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• recursive function's summation in MATLAB

  - by lucky

  B=[1 1 1 1 1 1....1] % vector of length N elements Xk= sin(2*pi/16) i need to find function alpha(l,k) which is having two variables l and k and a condition given that alpha(l,0)=alpha(l,-1)=alpha(l,-2)......=alpha(l,-(N-1))=0 i.e no matter what value of l ,alpha = 0 for past values A= input('no of iterations'); % no. of iterations user want N=input('N values of alpha:') alpha1=[]; for k=0:A-1 l=0:N-1 % need 10 separate alpha values for every k, which goes from 0 to A-1 alpha(l,k)= Xk + summation( B(j)*alpha(l,k-j)) % as summation goes from j=1 to N alpha1=[alpha1 alpha] end; could anyone please help me to solve this recursive function, i am new to matlab. alpha

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• How do I implement a fibonacci sequence in java using try/catch logic?

  - by Lars Flyger

  I know how to do it using simple recursion, but in order to complete this particular assignment I need to be able to accumulate on the stack and throw an exception that holds the answer in it. So far I have: public static int fibo(int index) { int sum = 0; try { fibo_aux(index, 1, 1); } catch (IntegerException me) { sum = me.getIntValue(); } return sum; } fibo_aux is supposed to throw an IntegerException (which holds the value of the answer that is retireved via getIntValue) and accumulates the answer on the stack, but so far I can't figure it out. Can anyone help?

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• Available message types in JMS?

  - by Caylem

  This is based on a past exam question. The question is asking to describe the four types of message available using JMS. The problem is it says the four, not just four. So it assumes their is only four, no more no less. However according to this site their seems to be five; streams maps text objects bytes *Another book states that XML is another potential type in future versions of JMS. Is XML already available? Am I missing something or is the question just wrong? Thanks.

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• Enhanced for loop problem

  - by Antony Kelt

  Why is my enhanced loop not working? Vector<String> v = new Vector<String>(); v.add("one"); v.add("two"); v.add("three"); for(String str : v){ System.out.println(v); }

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• NoClassDefFoundError when running Netbeans 6.8 application

  - by Malachi

  My application has recently started throwing NoClassDefFoundError errors when I am running my application from within NetBeans. It never used to do this. And when I run this outside of NetBeans using the command line argument that it provides once built, these errors are not produced Is this a bug with Netbeans? I have reinstalled NetBeans as I recently had another that was resolved by reinstalling, however the problem still persists. http://stackoverflow.com/questions/2654892/symbol-error-in-java-application-using-netbeans-6-8-when-adding-a-shared-project Within Netbeans Outside NetBeans

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• Dynamic programming: Find largest diamond (rhombus)

  - by Darksody

  I have a small program to do in Java. I have a 2D array filled with 0 and 1, and I must find the largest rhombus (as in square rotated by 90 degrees) and their numbers. Example: 0 1 0 0 0 1 1 0 1 1 1 0 1 0 1 1 1 1 0 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 Result: 1 1 1 1 1 1 1 1 1 1 1 1 1 The problem is similar to this SO question. If you have any idea, post it here.

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• Implementing Dijkstra's Algorithm

  - by DeadMG

  I've been tasked (coursework @ university) to implement a form of path-finding. Now, in-spec, I could just implement a brute force, since there's a limit on the number of nodes to search (begin, two in the middle, end), but I want to re-use this code and came to implement Dijkstra's algorithm. I've seen the pseudo on Wikipedia and a friend wrote some for me as well, but it flat out doesn't make sense. The algorithm seems pretty simple and it's not a problem for me to understand it, but I just can't for the life of me visualize the code that would realize such a thing. Any suggestions/tips?

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• Prolog - generate correct bracketing

  - by Henrik Bak

  I'd like to get some help in the following exam problem, i have no idea how to do this: Input: a list of numbers, eg.: [1,2,3,4] Output: every possible correct bracketing. Eg.: (in case of input [1,2,3,4]): ((1 2) (3 4)) ((1 (2 3)) 4) (1 ((2 3) 4)) (1 (2 (3 4))) (((1 2) 3) 4) Bracketing here is like a method with two arguments, for example multiplication - then the output is the possible multiplication orders. Please help, i'm stuck with this one. Any help is appreciated, thanks!

