) and the projectiles exploding, though I plan on having looping background music. I have found two different methods to accomplish this, but both don't work how I want. The first method is literally a method: public void playSoundFile(File file) {//http://java.ittoolbox.com/groups/technical-functional/java-l/sound-in-an-application-90681 try { //get an AudioInputStream AudioInputStream ais = AudioSystem.getAudioInputStream(file); //get the AudioFormat for the AudioInputStream AudioFormat audioformat = ais.getFormat(); System.out.println("Format: " + audioformat.toString()); System.out.println("Encoding: " + audioformat.getEncoding()); System.out.println("SampleRate:" + audioformat.getSampleRate()); System.out.println("SampleSizeInBits: " + audioformat.getSampleSizeInBits()); System.out.println("Channels: " + audioformat.getChannels()); System.out.println("FrameSize: " + audioformat.getFrameSize()); System.out.println("FrameRate: " + audioformat.getFrameRate()); System.out.println("BigEndian: " + audioformat.isBigEndian()); //ULAW format to PCM format conversion if ((audioformat.getEncoding() == AudioFormat.Encoding.ULAW) || (audioformat.getEncoding() == AudioFormat.Encoding.ALAW)) { AudioFormat newformat = new AudioFormat(AudioFormat.Encoding.PCM_SIGNED, audioformat.getSampleRate(), audioformat.getSampleSizeInBits() * 2, audioformat.getChannels(), audioformat.getFrameSize() * 2, audioformat.getFrameRate(), true); ais = AudioSystem.getAudioInputStream(newformat, ais); audioformat = newformat; } //checking for a supported output line DataLine.Info datalineinfo = new DataLine.Info(SourceDataLine.class, audioformat); if (!AudioSystem.isLineSupported(datalineinfo)) { //System.out.println("Line matching " + datalineinfo + " is not supported."); } else { //System.out.println("Line matching " + datalineinfo + " is supported."); //opening the sound output line SourceDataLine sourcedataline = (SourceDataLine) AudioSystem.getLine(datalineinfo); sourcedataline.open(audioformat); sourcedataline.start(); //Copy data from the input stream to the output data line int framesizeinbytes = audioformat.getFrameSize(); int bufferlengthinframes = sourcedataline.getBufferSize() / 8; int bufferlengthinbytes = bufferlengthinframes * framesizeinbytes; byte[] sounddata = new byte[bufferlengthinbytes]; int numberofbytesread = 0; while ((numberofbytesread = ais.read(sounddata)) != -1) { int numberofbytesremaining = numberofbytesread; sourcedataline.write(sounddata, 0, numberofbytesread); } } } catch (Exception e) { e.printStackTrace(); } } The problem with this is that my entire program stops until the sound file is finished, or at least nearly finished. The second method is this: File file = new File("Launch1.wav"); AudioClip clip; try { clip = JApplet.newAudioClip(file.toURL()); clip.play(); } catch (Exception e) { e.getMessage(); } The problem I have here is that every time the sound file ends early or doesn't play at all depending on where I place the code. Is their any way to play sound without the above mentioned problems? Am I doing something wrong? Any help is greatly appreciated.

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• count specific things within a code in c++

  - by shap

  can anyone help me make this more generalised and more pro? #include <fstream> #include <iostream> #include <string> #include <vector> using namespace std; int main() { // open text file for input: string file_name; cout << "please enter file name: "; cin >> file_name; // associate the input file stream with a text file ifstream infile(file_name.c_str()); // error checking for a valid filename if ( !infile ) { cerr << "Unable to open file " << file_name << " -- quitting!\n"; return( -1 ); } else cout << "\n"; // some data structures to perform the function vector<string> lines_of_text; string textline; // read in text file, line by while (getline( infile, textline, '\n' )) { // add the new element to the vector lines_of_text.push_back( textline ); // print the 'back' vector element - see the STL documentation cout << lines_of_text.back() << "\n"; } cout<<"OUTPUT BEGINS HERE: "<<endl<<endl; cout<<"the total capacity of vector: lines_of_text is: "<<lines_of_text.capacity()<<endl; int PLOC = (lines_of_text.size()+1); int numbComments =0; int numbClasses =0; cout<<"\nThe total number of physical lines of code is: "<<PLOC<<endl; for (int i=0; i<(PLOC-1); i++) //reads through each part of the vector string line-by-line and triggers if the //it registers the "//" which will output a number lower than 100 (since no line is 100 char long and if the function does not //register that character within the string, it outputs a public status constant that is found in the class string and has a huge value //alot more than 100. { string temp(lines_of_text [i]); if (temp.find("//")<100) numbComments +=1; } cout<<"The total number of comment lines is: "<<numbComments<<endl; for (int j=0; j<(PLOC-1); j++) { string temp(lines_of_text [j]); if (temp.find("};")<100) numbClasses +=1; } cout<<"The total number of classes is: "<<numbClasses<<endl;

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• Java: immutability, overuse of stack -- better data structure?

  - by HH

  I overused hashSets but it was slow, then changed to Stacks, speed boost-up. Poly's reply uses Collections.emptyList() as immutable list, cutting out excess null-checkers. No Collections.emptyStack(). Combining the words stack and immutability, from the last experiences, gets "immutable stack" (probably not related to functional prog). Java Api 5 for list interface shows that Stack is an implementing class for list and arraylist, here. The java.coccurrent pkg does not have any immutable Stack data structure. The first hinted of misusing stack. The lack of immutabily in the last and poly's book recommendation leads way to list. Something very primitive, fast, no extra layers, with methods like emptyThing(). Overuse of stack and where I use it DataFile.java: public Stack<DataFile> files; FileObject.java: public Stack<String> printViews = new Stack<String>(); FileObject.java:// private static Stack<Object> getFormat(File f){return (new Format(f)).getFormat();} Format.java: private Stack<Object> getLine(File[] fs,String s){return wF;} Format.java: private Stack<Object> getFormat(){return format;} Positions.java: public static Stack<Integer[]> getPrintPoss(String s,File f,Integer maxViewPerF) Positions.java: Stack<File> possPrint = new Stack<File>(); Positions.java: Stack<Integer> positions=new Stack<Integer>(); Record.java: private String getFormatLine(Stack<Object> st) Record.java: Stack<String> lines=new Stack<String>(); SearchToUser.java: public static final Stack<File> allFiles = findf.getFs(); SearchToUser.java: public static final Stack<File> allDirs = findf.getDs(); SearchToUser.java: private Stack<Integer[]> positionsPrint=new Stack<Integer[]>(); SearchToUser.java: public Stack<String> getSearchResults(String s, Integer countPerFile, Integer resCount) SearchToUser.java: Stack<File> filesToS=Fs2Word.getFs2W(s,50); SearchToUser.java: Stack<String> rs=new Stack<String>(); View.java: public Stack<Integer[]> poss = new Stack<Integer[4]>(); View.java: public static Stack<String> getPrintViewsFileWise(String s,Object[] df,Integer maxViewsPerF) View.java: Stack<String> substrings = new Stack<String>(); View.java: private Stack<String> printViews=new Stack<String>(); View.java: MatchView(Stack<Integer> pss,File f,Integer maxViews) View.java: Stack<String> formatFile; View.java: private Stack<Search> files; View.java: private Stack<File> matchingFiles; View.java: private Stack<String> matchViews; View.java: private Stack<String> searchMatches; View.java: private Stack<String> getSearchResults(Integer numbResults) Easier with List: AllDirs and AllFs, now looping with push, but list has more pow. methods such as addAll [OLD] From Stack to some immutable data structure How to get immutable Stack data structure? Can I box it with list? Should I switch my current implementatios from stacks to Lists to get immutable? Which immutable data structure is Very fast with about similar exec time as Stack? No immutability to Stack with Final import java.io.*; import java.util.*; public class TestStack{ public static void main(String[] args) { final Stack<Integer> test = new Stack<Integer>(); Stack<Integer> test2 = new Stack<Integer>(); test.push(37707); test2.push(80437707); //WHY is there not an error to remove an elment // from FINAL stack? System.out.println(test.pop()); System.out.println(test2.pop()); } }

