Search Results

Search found 13969 results on 559 pages for 'word count'.

Page 87/559 | < Previous Page | 83 84 85 86 87 88 89 90 91 92 93 94  | Next Page >

• python-McNuggets

  - by challarao

  I have created some program for this.But printed a,b,c values are not correct.Please check this weather it is correct or not? n=input("Enter the no.of McNuggets:") a,b,c=0,0,0 count=0 for a in range(n): if 6*a+9*b+20*c==n: count=count+1 break else: for b in range(n): if 6*a+9*b+20*c==n: count=count+1 break else: for c in range(n): if 6*a+9*b+20*c==n: count=count+1 break if count>0: print "It is possible to buy exactly",n,"packs of McNuggetss",a,b,c else: print "It is not possible to buy"

  Read the article

• How do you make a regular expression that matches a word with one randomly inserted character?

  - by Dfowj

  Hey all, i want to use a regular expression to match a word with one specified character randomly placed within it. I also want to keep that 'base' word's characters in their original order. For example, with the 'base' word of test and the specified character of 'y', i want the regular expression to match all the following, and ONLY the following: ytest, tyest, teyst, tesyt, testy Incase it matters, im working in javascript and using the dojo toolkit. Thanks!

  Read the article

• [AS 3.0] How to use the string.match method to find multiple occurrences of the same word in a strin

  - by Steven

  In Actionscript and Adobe Flex, I'm using a pattern and regexp (with the global flag) with the string.match method and it works how I'd like except when the match returns multiple occurrences of the same word in the text. In that case, all the matches for that word point only to the index for the first occurrence of that word. For example, if the text is "cat dog cat cat cow" and the pattern is a search for cat*, the match method returns an array of three occurrences of "cat", however, they all point to only the index of the first occurrence of cat when i use indexOf on a loop through the array. I'm assuming this is just how the string.match method is (although please let me know if i'm doing something wrong or missing something!). I want to find the specific indices of every occurrence of a match, even if it is of a word that was already previously matched. I'm wondering if that is just how the string.match method is and if so, if anyone has any idea what the best way to do this would be. Thanks.

  Read the article

• Replacing the last word of a path using sed.

  - by yael

  I have the following: param="/var/tmp/test" I need to replace the word test with another word such as new_test need a smart way to replace the last word after "/" with sed

  Read the article

• debugging Python program

  - by challarao

  I have created some program for this.But printed a,b,c values are not correct.Please check this whether it is correct or not? n=input("Enter the no.of McNuggets:") a,b,c=0,0,0 count=0 for a in range(n): if 6*a+9*b+20*c==n: count=count+1 break else: for b in range(n): if 6*a+9*b+20*c==n: count=count+1 break else: for c in range(n): if 6*a+9*b+20*c==n: count=count+1 break if count>0: print "It is possible to buy exactly",n,"packs of McNuggetss",a,b,c else: print "It is not possible to buy"

  Read the article

• How can i optimize this python code

  - by RandomVector

  def maxVote(nLabels): count = {} maxList = [] maxCount = 0 for nLabel in nLabels: if nLabel in count: count[nLabel] += 1 else: count[nLabel] = 1 #Check if the count is max if count[nLabel] > maxCount: maxCount = count[nLabel] maxList = [nLabel,] elif count[nLabel]==maxCount: maxList.append(nLabel) return random.choice(maxList) nLabels contains a list of integers. The above function returns the integer with highest frequency, if more than one have same frequency then a randomly selected integer from them is returned. E.g. maxVote([1,3,4,5,5,5,3,12,11]) is 5

  Read the article

• Some bold and some simple word within the same statement.

  - by Harikrishna

  I want to show some bold and some simple word within the same sentence.If I use the textbox then every word will be simple, not some bold and some simple so I want a control's text property in that I can write some bold and some simple word within the same statement

  Read the article

• MapReduce

  - by kaleidoscope

  MapReduce is a programming model and an associated implementation for processing and generating large data sets. Users specify a map function that processes a key/value pair to generate a set of  intermediate key/value pairs, and a reduce function that merges all intermediate values associated with the same intermediate key. Many real world tasks are expressible in this model, as shown in the paper. Programs written in this functional style are automatically parallelized and executed on a large cluster of commodity machines. The run-time system takes care of the details of partitioning the input data,  scheduling the program's execution across a set of machines, handling machine failures, and managing the required inter-machine communication. This allows programmers without any experience with parallel and distributed systems to easily utilize the resources of a large distributed system. Example: A process to count the appearances of each different word in a set of documents void map(String name, String document):   // name: document name   // document: document contents   for each word w in document:     EmitIntermediate(w, 1); void reduce(String word, Iterator partialCounts):   // word: a word   // partialCounts: a list of aggregated partial counts   int result = 0;   for each pc in partialCounts:     result += ParseInt(pc);   Emit(result); Here, each document is split in words, and each word is counted initially with a "1" value by the Map function, using the word as the result key. The framework puts together all the pairs with the same key and feeds them to the same call to Reduce, thus this function just needs to sum all of its input values to find the total appearances of that word.   Sarang, K