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• Finding height in Binary Search Tree

  - by mike

  Hey I was wondering if anybody could help me rework this method to find the height of a binary search tree. So far my code looks like this however the answer im getting is larger than the actual height by 1, but when I remove the +1 from my return statements its less than the actual height by 1? I'm still trying to wrap my head around recursion with these BST any help would be much appreciated. public int findHeight(){ if(this.isEmpty()){ return 0; } else{ TreeNode<T> node = root; return findHeight(node); } } private int findHeight(TreeNode<T> aNode){ int heightLeft = 0; int heightRight = 0; if(aNode.left!=null) heightLeft = findHeight(aNode.left); if(aNode.right!=null) heightRight = findHeight(aNode.right); if(heightLeft > heightRight){ return heightLeft+1; } else{ return heightRight+1; } }

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• How do I improve my performance with this singly linked list struct within my program?

  - by Jesus

  Hey guys, I have a program that does operations of sets of strings. We have to implement functions such as addition and subtraction of two sets of strings. We are suppose to get it down to the point where performance if of O(N+M), where N,M are sets of strings. Right now, I believe my performance is at O(N*M), since I for each element of N, I go through every element of M. I'm particularly focused on getting the subtraction to the proper performance, as if I can get that down to proper performance, I believe I can carry that knowledge over to the rest of things I have to implement. The '-' operator is suppose to work like this, for example. Declare set1 to be an empty set. Declare set2 to be a set with { a b c } elements Declare set3 to be a set with ( b c d } elements set1 = set2 - set3 And now set1 is suppose to equal { a }. So basically, just remove any element from set3, that is also in set2. For the addition implementation (overloaded '+' operator), I also do the sorting of the strings (since we have to). All the functions work right now btw. So I was wondering if anyone could a) Confirm that currently I'm doing O(N*M) performance b) Give me some ideas/implementations on how to improve the performance to O(N+M) Note: I cannot add any member variables or functions to the class strSet or to the node structure. The implementation of the main program isn't very important, but I will post the code for my class definition and the implementation of the member functions: strSet2.h (Implementation of my class and struct) // Class to implement sets of strings // Implements operators for union, intersection, subtraction, // etc. for sets of strings // V1.1 15 Feb 2011 Added guard (#ifndef), deleted using namespace RCH #ifndef _STRSET_ #define _STRSET_ #include <iostream> #include <vector> #include <string> // Deleted: using namespace std; 15 Feb 2011 RCH struct node { std::string s1; node * next; }; class strSet { private: node * first; public: strSet (); // Create empty set strSet (std::string s); // Create singleton set strSet (const strSet &copy); // Copy constructor ~strSet (); // Destructor int SIZE() const; bool isMember (std::string s) const; strSet operator + (const strSet& rtSide); // Union strSet operator - (const strSet& rtSide); // Set subtraction strSet& operator = (const strSet& rtSide); // Assignment }; // End of strSet class #endif // _STRSET_ strSet2.cpp (implementation of member functions) #include <iostream> #include <vector> #include <string> #include "strset2.h" using namespace std; strSet::strSet() { first = NULL; } strSet::strSet(string s) { node *temp; temp = new node; temp->s1 = s; temp->next = NULL; first = temp; } strSet::strSet(const strSet& copy) { if(copy.first == NULL) { first = NULL; } else { node *n = copy.first; node *prev = NULL; while (n) { node *newNode = new node; newNode->s1 = n->s1; newNode->next = NULL; if (prev) { prev->next = newNode; } else { first = newNode; } prev = newNode; n = n->next; } } } strSet::~strSet() { if(first != NULL) { while(first->next != NULL) { node *nextNode = first->next; first->next = nextNode->next; delete nextNode; } } } int strSet::SIZE() const { int size = 0; node *temp = first; while(temp!=NULL) { size++; temp=temp->next; } return size; } bool strSet::isMember(string s) const { node *temp = first; while(temp != NULL) { if(temp->s1 == s) { return true; } temp = temp->next; } return false; } strSet strSet::operator + (const strSet& rtSide) { strSet newSet; newSet = *this; node *temp = rtSide.first; while(temp != NULL) { string newEle = temp->s1; if(!isMember(newEle)) { if(newSet.first==NULL) { node *newNode; newNode = new node; newNode->s1 = newEle; newNode->next = NULL; newSet.first = newNode; } else if(newSet.SIZE() == 1) { if(newEle < newSet.first->s1) { node *tempNext = newSet.first; node *newNode; newNode = new node; newNode->s1 = newEle; newNode->next = tempNext; newSet.first = newNode; } else { node *newNode; newNode = new node; newNode->s1 = newEle; newNode->next = NULL; newSet.first->next = newNode; } } else { node *prev = NULL; node *curr = newSet.first; while(curr != NULL) { if(newEle < curr->s1) { if(prev == NULL) { node *newNode; newNode = new node; newNode->s1 = newEle; newNode->next = curr; newSet.first = newNode; break; } else { node *newNode; newNode = new node; newNode->s1 = newEle; newNode->next = curr; prev->next = newNode; break; } } if(curr->next == NULL) { node *newNode; newNode = new node; newNode->s1 = newEle; newNode->next = NULL; curr->next = newNode; break; } prev = curr; curr = curr->next; } } } temp = temp->next; } return newSet; } strSet strSet::operator - (const strSet& rtSide) { strSet newSet; newSet = *this; node *temp = rtSide.first; while(temp != NULL) { string element = temp->s1; node *prev = NULL; node *curr = newSet.first; while(curr != NULL) { if( element < curr->s1 ) break; if( curr->s1 == element ) { if( prev == NULL) { node *duplicate = curr; newSet.first = newSet.first->next; delete duplicate; break; } else { node *duplicate = curr; prev->next = curr->next; delete duplicate; break; } } prev = curr; curr = curr->next; } temp = temp->next; } return newSet; } strSet& strSet::operator = (const strSet& rtSide) { if(this != &rtSide) { if(first != NULL) { while(first->next != NULL) { node *nextNode = first->next; first->next = nextNode->next; delete nextNode; } } if(rtSide.first == NULL) { first = NULL; } else { node *n = rtSide.first; node *prev = NULL; while (n) { node *newNode = new node; newNode->s1 = n->s1; newNode->next = NULL; if (prev) { prev->next = newNode; } else { first = newNode; } prev = newNode; n = n->next; } } } return *this; }