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• Convert C++Builder AnsiString to std::string via boost::lexical_cast

  - by David Klein

  For a school assignment I have to implement a project in C++ using Borland C++ Builder. As the VCL uses AnsiString for all GUI Components I have to convert all of my std::strings to AnsiString for the sake of displaying. std::string inp = "Hello world!"; AnsiString outp(inp.c_str()); works of course but is a bit tedious to write and code duplication I want to avoid. As we use Boost in other contexts I decided to provide some helper functions go get boost::lexical_cast to work with AnsiString. Here is my implementation so far: std::istream& operator>>(std::istream& istr, AnsiString& str) { istr.exceptions(std::ios::badbit | std::ios::failbit | std::ios::eofbit); std::string s; std::getline(istr,s); str = AnsiString(s.c_str()); return istr; } In the beginning I got Access Violation after Access Violation but since I added the .exceptions() stuff the picture gets clearer. When the conversion is performed I get the following Exception: ios_base::eofbit set [Runtime Error/std::ios_base::failure] Does anyone have an idea how to fix it and can explain why the error occurs? My C++ experience is very limited. The conversion routine the other way round would be: std::ostream& operator<<(std::ostream& ostr,const AnsiString& str) { ostr << (str.c_str()); return ostr; } Maybe someone will spot an error here too :) With best regards! Edit: At the moment I'm using the edited version of Jem, it works in the beginning. After a while of using the programm the Borland Codeguard mentions some pointer arithmetic in already freed regions. Any ideas how this could be related? The Codeguard log (I'm using the german version, translations marked with stars): ------------------------------------------ Fehler 00080. 0x104230 (r) (Thread 0x07A4): Zeigerarithmetik in freigegebenem Speicher: 0x0241A238-0x0241A258. **(pointer arithmetic in freed region)** | d:\program files\borland\bds\4.0\include\dinkumware\sstream Zeile 126: | { // not first growth, adjust pointers | _Seekhigh = _Seekhigh - _Mysb::eback() + _Ptr; |> _Mysb::setp(_Mysb::pbase() - _Mysb::eback() + _Ptr, | _Mysb::pptr() - _Mysb::eback() + _Ptr, _Ptr + _Newsize); | if (_Mystate & _Noread) Aufrufhierarchie: **(stack-trace)** 0x00411731(=FOSChampion.exe:0x01:010731) d:\program files\borland\bds\4.0\include\dinkumware\sstream#126 0x00411183(=FOSChampion.exe:0x01:010183) d:\program files\borland\bds\4.0\include\dinkumware\streambuf#465 0x0040933D(=FOSChampion.exe:0x01:00833D) d:\program files\borland\bds\4.0\include\dinkumware\streambuf#151 0x00405988(=FOSChampion.exe:0x01:004988) d:\program files\borland\bds\4.0\include\dinkumware\ostream#679 0x00405759(=FOSChampion.exe:0x01:004759) D:\Projekte\Schule\foschamp\src\Server\Ansistringkonverter.h#31 0x004080C9(=FOSChampion.exe:0x01:0070C9) D:\Projekte\Schule\foschamp\lib\boost_1_34_1\boost/lexical_cast.hpp#151 Objekt (0x0241A238) [Größe: 32 Byte] war erstellt mit new **(Object was created with new)** | d:\program files\borland\bds\4.0\include\dinkumware\xmemory Zeile 28: | _Ty _FARQ *_Allocate(_SIZT _Count, _Ty _FARQ *) | { // allocate storage for _Count elements of type _Ty |> return ((_Ty _FARQ *)::operator new(_Count * sizeof (_Ty))); | } | Aufrufhierarchie: **(stack-trace)** 0x0040ED90(=FOSChampion.exe:0x01:00DD90) d:\program files\borland\bds\4.0\include\dinkumware\xmemory#28 0x0040E194(=FOSChampion.exe:0x01:00D194) d:\program files\borland\bds\4.0\include\dinkumware\xmemory#143 0x004115CF(=FOSChampion.exe:0x01:0105CF) d:\program files\borland\bds\4.0\include\dinkumware\sstream#105 0x00411183(=FOSChampion.exe:0x01:010183) d:\program files\borland\bds\4.0\include\dinkumware\streambuf#465 0x0040933D(=FOSChampion.exe:0x01:00833D) d:\program files\borland\bds\4.0\include\dinkumware\streambuf#151 0x00405988(=FOSChampion.exe:0x01:004988) d:\program files\borland\bds\4.0\include\dinkumware\ostream#679 Objekt (0x0241A238) war Gelöscht mit delete **(Object was deleted with delete)** | d:\program files\borland\bds\4.0\include\dinkumware\xmemory Zeile 138: | void deallocate(pointer _Ptr, size_type) | { // deallocate object at _Ptr, ignore size |> ::operator delete(_Ptr); | } | Aufrufhierarchie: **(stack-trace)** 0x004044C6(=FOSChampion.exe:0x01:0034C6) d:\program files\borland\bds\4.0\include\dinkumware\xmemory#138 0x00411628(=FOSChampion.exe:0x01:010628) d:\program files\borland\bds\4.0\include\dinkumware\sstream#111 0x00411183(=FOSChampion.exe:0x01:010183) d:\program files\borland\bds\4.0\include\dinkumware\streambuf#465 0x0040933D(=FOSChampion.exe:0x01:00833D) d:\program files\borland\bds\4.0\include\dinkumware\streambuf#151 0x00405988(=FOSChampion.exe:0x01:004988) d:\program files\borland\bds\4.0\include\dinkumware\ostream#679 0x00405759(=FOSChampion.exe:0x01:004759) D:\Projekte\Schule\foschamp\src\Server\Ansistringkonverter.h#31 ------------------------------------------ Ansistringkonverter.h is the file with the posted operators and line 31 is: std::ostream& operator<<(std::ostream& ostr,const AnsiString& str) { ostr << (str.c_str()); **(31)** return ostr; } Thanks for your help :)