  Read the article

• Project Euler 17: (Iron)Python

  - by Ben Griswold

  In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 17.  As always, any feedback is welcome. # Euler 17 # http://projecteuler.net/index.php?section=problems&id=17 # If the numbers 1 to 5 are written out in words: # one, two, three, four, five, then there are # 3 + 3 + 5 + 4 + 4 = 19 letters used in total. # If all the numbers from 1 to 1000 (one thousand) # inclusive were written out in words, how many letters # would be used? # # NOTE: Do not count spaces or hyphens. For example, 342 # (three hundred and forty-two) contains 23 letters and # 115 (one hundred and fifteen) contains 20 letters. The # use of "and" when writing out numbers is in compliance # with British usage. import time start = time.time() def to_word(n): h = { 1 : "one", 2 : "two", 3 : "three", 4 : "four", 5 : "five", 6 : "six", 7 : "seven", 8 : "eight", 9 : "nine", 10 : "ten", 11 : "eleven", 12 : "twelve", 13 : "thirteen", 14 : "fourteen", 15 : "fifteen", 16 : "sixteen", 17 : "seventeen", 18 : "eighteen", 19 : "nineteen", 20 : "twenty", 30 : "thirty", 40 : "forty", 50 : "fifty", 60 : "sixty", 70 : "seventy", 80 : "eighty", 90 : "ninety", 100 : "hundred", 1000 : "thousand" } word = "" # Reverse the numbers so position (ones, tens, # hundreds,...) can be easily determined a = [int(x) for x in str(n)[::-1]] # Thousands position if (len(a) == 4 and a[3] != 0): # This can only be one thousand based # on the problem/method constraints word = h[a[3]] + " thousand " # Hundreds position if (len(a) >= 3 and a[2] != 0): word += h[a[2]] + " hundred" # Add "and" string if the tens or ones # position is occupied with a non-zero value. # Note: routine is broken up this way for [my] clarity. if (len(a) >= 2 and a[1] != 0): # catch 10 - 99 word += " and" elif len(a) >= 1 and a[0] != 0: # catch 1 - 9 word += " and" # Tens and ones position tens_position_value = 99 if (len(a) >= 2 and a[1] != 0): # Calculate the tens position value per the # first and second element in array # e.g. (8 * 10) + 1 = 81 tens_position_value = int(a[1]) * 10 + a[0] if tens_position_value <= 20: # If the tens position value is 20 or less # there's an entry in the hash. Use it and there's # no need to consider the ones position word += " " + h[tens_position_value] else: # Determine the tens position word by # dividing by 10 first. E.g. 8 * 10 = h[80] # We will pick up the ones position word later in # the next part of the routine word += " " + h[(a[1] * 10)] if (len(a) >= 1 and a[0] != 0 and tens_position_value > 20): # Deal with ones position where tens position is # greater than 20 or we have a single digit number word += " " + h[a[0]] # Trim the empty spaces off both ends of the string return word.replace(" ","") def to_word_length(n): return len(to_word(n)) print sum([to_word_length(i) for i in xrange(1,1001)]) print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