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• Using recursion to to trim a binary tree based on a given min and max value

  - by Justin

  As the title says, I have to trim a binary tree based on a given min and max value. Each node stores a value, and a left/right node. I may define private helper methods to solve this problem, but otherwise I may not call any other methods of the class nor create any data structures such as arrays, lists, etc. An example would look like this: overallRoot _____[50]____________________ / \ __________[38] _______________[90] / \ / _[14] [42] [54]_____ / \ \ [8] [20] [72] \ / \ [26] [61] [83] trim(52, 65); should return: overallRoot [54] \ [61] My attempted solution has three methods: public void trim(int min, int max) { rootFinder(overallRoot, min, max); } First recursive method finds the new root perfectly. private void rootFinder(IntTreeNode node, int min, int max) { if (node == null) return; if (overallRoot.data < min) { node = overallRoot = node.right; rootFinder(node, min, max); } else if (overallRoot.data > max) { node = overallRoot = node.left; rootFinder(node, min, max); } else cutter(overallRoot, min, max); } This second method should eliminate any further nodes not within the min/max, but it doesn't work as I would hope. private void cutter(IntTreeNode node, int min, int max) { if (node == null) return; if (node.data <= min) { node.left = null; } if (node.data >= max) { node.right = null; } if (node.data < min) { node = node.right; } if (node.data > max) { node = node.left; } cutter(node.left, min, max); cutter(node.right, min, max); } This returns: overallRoot [54]_____ \ [72] / [61] Any help is appreciated. Feel free to ask for further explanation as needed.