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• Referenced vector does not pass through functions

  - by kylepayne

  The referenced vector to functions does not hold the information in memory. Do I have to use pointers? Thanks. #include <iostream> #include <cstdlib> #include <vector> #include <string> using namespace std; void menu(); void addvector(vector<string>& vec); void subvector(vector<string>& vec); void vectorsize(const vector<string>& vec); void printvec(const vector<string>& vec); void printvec_bw(const vector<string>& vec); int main() { vector<string> svector; menu(); return 0; } //functions definitions void menu() { vector<string> svector; int choice = 0; cout << "Thanks for using this program! \n" << "Enter 1 to add a string to the vector \n" << "Enter 2 to remove the last string from the vector \n" << "Enter 3 to print the vector size \n" << "Enter 4 to print the contents of the vector \n" << "Enter 5 ----------------------------------- backwards \n" << "Enter 6 to end the program \n"; cin >> choice; switch(choice) { case 1: addvector(svector); menu(); break; case 2: subvector(svector); menu(); break; case 3: vectorsize(svector); menu(); break; case 4: printvec(svector); menu(); break; case 5: printvec_bw(svector); menu(); break; case 6: exit(1); default: cout << "not a valid choice \n"; // menu is structured so that all other functions are called from it. } } void addvector(vector<string>& vec) { //string line; //int i = 0; //cin.ignore(1, '\n'); //cout << "Enter the string please \n"; //getline(cin, line); vec.push_back("the police man's beard is half-constructed"); } void subvector(vector<string>& vec) { vec.pop_back(); return; } void vectorsize(const vector<string>& vec) { if (vec.empty()) { cout << "vector is empty"; } else { cout << vec.size() << endl; } return; } void printvec(const vector<string>& vec) { for(int i = 0; i < vec.size(); i++) { cout << vec[i] << endl; } return; } void printvec_bw(const vector<string>& vec) { for(int i = vec.size(); i > 0; i--) { cout << vec[i] << endl; } return; }

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• Memory leak in C++ program.

  - by lampshade

  What I have is a very crude linked list..THe problem for me is that I am getting a memory leak in the constructor or main. I think it is the constructor. I have not yet deleted the eventName varaible that I have allocated memory for. Could someone help please? :/ (This is not a homework question) class Event { private: char * eventName ; string userEvent; struct node { node(); node * nextByName; const char * eventName; }; node * headByName; public: Event(const char * eventName, const Date &myDate); Event(); virtual ~Event(); void insert(const char * eventName, const Date &myDate, const Time &myTime); void setEvent(); const char * const getEvent() const { return userEvent.c_str(); }; void displayByName(ostream& out) const; }; Event::Event(const char * eventName, const Date &myDate) : eventName(new char[strlen(eventName)+1]), headByName(NULL), userEvent("") { if (eventName) { size_t length = strlen(eventName) +1; strcpy_s(this->eventName, length, eventName); } else eventName = NULL; } Event::Event() : eventName(NULL), userEvent(NULL), headByName(NULL) { } Event::~Event() { node * temp_node = NULL; node * current_node = headByName; while ( current_node ) { temp_node = current_node->nextByName; delete current_node; current_node = temp_node; } } void Event::insert(const char * eventName, const Date &myDate, const Time &myTime) // when we insert we dont care about the time, just the name and the date { node * current_node = new node(); if ( headByName == NULL ) { headByName = current_node; headByName->eventName = eventName; } else { node * search_node = headByName; node * prev_node = NULL; while ( search_node != NULL ) { prev_node = search_node; search_node = search_node->nextByName; } if ( NULL == prev_node ) { headByName = current_node; } else { prev_node->nextByName = current_node; } current_node->nextByName = search_node; current_node->eventName = eventName ; } } void Event::displayByName(ostream& out) const { cout << "Scheduled Events are: " << endl << endl; node * current_node = headByName; while ( current_node ) { (char*)eventName = (char*)current_node->eventName; out << eventName << endl; current_node = current_node->nextByName; } } Event::node::node() : nextByName(NULL), eventName(NULL) { } void Event::setEvent() { cout << "\n\nEnter a new event! "; cin.getline((char*)userEvent.c_str(), 256); size_t length = strlen(userEvent.c_str()) +1; strcpy_s((char*)this->userEvent.c_str(), length, userEvent.c_str()); } /********************************************************************************* **********************************************************************************/ int main() { Date * dPtr = new Date("March", 21, 2010); // instaintiate our Date class object by allocating default date paramateres. Event * ePtr = new Event("First Day of Spring", *dPtr); Time * tPtr = new Time(10,12,"PM"); cout << "default Time is: " << tPtr << endl; cout << "default Date is: " << dPtr << endl; ePtr->insert("First Day of Spring",*dPtr, *tPtr); ePtr->insert("Valentines Day", Date("February",14,2010), *tPtr); ePtr->insert("New Years Day", Date("Janurary",1,2011), *tPtr); ePtr->insert("St. Patricks Day", Date("March",17,2010), *tPtr); ePtr->displayByName(cout); ePtr->setEvent(); const char * const theEvent = ePtr->getEvent(); dPtr->setDate(); ePtr->insert(theEvent, *dPtr, *tPtr); tPtr->setTime(); cout << "Your event: " << theEvent << " is scheduled for: " << endl << dPtr << "at" << tPtr; ePtr->displayByName(cout); delete tPtr; delete dPtr; delete ePtr; cin.ignore(); return 0; }

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• Loop crashing program having to do with 2D arrays