  Read the article

• Checking if an Unloaded Collection Contains Elements

  - by Ricardo Peres

  If you want to know if an unloaded collection in an entity contains elements, or count them, without actually loading them, you need to use a custom query; that is because the Count property (if the collection is not mapped with lazy=”extra”) and the LINQ Count() and Any() methods force the whole collection to be loaded. You can use something like these two methods, one for checking if there are any values, the other for actually counting them: 1: public static Boolean Exists(this ISession session, IEnumerable collection) 2: { 3: if (collection is IPersistentCollection) 4: { 5: IPersistentCollection col = collection as IPersistentCollection; 6:  7: if (col.WasInitialized == false) 8: { 9: String[] roleParts = col.Role.Split('.'); 10: String ownerTypeName = String.Join(".", roleParts, 0, roleParts.Length - 1); 11: String ownerCollectionName = roleParts.Last(); 12: String hql = "select 1 from " + ownerTypeName + " it where it.id = :id and exists elements(it." + ownerCollectionName + ")"; 13: Boolean exists = session.CreateQuery(hql).SetParameter("id", col.Key).List().Count == 1; 14:  15: return (exists); 16: } 17: } 18:  19: return ((collection as IEnumerable).OfType<Object>().Any()); 20: } 21:  22: public static Int64 Count(this ISession session, IEnumerable collection) 23: { 24: if (collection is IPersistentCollection) 25: { 26: IPersistentCollection col = collection as IPersistentCollection; 27:  28: if (col.WasInitialized == false) 29: { 30: String[] roleParts = col.Role.Split('.'); 31: String ownerTypeName = String.Join(".", roleParts, 0, roleParts.Length - 1); 32: String ownerCollectionName = roleParts.Last(); 33: String hql = "select count(elements(it." + ownerCollectionName + ")) from " + ownerTypeName + " it where it.id = :id"; 34: Int64 count = session.CreateQuery(hql).SetParameter("id", col.Key).UniqueResult<Int64>(); 35:  36: return (count); 37: } 38: } 39:  40: return ((collection as IEnumerable).OfType<Object>().Count()); 41: } Here’s how: 1: MyEntity entity = session.Load(100); 2:  3: if (session.Exists(entity.SomeCollection)) 4: { 5: Int32 count = session.Count(entity.SomeCollection); 6: //... 7: }

  Read the article

• How to get Messages Pending Count from a Queue using WLST?

  - by lmestre

  WLST is a scripting Language that helps to achieve similar functionality as the ones you have in WebLogic console, but in a command line fashion.You can develop your WLST Scripts using Eclipse OEPE, read more here:https://blogs.oracle.com/oepe/entry/new_oracle_enterprise_pack_forFinally, here is an example to get Messages Pending Count using WLST: . ./setDomainEnv.sh java weblogic.WLST connect('weblogic','welcome1','t3://localhost:7001') domainRuntime() jms= getMBean ('ServerRuntimes/MyManagedServer/JMSRuntime/MyManagedServer.jms/JMSServers/MyJMSServer/Destinations/MyModule!MyQueue') jms.getMessagesPendingCount() Enjoy!WLST documentation:http://docs.oracle.com/middleware/1212/wls/WLSTG/index.html

  Read the article

• OpenOffice Calc: How can I count the number of different items with data pilot?

  - by manu

  Hi all, I have a rather long spreadsheet with historical information of issues solved by some user on a collaborative environment. The spreadsheet have the following (relevant) columns date, week no., project, author id, etc... The week no. is calculated from the date, is basically the year concatenated with the week number within that year; for instance, both 2009-02-18 and 2009-02-20 yield the week number 200908 - the 8th week of year 2009; and 2009-02-23 yields 200909 - the 9th week of year 2009. I need to count how many different users (given by author id) contributed to some project, on a weekly basis. I have setup a data pilot with the week as Row Field, the project as the Column Field, and count-author as the Data Field. However, this counts the author id as different instances. This is not what I need. I need to count how many different users contributed to each project on a weekly basis. I expect to get something like: projects week Project1 Project2 Project3 200901 10 2 200902 2 7 Each inner cell containing how many different users contributed. With the count-author configuration, what I get is how many contributions (total) got the project on that week. Is there a way to tell OpenOffice Calc to do what I want?

  Read the article

• Invalid conversion from int to int

  - by FOXMULDERIZE

  #include <iostream> #include<fstream> using namespace std; void showvalues(int,int,int []); void showvalues2(int,int); void sumtotal(int,int); int main() { const int SIZE_A= 9; int arreglo[SIZE_A]; ifstream archivo_de_entrada; archivo_de_entrada.open("numeros.txt"); int count,suma,total,a,b,c,d,e,f; int total1=0; int total2=0; //lee/// for(count =0 ;count < SIZE_A;count++) archivo_de_entrada>>arreglo[count] ; archivo_de_entrada.close(); showvalues(0,3,9); HERE IS THE PROBLEM showvalues2(5,8); sumtotal(total1,total2); system("pause"); return 0; } void showvalues(int a,int b,int v) { //muestra//////////////////////// cout<< "los num son "; for(count = a ;count <= b;count++) total1 = total1 + arreglo[count]; cout <<total1<<" "; cout <<endl; } void showvalues2(int c,int d) { ////////////////////////////// cout<< "los num 2 son "; for(count =5 ;count <=8;count++) total2 = total2 + arreglo[count]; cout <<total2<<" "; cout <<endl; } void sumtotal(int e,int f) { ///////////////////////////////// cout<<"la suma de t1 y t2 es "; total= total1 + total2; cout<<total; cout <<endl; }