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• String operations

  - by NEO

  I am trying to find the most repeated word in a string. My code is as follows: public class Word { private String toWord; private int Count; public Word(int count, String word){ toWord = word; Count = count; } public static void main(String args[]){ String str="my name is neo and my other name is also neo because I am neo"; String []str1=str.split(" "); Word w1=new Word(0,str1[0]); LinkedList<Word> list = new LinkedList<Word>(); list.add(w1); ListIterator itr = list.listIterator(); for(int i=1;i<str1.length;i++){ while(itr.hasNext()){ if(str1[i].equalsTO(????)); else list.add(new Word(0,str1[i])); } How do I compare the string from string array str1 to the string stored in the linked list and then how do i increase the respective count. I will then print the string with the highest count, I dont know how to do that either.

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• 47 memory leaks. STL pointers.

  - by icelated

  I have a major amount of memory leaks. I know that the Sets have pointers and i cannot change that! I cannot change anything, but clean up the mess i have... I am creating memory with new in just about every function to add information to the sets. I have a Cd/ DVD/book: super classes of ITEM class and a library class.. In the library class i have 2 functions for cleaning up the sets.. Also, the CD, DVD, book destructors are not being called.. here is my potential leaks.. library.h #pragma once #include <ostream> #include <map> #include <set> #include <string> #include "Item.h" using namespace std; typedef set<Item*> ItemSet; typedef map<string,Item*> ItemMap; typedef map<string,ItemSet*> ItemSetMap; class Library { public: // general functions void addKeywordForItem(const Item* const item, const string& keyword); const ItemSet* itemsForKeyword(const string& keyword) const; void printItem(ostream& out, const Item* const item) const; // book-related functions const Item* addBook(const string& title, const string& author, int const nPages); const ItemSet* booksByAuthor(const string& author) const; const ItemSet* books() const; // music-related functions const Item* addMusicCD(const string& title, const string& band, const int nSongs); void addBandMember(const Item* const musicCD, const string& member); const ItemSet* musicByBand(const string& band) const; const ItemSet* musicByMusician(const string& musician) const; const ItemSet* musicCDs() const; // movie-related functions const Item* addMovieDVD(const string& title, const string& director, const int nScenes); void addCastMember(const Item* const movie, const string& member); const ItemSet* moviesByDirector(const string& director) const; const ItemSet* moviesByActor(const string& actor) const; const ItemSet* movies() const; ~Library(); void Purge(ItemSet &set); void Purge(ItemSetMap &map); }; here is some functions for adding info using new in library. Keep in mind i am cutting out alot of code to keep this post short. library.cpp #include "Library.h" #include "book.h" #include "cd.h" #include "dvd.h" #include <iostream> // general functions ItemSet allBooks; ItemSet allCDS; ItemSet allDVDs; ItemSetMap allBooksByAuthor; ItemSetMap allmoviesByDirector; ItemSetMap allmoviesByActor; ItemSetMap allMusicByBand; ItemSetMap allMusicByMusician; const ItemSet* Library::itemsForKeyword(const string& keyword) const { const StringSet* kw; ItemSet* obj = new ItemSet(); return obj; const Item* Library::addBook(const string& title, const string& author, const int nPages) { ItemSet* obj = new ItemSet(); Book* item = new Book(title,author,nPages); allBooks.insert(item); // add to set of all books obj->insert(item); return item; const Item* Library::addMusicCD(const string& title, const string& band, const int nSongs) { ItemSet* obj = new ItemSet(); CD* item = new CD(title,band,nSongs); return item; void Library::addBandMember(const Item* musicCD, const string& member) { ItemSet* obj = new ItemSet(); (((CD*) musicCD)->addBandMember(member)); obj->insert((CD*) musicCD); here is the library destructor..... Library::~Library() { Purge(allBooks); Purge(allCDS); Purge(allDVDs); Purge(allBooksByAuthor); Purge(allmoviesByDirector); Purge(allmoviesByActor); Purge(allMusicByBand); Purge(allMusicByMusician); } void Library::Purge(ItemSet &set) { for (ItemSet::iterator it = set.begin(); it != set.end(); ++it) delete *it; set.clear(); } void Library::Purge(ItemSetMap &map) { for (ItemSetMap::iterator it = map.begin(); it != map.end(); ++it) delete it->second; map.clear(); } so, basically item, cd, dvd class all have a set like this: typedef set<string> StringSet; class CD : public Item StringSet* music; and i am deleting it like: but those superclasses are not being called.. Item destructor is. CD::~CD() { delete music; } Do, i need a copy constructor? and how do i delete those objects i am creating in the library class? and how can i get the cd,dvd, destructor called? would the addbandmember function located in the library.cpp cause me to have a copy constructor? Any real help you can provide me to help me clean up this mess instead of telling me not to use pointers in my sets i would really appreciate. How can i delete the memory i am creating in those functions? I cannot delete them in the function!!