  - by user450062

  I am creating an encoding program and when I instruct the program to create a 5X5 grid based on the alphabet while skipping over letters that match up to certain pre-defined variables(which are given values by user input during runtime). I have a loop that instructs the loop to keep running until the values that access the array are out of bounds, the loop seems to cause the problem. This code is standardized so there shouldn't be much trouble compiling it in another compiler. Also would it be better to seperate my program into functions? here is the code: #include<iostream> #include<fstream> #include<cstdlib> #include<string> #include<limits> using namespace std; int main(){ while (!cin.fail()) { char type[81]; char filename[20]; char key [5]; char f[2] = "q"; char g[2] = "q"; char h[2] = "q"; char i[2] = "q"; char j[2] = "q"; char k[2] = "q"; char l[2] = "q"; int a = 1; int b = 1; int c = 1; int d = 1; int e = 1; string cipherarraytemplate[5][5]= { {"a","b","c","d","e"}, {"f","g","h","i","j"}, {"k","l","m","n","o"}, {"p","r","s","t","u"}, {"v","w","x","y","z"} }; string cipherarray[5][5]= { {"a","b","c","d","e"}, {"f","g","h","i","j"}, {"k","l","m","n","o"}, {"p","r","s","t","u"}, {"v","w","x","y","z"} }; cout<<"Enter the name of a file you want to create.\n"; cin>>filename; ofstream outFile; outFile.open(filename); outFile<<fixed; outFile.precision(2); outFile.setf(ios_base::showpoint); cin.ignore(std::numeric_limits<int>::max(),'\n'); cout<<"enter your codeword(codeword can have no repeating letters)\n"; cin>>key; while (key[a] != '\0' ){ while(b < 6){ cipherarray[b][c] = key[a]; if ( f == "q" ) { cipherarray[b][c] = f; } if ( f != "q" && g == "q" ) { cipherarray[b][c] = g; } if ( g != "q" && h == "q" ) { cipherarray[b][c] = h; } if ( h != "q" && i == "q" ) { cipherarray[b][c] = i; } if ( i != "q" && j == "q" ) { cipherarray[b][c] = j; } if ( j != "q" && k == "q" ) { cipherarray[b][c] = k; } if ( k != "q" && l == "q" ) { cipherarray[b][c] = l; } a++; b++; } c++; b = 1; } while (c < 6 || b < 6){ if (cipherarraytemplate[d][e] == f || cipherarraytemplate[d][e] == g || cipherarraytemplate[d][e] == h || cipherarraytemplate[d][e] == i || cipherarraytemplate[d][e] == j || cipherarraytemplate[d][e] == k || cipherarraytemplate[d][e] == l){ d++; } else { cipherarray[b][c] = cipherarraytemplate[d][e]; d++; b++; } if (d == 6){ d = 1; e++; } if (b == 6){ c++; b = 1; } } cout<<"now enter some text."<<endl<<"To end this program press Crtl-Z\n"; while(!cin.fail()){ cin.getline(type,81); outFile<<type<<endl; } outFile.close(); } } I know there is going to be some mid-forties guy out there who is going to stumble on to this post, he's have been programming for 20-some years and he's going to look at my code and say: "what is this guy doing".

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• problem with piping in my own implementation of shell

  - by codemax

  Hey guys, i am implementing my own shell. I want to involve piping. i searched here and i got a code. But it is not working.Can any one help me? this is my code #include <sys/types.h> #include <sys/wait.h> #include <sys/ipc.h> #include <fcntl.h> #include <unistd.h> #include <string.h> #include <iostream> #include <cstdlib> using namespace std; char temp1[81][81],temp2[81][81] ,*cmdptr1[40], *cmdptr2[40]; void process(char**,int); int arg_count, count; int arg_cnt[2]; int pip,tok; char input[81]; int fds[2]; void process( char* cmd[])//, int arg_count ) { pid_t pid; pid = fork(); //char path[81]; //getcwd(path,81); //strcat(path,"/"); //strcat(path,cmd[0]); if(pid < 0) { cout << "Fork Failed" << endl; exit(-1); } else if( pid == 0 ) { execvp( cmd[0] , cmd ); } else { wait(NULL); } } void pipe(char **cmd1, char**cmd2) { cout<<endl<<endl<<"in pipe"<<endl; for(int i=0 ; i<arg_cnt[0] ; i++) { cout<<cmdptr1[i]<<" "; } cout<<endl; for(int i=0 ; i<arg_cnt[1] ; i++) { cout<<cmdptr2[i]<<" "; } pipe(fds); if (fork() == 0 ) { dup2(fds[1], 1); close(fds[0]); close(fds[1]); process(cmd1); } if (fork() == 0) { dup2(fds[0], 0); close(fds[0]); close(fds[1]); process(cmd2); } close(fds[0]); close(fds[1]); wait(NULL); } void pipecommand(char** cmd1, char** cmd2) { cout<<endl<<endl; for(int i=0 ; i<arg_cnt[0] ; i++) { cout<<cmd1[i]<<" "; } cout<<endl; for(int i=0 ; i<arg_cnt[1] ; i++) { cout<<cmd2[i]<<" "; } int fds[2]; // file descriptors pipe(fds); // child process #1 if (fork() == 0) { // Reassign stdin to fds[0] end of pipe. dup2(fds[0], STDIN_FILENO); close(fds[1]); close(fds[0]); process(cmd2); // child process #2 if (fork() == 0) { // Reassign stdout to fds[1] end of pipe. dup2(fds[1], STDOUT_FILENO); close(fds[0]); close(fds[1]); // Execute the first command. process(cmd1); } wait(NULL); } close(fds[1]); close(fds[0]); wait(NULL); } void splitcommand1() { tok++; int k,done=0,no=0; arg_count = 0; for(int i=count ; input[i] != '\0' ; i++) { k=0; while(1) { count++; if(input[i] == ' ') { break; } if((input[i] == '\0')) { done = 1; break; } if(input[i] == '|') { pip = 1; done = 1; break; } temp1[arg_count][k++] = input[i++]; } temp1[arg_count][k++] = '\0'; arg_count++; if(done == 1) { break; } } for(int i=0 ; i<arg_count ; i++) { cmdptr1[i] = temp1[i]; } arg_cnt[tok] = arg_count; } void splitcommand2() { tok++; cout<<"count is :"<<count<<endl; int k,done=0,no=0; arg_count = 0; for(int i=count ; input[i] != '\0' ; i++) { k=0; while(1) { count++; if(input[i] == ' ') { break; } if((input[i] == '\0')) { done = 1; break; } if(input[i] == '|') { pip = 1; done = 1; cout<<"PIP"; break; } temp2[arg_count][k++] = input[i++]; } temp2[arg_count][k++] = '\0'; arg_count++; if(done == 1) { break; } } for(int i=0 ; i<arg_count ; i++) { cmdptr2[i] = temp2[i]; } arg_cnt[tok] = arg_count; } int main() { cout<<endl<<endl<<"Welcome to unique shell !!!!!!!!!!!"<<endl; tok=-1; while(1) { cout<<endl<<"***********UNIQUE**********"<<endl; cin.getline(input,81); count = 0,pip=0; splitcommand1(); if(pip == 1) { count++; splitcommand2(); } cout<<endl<<endl; if(strcmp(cmdptr1[0], "exit") == 0 ) { cout<<endl<<"EXITING UNIQUE SHELL"<<endl; exit(0); } //cout<<endl<<"Arg count is :"<<arg_count<<endl; if(pip == 1) { cout<<endl<<endl<<"in main :"; for(int i=0 ; i<arg_cnt[0] ; i++) { cout<<cmdptr1[i]<<" "; } cout<<endl; for(int i=0 ; i<arg_cnt[1] ; i++) { cout<<cmdptr2[i]<<" "; } pipe(cmdptr1, cmdptr2); } else { process (cmdptr1);//,arg_count); } } } I know it is not well coded. But try to help me :(

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• Visual Studio C++ list iterator not decementable