  Read the article

• How to draw multiple rectangles using c#

  - by Nivas

  I have drawn and saved the Rectangle on the image i loaded in the picture box. How i like to draw multiple rectangles for that i tried array in the rectangle but it gives error ("Object reference not set to an instance of an object." (Null reference Exception was unhandled). private void pictureBox1_MouseDown(object sender, MouseEventArgs e) { if (mybitmap == null) { mybitmap = new Bitmap(sz.Width, sz.Height); } rect[count] = new Rectangle(e.X, e.Y, 0, 0); this.Invalidate(); } private void pictureBox1_MouseMove(object sender, MouseEventArgs e) { if (stayToolStripMenuItem.Checked == true) { switch (e.Button) { case MouseButtons.Left: { rect[count] = new Rectangle(rect[count].Left, rect[count].Top, e.X - rect[count].Left, e.Y - rect[count].Top); pictureBox1.Invalidate(); count++: break; } } } } private void pictureBox1_Paint(object sender, PaintEventArgs e) { if (stayToolStripMenuItem.Checked == true) { button1.Visible = true; button2.Visible = true; if (mybitmap == null) { return; } using (g = Graphics.FromImage(mybitmap)) { using (Pen pen = new Pen(Color.Red, 2)) { //g.Clear(Color.Transparent); e.Graphics.DrawRectangle(pen, rect); label1.Top = rect[count].Top; label1[count].Left = rect[count].Left; label1.Width = rect[count].Width; label1.Height = rect[count].Height; if (label1.TextAlign == ContentAlignment.TopLeft) { e.Graphics.DrawString(label1.Text, label1.Font, new SolidBrush(label1.ForeColor), rect); g.DrawString(label1.Text, label1.Font, new SolidBrush(label1.ForeColor), rect); g.DrawRectangle(pen, rect[count]); } } } } } How can i do this.....