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• Very weird C file-handling anomaly

  - by KáGé

  Hello, I got a very weird issue that I cant figure out in my school project, which is the simulation of a simple filesystem in a human-readable textfile. Unfortunately I don't yet have enough time to translate the comments in my code or make it less gibberish, so if you are bothered by that, you don't have to help, I understand. See the code HERE. Now in drive.h, at line 574 is this part: i = getline(); #ifdef DEBUG printf("Free space in all found at %d.\n\n", i); if(drive.disk != NULL){ printf("Disk OK\n\n"); } #endif //write in data state = seekline(i); Before this it finds place for the allocation database entry in the ALL sector (see the "image files" in the mounts folder, this issue was tested on mount_30.efs-dbf), then gets the line with i = getline() fine (getline is in lglobal.h, line 39), but after that any file manipulation (in this case seekline's fseek, but if I comment that out, then the first fprintf after that) crashes the program straight away. I think the file gets somehow corrupted (though the Disk OK message appears) but can't figure out how. I've tried putting i = getline(); into comment, but it didn't make any difference. I've also tried asking at local programming forums but they didn't really help either. The last few lines of the output before it crashes: Dir written. (drive.h line 562) Seekline entered: 268 (called at drive.h line 564) Getline entered. (called at drive.h line 574) Line got: 268. Free space in all found at 268. (drive.h line 576) Seekline entered: 268 (called at drive.h line 582, note that this exact call was run successfully less than 20 lines back. This one should set the pointer to the beginning of the line it is currently in) After this it crashes. Does anyone has any idea of what causes this and how could I fix it? Thank you.

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• Count the number of lines in a Java String

  - by Simon Guo

  Need some compact code for counting the number of lines in a string in Java. The string is to be separated by \r or \n. Each instance of those newline characters will be considered as a separate line. For example, "Hello\nWorld\nThis\nIs\t" should return 4. The prototype is private static int countLines(String str) {...} Can someone provide a compact set of statements? I have solution at here but it is too long, I think. Thank you.

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• JAVA: storing input into array

  - by Jann

  I need to write a program where the program would generate random letter and i would need to store this random character into an array char[] arrayRandom = new char[10]; for (int i = 0; i < 8; i++) { randomNumLet = (generator.nextInt(20) + 1); System.out.print(arrayRandomLetter[randomNumLet] + " "); arrayRandomLetter[randomNumLet] = arrayRandom[i]; } is there anything wrong with my code? because when i run this and printed the array i get boxes for all the values in the array and there are some letter that this line of code cannot print System.out.print(arrayRandomLetter[randomNumLet] + " "); Thanks

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• Java swing app can't find image

  - by KáGé

  Hello, I'm making a torpedo game for school in java with swing gui, please see the zipped source HERE. I use custom button icons and mouse cursors of images stored in the /bin/resource/graphics/default folder's subfolders, where the root folder is the program's root folder (it will be the root in the final .jar as well I suppose) which apart from "bin" contains a "main" folder with all the classes. The relative path of the resources is stored in MapStruct.java's shipPath and mapPath variables. Now Battlefield.java's PutPanel class finds them all right and sets up its buttons' icons fine, but every other class fail to get their icons, e.g. Table.java's setCursor, which should set the mouse cursor for all its elements for the selected ship's image or Field.java's this.button.setIcon(icon); in the constructor, which should set the icon for the buttons of the "water". I watched with debug what happens, and the images stay null after loading, though the paths seem to be correct. I've also tried to write a test file in the image folder but the method returns a filenotfound exception. I've tried to get the path of the class to see if it runs from the supposed place and it seems it does, so I really can't find the problem now. Could anyone please help me? Thank you.