  - by user69514

  I keep getting an error on visual studio that says list iterator not decrementable: line 256 My program works fine on Linux, but visual studio compiler throws this error. Dammit this is why I hate windows. Why can't the world run on Linux? Anyway, do you see what my problem is? #include <iostream> #include <fstream> #include <sstream> #include <list> using namespace std; int main(){ /** create the list **/ list<int> l; /** create input stream to read file **/ ifstream inputstream("numbers.txt"); /** read the numbers and add them to list **/ if( inputstream.is_open() ){ string line; istringstream instream; while( getline(inputstream, line) ){ instream.clear(); instream.str(line); /** get he five int's **/ int one, two, three, four, five; instream >> one >> two >> three >> four >> five; /** add them to the list **/ l.push_back(one); l.push_back(two); l.push_back(three); l.push_back(four); l.push_back(five); }//end while loop }//end if /** close the stream **/ inputstream.close(); /** display the list **/ cout << "List Read:" << endl; list<int>::iterator i; for( i=l.begin(); i != l.end(); ++i){ cout << *i << " "; } cout << endl << endl; /** now sort the list **/ l.sort(); /** display the list **/ cout << "Sorted List (head to tail):" << endl; for( i=l.begin(); i != l.end(); ++i){ cout << *i << " "; } cout << endl; list<int> lReversed; for(i=l.begin(); i != l.end(); ++i){ lReversed.push_front(*i); } cout << "Sorted List (tail to head):" << endl; for(i=lReversed.begin(); i!=lReversed.end(); ++i){ cout << *i << " "; } cout << endl << endl; /** remove first biggest element and display **/ l.pop_back(); cout << "List after removing first biggest element:" << endl; cout << "Sorted List (head to tail):" << endl; for( i=l.begin(); i != l.end(); ++i){ cout << *i << " "; } cout << endl; cout << "Sorted List (tail to head):" << endl; lReversed.pop_front(); for(i=lReversed.begin(); i!=lReversed.end(); ++i){ cout << *i << " "; } cout << endl << endl; /** remove second biggest element and display **/ l.pop_back(); cout << "List after removing second biggest element:" << endl; cout << "Sorted List (head to tail):" << endl; for( i=l.begin(); i != l.end(); ++i){ cout << *i << " "; } cout << endl; lReversed.pop_front(); cout << "Sorted List (tail to head):" << endl; for(i=lReversed.begin(); i!=lReversed.end(); ++i){ cout << *i << " "; } cout << endl << endl; /** remove third biggest element and display **/ l.pop_back(); cout << "List after removing third biggest element:" << endl; cout << "Sorted List (head to tail):" << endl; for( i=l.begin(); i != l.end(); ++i){ cout << *i << " "; } cout << endl; cout << "Sorted List (tail to head):" << endl; lReversed.pop_front(); for(i=lReversed.begin(); i!=lReversed.end(); ++i){ cout << *i << " "; } cout << endl << endl; /** create frequency table **/ const int biggest = 1000; //create array size of biggest element int arr[biggest]; //set everything to zero for(int j=0; j<biggest+1; j++){ arr[j] = 0; } //now update number of occurences for( i=l.begin(); i != l.end(); i++){ arr[*i]++; } //now print the frequency table. only print where occurences greater than zero cout << "Final list frequency table: " << endl; for(int j=0; j<biggest+1; j++){ if( arr[j] > 0 ){ cout << j << ": " << arr[j] << " occurences" << endl; } } return 0; }//end main

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• Using R to Analyze G1GC Log Files