  Read the article

• Strange (Undefined?) Behavior of Free in C

  - by Chris Cirefice

  This is really strange... and I can't debug it (tried for about two hours, debugger starts going haywire after a while...). Anyway, I'm trying to do something really simple: Free an array of strings. The array is in the form: char **myStrings. The array elements are initialized as: myString[index] = malloc(strlen(word)); myString[index] = word; and I'm calling a function like this: free_memory(myStrings, size); where size is the length of the array (I know this is not the problem, I tested it extensively and everything except this function is working). free_memory looks like this: void free_memory(char **list, int size) { for (int i = 0; i < size; i ++) { free(list[i]); } free(list); } Now here comes the weird part. if (size> strlen(list[i])) then the program crashes. For example, imagine that I have a list of strings that looks something like this: myStrings[0] = "Some"; myStrings[1] = "random"; myStrings[2] = "strings"; And thus the length of this array is 3. If I pass this to my free_memory function, strlen(myStrings[0]) > 3 (4 3), and the program crashes. However, if I change myStrings[0] to be "So" instead, then strlen(myStrings[0]) < 3 (2 < 3) and the program does not crash. So it seems to me that free(list[i]) is actually going through the char[] that is at that location and trying to free each character, which I imagine is undefined behavior. The only reason I say this is because I can play around with the size of the first element of myStrings and make the program crash whenever I feel like it, so I'm assuming that this is the problem area. Note: I did try to debug this by stepping through the function that calls free_memory, noting any weird values and such, but the moment I step into the free_memory function, the debugger crashes, so I'm not really sure what is going on. Nothing is out of the ordinary until I enter the function, then the world explodes. Another note: I also posted the shortened version of the source for this program (not too long; Pastebin) here. I am compiling on MinGW with the c99 flag on. PS - I just thought of this. I am indeed passing numUniqueWords to the free function, and I know that this does not actually free the entire piece of memory that I allocated. I've called it both ways, that's not the issue. And I left it how I did because that is the way that I will be calling it after I get it to work in the first place, I need to revise some of my logic in that function. Source, as per request (on-site): #include <stdio.h> #include <string.h> #include <ctype.h> #include <stdlib.h> #include "words.h" int getNumUniqueWords(char text[], int size); int main(int argc, char* argv[]) { setvbuf(stdout, NULL, 4, _IONBF); // For Eclipse... stupid bug. --> does NOT affect the program, just the output to console! int nbr_words; char text[] = "Some - \"text, a stdin\". We'll have! also repeat? We'll also have a repeat!"; int length = sizeof(text); nbr_words = getNumUniqueWords(text, length); return 0; } void free_memory(char **list, int size) { for (int i = 0; i < size; i ++) { // You can see that printing the values is fine, as long as free is not called. // When free is called, the program will crash if (size > strlen(list[i])) //printf("Wanna free value %d w/len of %d: %s\n", i, strlen(list[i]), list[i]); free(list[i]); } free(list); } int getNumUniqueWords(char text[], int length) { int numTotalWords = 0; char *word; printf("Length: %d characters\n", length); char totalWords[length]; strcpy(totalWords, text); word = strtok(totalWords, " ,.-!?()\"0123456789"); while (word != NULL) { numTotalWords ++; printf("%s\n", word); word = strtok(NULL, " ,.-!?()\"0123456789"); } printf("Looks like we counted %d total words\n\n", numTotalWords); char *uniqueWords[numTotalWords]; char *tempWord; int wordAlreadyExists = 0; int numUniqueWords = 0; char totalWordsCopy[length]; strcpy(totalWordsCopy, text); for (int i = 0; i < numTotalWords; i++) { uniqueWords[i] = NULL; } // Tokenize until all the text is consumed. word = strtok(totalWordsCopy, " ,.-!?()\"0123456789"); while (word != NULL) { // Look through the word list for the current token. for (int j = 0; j < numTotalWords; j ++) { // Just for clarity, no real meaning. tempWord = uniqueWords[j]; // The word list is either empty or the current token is not in the list. if (tempWord == NULL) { break; } //printf("Comparing (%s) with (%s)\n", tempWord, word); // If the current token is the same as the current element in the word list, mark and break if (strcmp(tempWord, word) == 0) { printf("\nDuplicate: (%s)\n\n", word); wordAlreadyExists = 1; break; } } // Word does not exist, add it to the array. if (!wordAlreadyExists) { uniqueWords[numUniqueWords] = malloc(strlen(word)); uniqueWords[numUniqueWords] = word; numUniqueWords ++; printf("Unique: %s\n", word); } // Reset flags and continue. wordAlreadyExists = 0; word = strtok(NULL, " ,.-!?()\"0123456789"); } // Print out the array just for funsies - make sure it's working properly. for (int x = 0; x <numUniqueWords; x++) { printf("Unique list %d: %s\n", x, uniqueWords[x]); } printf("\nNumber of unique words: %d\n\n", numUniqueWords); // Right below is where things start to suck. free_memory(uniqueWords, numUniqueWords); return numUniqueWords; }

  Read the article

• How can I view and sort after the page count for multiple PDF files in a Windows file explorer?

  - by grunwald2.0

  I unsuccessfully used the "pages" feature in Windows Explorer, as well as in Directory Opus 10 and Free Commander XT (which I installed just for that reason, to try it out) to display the page count of multiple PDFs in a folder. All my PDF's are free to edit, i.e. not write-protected. I don't understand why any PDF reader can display the (correct) page number, but none of the file explorers can? (In the "details" view of course.) The only documents whose page count is displayed are MS Word documents. Do I have to use Adobe Bridge? (I didn't try it.) On a side-note: Did that change in Windows 8? Initial research: Google search was unsuccessful, the only slightly related SE topic I found was "How to count pages in multiple PDF files?".