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• finding "distance" between two pixel's colors.

  - by igor

  Once more something relatively simple, but confused as to what they want. the method to find distance on cartesian coordinate system is distance=sqrt[(x2-x1)^2 + (y2-y1)^2] but how do i apply it here? //Requires: testColor to be a valid Color //Effects: returns the "distance" between the current Pixel's color and // the passed color // uses the standard method to calculate "distance" // uses the same formula as finding distance on a // Cartesian coordinate system double colorDistance(Color testColor) const;

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• Greatest of Two Numbers Without Using A Relational Operator

  - by ratty

  I want to find the greater of two numbers without using < > this relational operator. I tried this int a,b; if(a/b==0) { b is greater } else { a is greater } but this not working for negative numbers.

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• Information Gain and Entropy

  - by dhorn

  I recently read this question regarding information gain and entropy. I think I have a semi-decent grasp on the main idea, but I'm curious as what to do with situations such as follows: If we have a bag of 7 coins, 1 of which is heavier than the others, and 1 of which is lighter than the others, and we know the heavier coin + the lighter coin is the same as 2 normal coins, what is the information gain associated with picking two random coins and weighing them against each other? Our goal here is to identify the two odd coins. I've been thinking this problem over for a while, and can't frame it correctly in a decision tree, or any other way for that matter. Any help? EDIT: I understand the formula for entropy and the formula for information gain. What I don't understand is how to frame this problem in a decision tree format. EDIT 2: Here is where I'm at so far: Assuming we pick two coins and they both end up weighing the same, we can assume our new chances of picking H+L come out to 1/5 * 1/4 = 1/20 , easy enough. Assuming we pick two coins and the left side is heavier. There are three different cases where this can occur: HM: Which gives us 1/2 chance of picking H and a 1/4 chance of picking L: 1/8 HL: 1/2 chance of picking high, 1/1 chance of picking low: 1/1 ML: 1/2 chance of picking low, 1/4 chance of picking high: 1/8 However, the odds of us picking HM are 1/7 * 5/6 which is 5/42 The odds of us picking HL are 1/7 * 1/6 which is 1/42 And the odds of us picking ML are 1/7 * 5/6 which is 5/42 If we weight the overall probabilities with these odds, we are given: (1/8) * (5/42) + (1/1) * (1/42) + (1/8) * (5/42) = 3/56. The same holds true for option B. option A = 3/56 option B = 3/56 option C = 1/20 However, option C should be weighted heavier because there is a 5/7 * 4/6 chance to pick two mediums. So I'm assuming from here I weight THOSE odds. I am pretty sure I've messed up somewhere along the way, but I think I'm on the right path! EDIT 3: More stuff. Assuming the scale is unbalanced, the odds are (10/11) that only one of the coins is the H or L coin, and (1/11) that both coins are H/L Therefore we can conclude: (10 / 11) * (1/2 * 1/5) and (1 / 11) * (1/2) EDIT 4: Going to go ahead and say that it is a total 4/42 increase.

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• this program runs but not correctly numbers arent right, i read numbers from a file and then when i