  - by user12620111

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  Using R to Analyze G1GC Log Files   Using R to Analyze G1GC Log Files Introduction Working in Oracle Platform Integration gives an engineer opportunities to work on a wide array of technologies. My team’s goal is to make Oracle applications run best on the Solaris/SPARC platform. When looking for bottlenecks in a modern applications, one needs to be aware of not only how the CPUs and operating system are executing, but also network, storage, and in some cases, the Java Virtual Machine. I was recently presented with about 1.5 GB of Java Garbage First Garbage Collector log file data. If you’re not familiar with the subject, you might want to review Garbage First Garbage Collector Tuning by Monica Beckwith. The customer had been running Java HotSpot 1.6.0_31 to host a web application server. I was told that the Solaris/SPARC server was running a Java process launched using a commmand line that included the following flags: -d64 -Xms9g -Xmx9g -XX:+UseG1GC -XX:MaxGCPauseMillis=200 -XX:InitiatingHeapOccupancyPercent=80 -XX:PermSize=256m -XX:MaxPermSize=256m -XX:+PrintGC -XX:+PrintGCTimeStamps -XX:+PrintHeapAtGC -XX:+PrintGCDateStamps -XX:+PrintFlagsFinal -XX:+DisableExplicitGC -XX:+UnlockExperimentalVMOptions -XX:ParallelGCThreads=8 Several sources on the internet indicate that if I were to print out the 1.5 GB of log files, it would require enough paper to fill the bed of a pick up truck. Of course, it would be fruitless to try to scan the log files by hand. Tools will be required to summarize the contents of the log files. Others have encountered large Java garbage collection log files. There are existing tools to analyze the log files: IBM’s GC toolkit The chewiebug GCViewer gchisto HPjmeter Instead of using one of the other tools listed, I decide to parse the log files with standard Unix tools, and analyze the data with R. Data Cleansing The log files arrived in two different formats. I guess that the difference is that one set of log files was generated using a more verbose option, maybe -XX:+PrintHeapAtGC, and the other set of log files was generated without that option. Format 1 In some of the log files, the log files with the less verbose format, a single trace, i.e. the report of a singe garbage collection event, looks like this: {Heap before GC invocations=12280 (full 61): garbage-first heap total 9437184K, used 7499918K [0xfffffffd00000000, 0xffffffff40000000, 0xffffffff40000000) region size 4096K, 1 young (4096K), 0 survivors (0K) compacting perm gen total 262144K, used 144077K [0xffffffff40000000, 0xffffffff50000000, 0xffffffff50000000) the space 262144K, 54% used [0xffffffff40000000, 0xffffffff48cb3758, 0xffffffff48cb3800, 0xffffffff50000000) No shared spaces configured. 2014-05-14T07:24:00.988-0700: 60586.353: [GC pause (young) 7324M->7320M(9216M), 0.1567265 secs] Heap after GC invocations=12281 (full 61): garbage-first heap total 9437184K, used 7496533K [0xfffffffd00000000, 0xffffffff40000000, 0xffffffff40000000) region size 4096K, 0 young (0K), 0 survivors (0K) compacting perm gen total 262144K, used 144077K [0xffffffff40000000, 0xffffffff50000000, 0xffffffff50000000) the space 262144K, 54% used [0xffffffff40000000, 0xffffffff48cb3758, 0xffffffff48cb3800, 0xffffffff50000000) No shared spaces configured. } A simple grep can be used to extract a summary: $ grep "\[ GC pause (young" g1gc.log 2014-05-13T13:24:35.091-0700: 3.109: [GC pause (young) 20M->5029K(9216M), 0.0146328 secs] 2014-05-13T13:24:35.440-0700: 3.459: [GC pause (young) 9125K->6077K(9216M), 0.0086723 secs] 2014-05-13T13:24:37.581-0700: 5.599: [GC pause (young) 25M->8470K(9216M), 0.0203820 secs] 2014-05-13T13:24:42.686-0700: 10.704: [GC pause (young) 44M->15M(9216M), 0.0288848 secs] 2014-05-13T13:24:48.941-0700: 16.958: [GC pause (young) 51M->20M(9216M), 0.0491244 secs] 2014-05-13T13:24:56.049-0700: 24.066: [GC pause (young) 92M->26M(9216M), 0.0525368 secs] 2014-05-13T13:25:34.368-0700: 62.383: [GC pause (young) 602M->68M(9216M), 0.1721173 secs] But that format wasn't easily read into R, so I needed to be a bit more tricky. I used the following Unix command to create a summary file that was easy for R to read. $ echo "SecondsSinceLaunch BeforeSize AfterSize TotalSize RealTime" $ grep "\[GC pause (young" g1gc.log | grep -v mark | sed -e 's/[A-SU-z\(\),]/ /g' -e 's/->/ /' -e 's/: / /g' | more SecondsSinceLaunch BeforeSize AfterSize TotalSize RealTime 2014-05-13T13:24:35.091-0700 3.109 20 5029 9216 0.0146328 2014-05-13T13:24:35.440-0700 3.459 9125 6077 9216 0.0086723 2014-05-13T13:24:37.581-0700 5.599 25 8470 9216 0.0203820 2014-05-13T13:24:42.686-0700 10.704 44 15 9216 0.0288848 2014-05-13T13:24:48.941-0700 16.958 51 20 9216 0.0491244 2014-05-13T13:24:56.049-0700 24.066 92 26 9216 0.0525368 2014-05-13T13:25:34.368-0700 62.383 602 68 9216 0.1721173 Format 2 In some of the log files, the log files with the more verbose format, a single trace, i.e. the report of a singe garbage collection event, was more complicated than Format 1. Here is a text file with an example of a single G1GC trace in the second format. As you can see, it is quite complicated. It is nice that there is so much information available, but the level of detail can be overwhelming. I wrote this awk script (download) to summarize each trace on a single line. #!/usr/bin/env awk -f BEGIN { printf("SecondsSinceLaunch IncrementalCount FullCount UserTime SysTime RealTime BeforeSize AfterSize TotalSize\n") } ###################### # Save count data from lines that are at the start of each G1GC trace. # Each trace starts out like this: # {Heap before GC invocations=14 (full 0): # garbage-first heap total 9437184K, used 325496K [0xfffffffd00000000, 0xffffffff40000000, 0xffffffff40000000) ###################### /{Heap.*full/{ gsub ( "\\)" , "" ); nf=split($0,a,"="); split(a[2],b," "); getline; if ( match($0, "first") ) { G1GC=1; IncrementalCount=b[1]; FullCount=substr( b[3], 1, length(b[3])-1 ); } else { G1GC=0; } } ###################### # Pull out time stamps that are in lines with this format: # 2014-05-12T14:02:06.025-0700: 94.312: [GC pause (young), 0.08870154 secs] ###################### /GC pause/ { DateTime=$1; SecondsSinceLaunch=substr($2, 1, length($2)-1); } ###################### # Heap sizes are in lines that look like this: # [ 4842M->4838M(9216M)] ###################### /\[ .*]$/ { gsub ( "\\[" , "" ); gsub ( "\ \]" , "" ); gsub ( "->" , " " ); gsub ( "\\( " , " " ); gsub ( "\ \)" , " " ); split($0,a," "); if ( split(a[1],b,"M") > 1 ) {BeforeSize=b[1]*1024;} if ( split(a[1],b,"K") > 1 ) {BeforeSize=b[1];} if ( split(a[2],b,"M") > 1 ) {AfterSize=b[1]*1024;} if ( split(a[2],b,"K") > 1 ) {AfterSize=b[1];} if ( split(a[3],b,"M") > 1 ) {TotalSize=b[1]*1024;} if ( split(a[3],b,"K") > 1 ) {TotalSize=b[1];} } ###################### # Emit an output line when you find input that looks like this: # [Times: user=1.41 sys=0.08, real=0.24 secs] ###################### /\[Times/ { if (G1GC==1) { gsub ( "," , "" ); split($2,a,"="); UserTime=a[2]; split($3,a,"="); SysTime=a[2]; split($4,a,"="); RealTime=a[2]; print DateTime,SecondsSinceLaunch,IncrementalCount,FullCount,UserTime,SysTime,RealTime,BeforeSize,AfterSize,TotalSize; G1GC=0; } } The resulting summary is about 25X smaller that the original file, but still difficult for a human to digest. SecondsSinceLaunch IncrementalCount FullCount UserTime SysTime RealTime BeforeSize AfterSize TotalSize ... 