  - by user320950

  this is what i have to do: write a program that determines the grade dispersal for 100 students You are to read the exam scores into three arrays, one array for each exam. You must then calculate how many students scored A’s (90 or above), B’s (80 or above), C’s (70 or above), D’s (60 or above), and F’s (less than 60). Do this for each exam and write the distribution to the screen. // basic file operations #include <iostream> #include <fstream> using namespace std; int read_file_in_array(double exam[100][3]); double calculate_total(double exam1[], double exam2[], double exam3[]); // function that calcualates grades to see how many 90,80,70,60 //void display_totals(); double exam[100][3]; int main() { double go,go2,go3; double exam[100][3],exam1[100],exam2[100],exam3[100]; go=read_file_in_array(exam); go2=calculate_total(exam1,exam2,exam3); //go3=display_totals(); cout << go,go2,go3; return 0; } /* int display_totals() { int grade_total; grade_total=calculate_total(exam1,exam2,exam3); return 0; } */ double calculate_total(double exam1[],double exam2[],double exam3[]) { int calc_tot,above90=0, above80=0, above70=0, above60=0,i,j, fail=0; double exam[100][3]; calc_tot=read_file_in_array(exam); for(i=0;i<100;i++) { for (j=0; j<3; j++) { exam1[i]=exam[100][0]; exam2[i]=exam[100][1]; exam3[i]=exam[100][2]; if(exam[i][j] <=90 && exam[i][j] >=100) { above90++; { if(exam[i][j] <=80 && exam[i][j] >=89) { above80++; { if(exam[i][j] <=70 && exam[i][j] >=79) { above70++; { if(exam[i][j] <=60 && exam[i][j] >=69) { above60++; { if(exam[i][j] >=59) { fail++; } } } } } } } } } } } return 0; } int read_file_in_array(double exam[100][3]) { ifstream infile; int exam1[100]; int exam2[100]; int exam3[100]; infile.open("grades.txt");// file containing numbers in 3 columns if(infile.fail()) // checks to see if file opended { cout << "error" << endl; } int num, i=0,j=0; while(!infile.eof()) // reads file to end of line { for(i=0;i<100;i++) // array numbers less than 100 { for(j=0;j<3;j++) // while reading get 1st array or element infile >> exam[i][j]; infile >> exam[i][j]; infile >> exam[i][j]; cout << exam[i][j] << endl; { if (! (infile >> exam[i][j]) ) cout << exam[i][j] << endl; } exam[i][j]=exam1[i]; exam[i][j]=exam2[i]; exam[i][j]=exam3[i]; } infile.close(); } return 0; }

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• C++ Memory Allocation & Linked List Implementation

  - by pws5068

  I'm writing software to simulate the "first-fit" memory allocation schema. Basically, I allocate a large X megabyte chunk of memory and subdivide it into blocks when chunks are requested according to the schema. I'm using a linked list called "node" as a header for each block of memory (so that we can find the next block without tediously looping through every address value. head_ptr = (char*) malloc(total_size + sizeof(node)); if(head_ptr == NULL) return -1; // Malloc Error .. :-( node* head_node = new node; // Build block header head_node->next = NULL; head_node->previous = NULL; // Header points to next block (which doesn't exist yet) memset(head_ptr,head_node, sizeof(node)); ` But this last line returns: error: invalid conversion from 'node*' to 'int' I understand why this is invalid.. but how can I place my node into the pointer location of my newly allocated memory?

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• Relational algebra help?!

  - by Tom

  im new to relational algebra and finding it difficult. Ive answered a few questions, however they where relatively simple. Could do with help with these though. Database Patient (PatientCode, PatientSurname, PatientFirstname, PatientSex, PatientAge, PatientOccupation, PatientHeight, PatientWeight, PatientAddress) Doctor (DoctorCode, DoctorSurName, DoctorFirstName, DoctorPrivateAddress, MobileNo, Doctor Specilisim) Operation (Operation Code, PatientCode, DoctorCode, Date, Time, Result, OperationType) Is_Seen_By (PatientCode, DoctorCode, Date, Time). Q1. Find the surname and gender of the patients that have been operated on by doctor "DR333" and results have not been successful. Q2. Find the code of the operations that have been done on the 18th of November 2010 and have been successful, please also list the name of the doctors which were involved with the operation.

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• Merging some sorted lists with unknown order sequence

  - by Gabriel

  I've some sorted lists with variable number of elements. I wold like to merge the lists into one big list which contains all other lists in same order, without duplicates. Example: 1. XS,M,L,XL 2. S,M,XXL 3. XXS,XS,S,L Result: XXS,XS,S,M,L,XL,XXL The function should notify, if there are elements which have ambiguous positions. Here, it would be XXL and I need to specify its position after XL.

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• Pros and cons of cloud computing?

  - by Vimvq1987

  After 3 months of research, my thesis is nearly complete. Now I'm writing the report. Interesting parts are finished, now the boring and hard-to-write parts. I need to write about pros and cons of cloud computing. What it gives us and what it take us. I've searched much but there's only list, no explains. So I need your helps, to list and explains all of pros and cons of cloud computing. Thank you so much for this.

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