2014-05-12T18:36:34.669-0700: 3985.744 561 0 0.57 0.06 0.16 1724416 1720320 9437184 2014-05-12T18:36:34.839-0700: 3985.914 562 0 0.51 0.06 0.19 1724416 1720320 9437184 2014-05-12T18:36:35.069-0700: 3986.144 563 0 0.60 0.04 0.27 1724416 1721344 9437184 2014-05-12T18:36:35.354-0700: 3986.429 564 0 0.33 0.04 0.09 1725440 1722368 9437184 2014-05-12T18:36:35.545-0700: 3986.620 565 0 0.58 0.04 0.17 1726464 1722368 9437184 2014-05-12T18:36:35.726-0700: 3986.801 566 0 0.43 0.05 0.12 1726464 1722368 9437184 2014-05-12T18:36:35.856-0700: 3986.930 567 0 0.30 0.04 0.07 1726464 1723392 9437184 2014-05-12T18:36:35.947-0700: 3987.023 568 0 0.61 0.04 0.26 1727488 1723392 9437184 2014-05-12T18:36:36.228-0700: 3987.302 569 0 0.46 0.04 0.16 1731584 1724416 9437184 Reading the Data into R Once the GC log data had been cleansed, either by processing the first format with the shell script, or by processing the second format with the awk script, it was easy to read the data into R. g1gc.df = read.csv("summary.txt", row.names = NULL, stringsAsFactors=FALSE,sep="") str(g1gc.df) ## 'data.frame': 8307 obs. of 10 variables: ## $ row.names : chr "2014-05-12T14:00:32.868-0700:" "2014-05-12T14:00:33.179-0700:" "2014-05-12T14:00:33.677-0700:" "2014-05-12T14:00:35.538-0700:" ... ## $ SecondsSinceLaunch: num 1.16 1.47 1.97 3.83 6.1 ... ## $ IncrementalCount : int 0 1 2 3 4 5 6 7 8 9 ... ## $ FullCount : int 0 0 0 0 0 0 0 0 0 0 ... ## $ UserTime : num 0.11 0.05 0.04 0.21 0.08 0.26 0.31 0.33 0.34 0.56 ... ## $ SysTime : num 0.04 0.01 0.01 0.05 0.01 0.06 0.07 0.06 0.07 0.09 ... ## $ RealTime : num 0.02 0.02 0.01 0.04 0.02 0.04 0.05 0.04 0.04 0.06 ... ## $ BeforeSize : int 8192 5496 5768 22528 24576 43008 34816 53248 55296 93184 ... ## $ AfterSize : int 1400 1672 2557 4907 7072 14336 16384 18432 19456 21504 ... ## $ TotalSize : int 9437184 9437184 9437184 9437184 9437184 9437184 9437184 9437184 9437184 9437184 ... head(g1gc.df) ## row.names SecondsSinceLaunch IncrementalCount ## 1 2014-05-12T14:00:32.868-0700: 1.161 0 ## 2 2014-05-12T14:00:33.179-0700: 1.472 1 ## 3 2014-05-12T14:00:33.677-0700: 1.969 2 ## 4 2014-05-12T14:00:35.538-0700: 3.830 3 ## 5 2014-05-12T14:00:37.811-0700: 6.103 4 ## 6 2014-05-12T14:00:41.428-0700: 9.720 5 ## FullCount UserTime SysTime RealTime BeforeSize AfterSize TotalSize ## 1 0 0.11 0.04 0.02 8192 1400 9437184 ## 2 0 0.05 0.01 0.02 5496 1672 9437184 ## 3 0 0.04 0.01 0.01 5768 2557 9437184 ## 4 0 0.21 0.05 0.04 22528 4907 9437184 ## 5 0 0.08 0.01 0.02 24576 7072 9437184 ## 6 0 0.26 0.06 0.04 43008 14336 9437184 Basic Statistics Once the data has been read into R, simple statistics are very easy to generate. All of the numbers from high school statistics are available via simple commands. For example, generate a summary of every column: summary(g1gc.df) ## row.names SecondsSinceLaunch IncrementalCount FullCount ## Length:8307 Min. : 1 Min. : 0 Min. : 0.0 ## Class :character 1st Qu.: 9977 1st Qu.:2048 1st Qu.: 0.0 ## Mode :character Median :12855 Median :4136 Median : 12.0 ## Mean :12527 Mean :4156 Mean : 31.6 ## 3rd Qu.:15758 3rd Qu.:6262 3rd Qu.: 61.0 ## Max. :55484 Max. :8391 Max. :113.0 ## UserTime SysTime RealTime BeforeSize ## Min. :0.040 Min. :0.0000 Min. : 0.0 Min. : 5476 ## 1st Qu.:0.470 1st Qu.:0.0300 1st Qu.: 0.1 1st Qu.:5137920 ## Median :0.620 Median :0.0300 Median : 0.1 Median :6574080 ## Mean :0.751 Mean :0.0355 Mean : 0.3 Mean :5841855 ## 3rd Qu.:0.920 3rd Qu.:0.0400 3rd Qu.: 0.2 3rd Qu.:7084032 ## Max. :3.370 Max. :1.5600 Max. :488.1 Max. :8696832 ## AfterSize TotalSize ## Min. : 1380 Min. :9437184 ## 1st Qu.:5002752 1st Qu.:9437184 ## Median :6559744 Median :9437184 ## Mean :5785454 Mean :9437184 ## 3rd Qu.:7054336 3rd Qu.:9437184 ## Max. :8482816 Max. :9437184 Q: What is the total amount of User CPU time spent in garbage collection? sum(g1gc.df$UserTime) ## [1] 6236 As you can see, less than two hours of CPU time was spent in garbage collection. Is that too much? To find the percentage of time spent in garbage collection, divide the number above by total_elapsed_time*CPU_count. In this case, there are a lot of CPU’s and it turns out the the overall amount of CPU time spent in garbage collection isn’t a problem when viewed in isolation. When calculating rates, i.e. events per unit time, you need to ask yourself if the rate is homogenous across the time period in the log file. Does the log file include spikes of high activity that should be separately analyzed? Averaging in data from nights and weekends with data from business hours may alias problems. If you have a reason to suspect that the garbage collection rates include peaks and valleys that need independent analysis, see the “Time Series” section, below. Q: How much garbage is collected on each pass? The amount of heap space that is recovered per GC pass is surprisingly low: At least one collection didn’t recover any data. (“Min.=0”) 25% of the passes recovered 3MB or less. (“1st Qu.=3072”) Half of the GC passes recovered 4MB or less. (“Median=4096”) The average amount recovered was 56MB. (“Mean=56390”) 75% of the passes recovered 36MB or less. (“3rd Qu.=36860”) At least one pass recovered 2GB. (“Max.=2121000”) g1gc.df$Delta = g1gc.df$BeforeSize - g1gc.df$AfterSize summary(g1gc.df$Delta) ## Min. 1st Qu. Median Mean 3rd Qu. Max. ## 0 3070 4100 56400 36900 2120000 Q: What is the maximum User CPU time for a single collection? The worst garbage collection (“Max.”) is many standard deviations away from the mean. The data appears to be right skewed. summary(g1gc.df$UserTime) ## Min. 1st Qu. Median Mean 3rd Qu. Max. ## 0.040 0.470 0.620 0.751 0.920 3.370 sd(g1gc.df$UserTime) ## [1] 0.3966 Basic Graphics Once the data is in R, it is trivial to plot the data with formats including dot plots, line charts, bar charts (simple, stacked, grouped), pie charts, boxplots, scatter plots histograms, and kernel density plots. Histogram of User CPU Time per Collection I don't think that this graph requires any explanation. hist(g1gc.df$UserTime, main="User CPU Time per Collection", xlab="Seconds", ylab="Frequency") Box plot to identify outliers When the initial data is viewed with a box plot, you can see the one crazy outlier in the real time per GC. Save this data point for future analysis and drop the outlier so that it’s not throwing off our statistics. Now the box plot shows many outliers, which will be examined later, using times series analysis. Notice that the scale of the x-axis changes drastically once the crazy outlier is removed. par(mfrow=c(2,1)) boxplot(g1gc.df$UserTime,g1gc.df$SysTime,g1gc.df$RealTime, main="Box Plot of Time per GC\n(dominated by a crazy outlier)", names=c("usr","sys","elapsed"), xlab="Seconds per GC", ylab="Time (Seconds)", horizontal = TRUE, outcol="red") crazy.outlier.df=g1gc.df[g1gc.df$RealTime > 400,] g1gc.df=g1gc.df[g1gc.df$RealTime < 400,] boxplot(g1gc.df$UserTime,g1gc.df$SysTime,g1gc.df$RealTime, main="Box Plot of Time per GC\n(crazy outlier excluded)", names=c("usr","sys","elapsed"), xlab="Seconds per GC", ylab="Time (Seconds)", horizontal = TRUE, outcol="red") box(which = "outer", lty = "solid") Here is the crazy outlier for future analysis: crazy.outlier.df ## row.names SecondsSinceLaunch IncrementalCount ## 8233 2014-05-12T23:15:43.903-0700: 20741 8316 ## FullCount UserTime SysTime RealTime BeforeSize AfterSize TotalSize ## 8233 112 0.55 0.42 488.1 8381440 8235008 9437184 ## Delta ## 8233 146432 R Time Series Data To analyze the garbage collection as a time series, I’ll use Z’s Ordered Observations (zoo). “zoo is the creator for an S3 class of indexed totally ordered observations which includes irregular time series.” require(zoo) ## Loading required package: zoo ## ## Attaching package: 'zoo' ## ## The following objects are masked from 'package:base': ## ## as.Date, as.Date.numeric head(g1gc.df[,1]) ## [1] "2014-05-12T14:00:32.868-0700:" "2014-05-12T14:00:33.179-0700:" ## [3] "2014-05-12T14:00:33.677-0700:" "2014-05-12T14:00:35.538-0700:" ## [5] "2014-05-12T14:00:37.811-0700:" "2014-05-12T14:00:41.428-0700:" options("digits.secs"=3) times=as.POSIXct( g1gc.df[,1], format="%Y-%m-%dT%H:%M:%OS%z:") g1gc.z = zoo(g1gc.df[,-c(1)], order.by=times) head(g1gc.z) ## SecondsSinceLaunch IncrementalCount FullCount ## 2014-05-12 17:00:32.868 1.161 0 0 ## 2014-05-12 17:00:33.178 1.472 1 0 ## 2014-05-12 17:00:33.677 1.969 2 0 ## 2014-05-12 17:00:35.538 3.830 3 0 ## 2014-05-12 17:00:37.811 6.103 4 0 ## 2014-05-12 17:00:41.427 9.720 5 0 ## UserTime SysTime RealTime BeforeSize AfterSize ## 2014-05-12 17:00:32.868 0.11 0.04 0.02 8192 1400 ## 2014-05-12 17:00:33.178 0.05 0.01 0.02 5496 1672 ## 2014-05-12 17:00:33.677 0.04 0.01 0.01 5768 2557 ## 2014-05-12 17:00:35.538 0.21 0.05 0.04 22528 4907 ## 2014-05-12 17:00:37.811 0.08 0.01 0.02 24576 7072 ## 2014-05-12 17:00:41.427 0.26 0.06 0.04 43008 14336 ## TotalSize Delta ## 2014-05-12 17:00:32.868 9437184 6792 ## 2014-05-12 17:00:33.178 9437184 3824 ## 2014-05-12 17:00:33.677 9437184 3211 ## 2014-05-12 17:00:35.538 9437184 17621 ## 2014-05-12 17:00:37.811 9437184 17504 ## 2014-05-12 17:00:41.427 9437184 28672 Example of Two Benchmark Runs in One Log File The data in the following graph is from a different log file, not the one of primary interest to this article. I’m including this image because it is an example of idle periods followed by busy periods. It would be uninteresting to average the rate of garbage collection over the entire log file period. More interesting would be the rate of garbage collect in the two busy periods. Are they the same or different? Your production data may be similar, for example, bursts when employees return from lunch and idle times on weekend evenings, etc. Once the data is in an R Time Series, you can analyze isolated time windows. Clipping the Time Series data Flashing back to our test case… Viewing the data as a time series is interesting. You can see that the work intensive time period is between 9:00 PM and 3:00 AM. Lets clip the data to the interesting period:     par(mfrow=c(2,1)) plot(g1gc.z$UserTime, type="h", main="User Time per GC\nTime: Complete Log File", xlab="Time of Day", ylab="CPU Seconds per GC", col="#1b9e77") clipped.g1gc.z=window(g1gc.z, start=as.POSIXct("2014-05-12 21:00:00"), end=as.POSIXct("2014-05-13 03:00:00")) plot(clipped.g1gc.z$UserTime, type="h", main="User Time per GC\nTime: Limited to Benchmark Execution", xlab="Time of Day", ylab="CPU Seconds per GC", col="#1b9e77") box(which = "outer", lty = "solid") Cumulative Incremental and Full GC count Here is the cumulative incremental and full GC count. When the line is very steep, it indicates that the GCs are repeating very quickly. Notice that the scale on the Y axis is different for full vs. incremental. plot(clipped.g1gc.z[,c(2:3)], main="Cumulative Incremental and Full GC count", xlab="Time of Day", col="#1b9e77") GC Analysis of Benchmark Execution using Time Series data In the following series of 3 graphs: The “After Size” show the amount of heap space in use after each garbage collection. Many Java objects are still referenced, i.e. alive, during each garbage collection. This may indicate that the application has a memory leak, or may indicate that the application has a very large memory footprint. Typically, an application's memory footprint plateau's in the early stage of execution. One would expect this graph to have a flat top. The steep decline in the heap space may indicate that the application crashed after 2:00. The second graph shows that the outliers in real execution time, discussed above, occur near 2:00. when the Java heap seems to be quite full. The third graph shows that Full GCs are infrequent during the first few hours of execution. The rate of Full GC's, (the slope of the cummulative Full GC line), changes near midnight.   plot(clipped.g1gc.z[,c("AfterSize","RealTime","FullCount")], xlab="Time of Day", col=c("#1b9e77","red","#1b9e77")) GC Analysis of heap recovered Each GC trace includes the amount of heap space in use before and after the individual GC event. During garbage coolection, unreferenced objects are identified, the space holding the unreferenced objects is freed, and thus, the difference in before and after usage indicates how much space has been freed. The following box plot and bar chart both demonstrate the same point - the amount of heap space freed per garbage colloection is surprisingly low. par(mfrow=c(2,1)) boxplot(as.vector(clipped.g1gc.z$Delta), main="Amount of Heap Recovered per GC Pass", xlab="Size in KB", horizontal = TRUE, col="red") hist(as.vector(clipped.g1gc.z$Delta), main="Amount of Heap Recovered per GC Pass", xlab="Size in KB", breaks=100, col="red") box(which = "outer", lty = "solid") This graph is the most interesting. The dark blue area shows how much heap is occupied by referenced Java objects. This represents memory that holds live data. The red fringe at the top shows how much data was recovered after each garbage collection. barplot(clipped.g1gc.z[,c("AfterSize","Delta")], col=c("#7570b3","#e7298a"), xlab="Time of Day", border=NA) legend("topleft", c("Live Objects","Heap Recovered on GC"), fill=c("#7570b3","#e7298a")) box(which = "outer", lty = "solid") When I discuss the data in the log files with the customer, I will ask for an explaination for the large amount of referenced data resident in the Java heap. There are two are posibilities: There is a memory leak and the amount of space required to hold referenced objects will continue to grow, limited only by the maximum heap size. After the maximum heap size is reached, the JVM will throw an “Out of Memory” exception every time that the application tries to allocate a new object. If this is the case, the aplication needs to be debugged to identify why old objects are referenced when they are no longer needed. The application has a legitimate requirement to keep a large amount of data in memory. The customer may want to further increase the maximum heap size. Another possible solution would be to partition the application across multiple cluster nodes, where each node has responsibility for managing a unique subset of the data. Conclusion In conclusion, R is a very powerful tool for the analysis of Java garbage collection log files. The primary difficulty is data cleansing so that information can be read into an R data frame. Once the data has been read into R, a rich set of tools may be used for thorough evaluation.

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