  Read the article

• lnk2019 error in very simple c++ program

  - by Erin

  I have tried removing various parts and building, but nothing makes the lnk2019 error go away, or even produces any normal errors. Everything is in the one file at the moment (it won't be later when it is finished). The program has three lists of words and makes a jargon phrase out of them, and you are supposed to be able to add words, remove words, view the lists, restore defaults, save changes to file, and load changes from file. #include "stdafx.h" #include <iostream> #include <string.h> using namespace std; const int maxlist = 20; string adj1[maxlist], adj2[maxlist], noun[maxlist]; void defaultlist(int list) { if(list == 1) { adj1[0] = "green"; adj1[1] = "red"; adj1[2] = "yellow"; adj1[3] = "blue"; adj1[4] = "purple"; int i = 5; while(i != maxlist) { adj1[i] = ""; i = i + 1; } } if(list == 2) { adj2[0] = "shiny"; adj2[1] = "hard"; adj2[2] = "soft"; adj2[3] = "spiky"; adj2[4] = "furry"; int i = 5; while(i != maxlist) { adj2[i] = ""; i = i + 1; } } if(list == 3) { noun[0] = "cat"; noun[1] = "dog"; noun[2] = "desk"; noun[3] = "chair"; noun[4] = "door"; int i = 5; while(i != maxlist) { noun[i] = ""; i = i + 1; } } return; } void printlist(int list) { if(list == 1) { int i = 0; while(!(i == maxlist)) { cout << adj1[i] << endl; i = i + 1; } } if(list == 2) { int i = 0; while(!(i == maxlist)) { cout << adj2[i] << endl; i = i + 1; } } if(list == 3) { int i = 0; while(!(i == maxlist)) { cout << noun[i] << endl; i = i + 1; } } return; } string makephrase() { int num1 = rand()%maxlist; int num2 = rand()%maxlist; int num3 = rand()%maxlist; int num4 = rand()%1; string word1, word2, word3; if(num4 = 0) { word1 = adj1[num1]; word2 = adj2[num2]; } else { word1 = adj2[num1]; word2 = adj1[num2]; } word3 = noun[num3]; return word1 + " ," + word2 + " " + word3; } string addword(string word, int list) { string result; if(list == 1) { int i = 0; while(!(adj1[i] == "" || i == maxlist)) { i = i + 1; } if(i == maxlist) result = "List is full. Please try again."; if(adj1[i] == "") { adj1[i] = word; result = "Word was entered successfully."; } } if(list == 2) { int i = 0; while(!(adj2[i] == "" || i == maxlist)) { i = i + 1; } if(i == maxlist) result = "List is full. Please try again."; if(adj2[i] == "") { adj2[i] = word; result = "Word was entered successfully."; } } if(list == 3) { int i = 0; while(!(noun[i] == "" || i == maxlist)) { i = i + 1; } if(i == maxlist) result = "List is full. Please try again."; if(noun[i] == "") { noun[i] = word; result = "Word was entered successfully."; } } return result; } string removeword(string word, int list) { string result; if(list == 1) { int i = 0; while(!(adj1[i] == word || i == maxlist)) { i = i + 1; } if(i == maxlist) result = "Word is not on the list. Please try again."; if(adj1[i] == word) { adj1[i] = ""; result = "Word was removed successfully."; } } if(list == 2) { int i = 0; while(!(adj2[i] == word || i == maxlist)) { i = i + 1; } if(i == maxlist) result = "Word is not on the list. Please try again."; if(adj2[i] == word) { adj2[i] = ""; result = "Word was removed successfully."; } } if(list == 3) { int i = 0; while(!(noun[i] == word || i == maxlist)) { i = i + 1; } if(i == maxlist) result = "Word is not on the list. Please try again."; if(noun[i] == word) { noun[i] = ""; result = "Word was removed successfully."; } } return result; } /////////////////////////////main/////////////////////////////////// int main() { string mainselection; string makeselection; string phrase; defaultlist(1); defaultlist(2); defaultlist(3); cout << "This program generates jargon phrases made of two adjectives and one noun,"; cout << " on three lists. Each list may contain a maximum of " << maxlist << "elements."; cout << " Please choose from the following menu by typing the appropriate number "; cout << "and pressing enter." << endl; cout << endl; cout << "1. Make a jargon phrase." << endl; cout << "2. View a list." << endl; cout << "3. Add a word to a list." << endl; cout << "4. Remove a word from a list." << endl; cout << "5. Restore default lists." << endl; cout << "More options coming soon!." << endl; cin mainselection if(mainselection == 1) { phrase = makephrase(); cout << "Your phrase is " << phrase << "." << endl; cout << "To make another phrase, press 1. To go back to the main menu,"; cout << " press 2. To exit the program, press 3." << endl; cin makeselection; while(!(makeselection == "1" || makeselection == "2" || makeselection == "3")) { cout << "You have entered an invalid selection. Please try again." << endl; cin makeselection; } while(makeselection == "1") { phrase = makephrase(); cout << "To make another phrase, press 1. To go back to the main menu,"; cout << " press 2. To exit the program, press 3." << endl; } if(makeselection == "2") main(); if(makeselection == "3") return 0; } return 0; } //Rest of the options coming soon!

  Read the article

• Why is my Internet connection randomly dropping?

  - by Jeanno

  Ever since I have installed 12.04 (clean install not an upgrade), i have been having a drop in the Internet connection. The drop in the connection can be anything from 15 seconds to about 3 mins, and then the connection comes back. This behaviour happens while I am actively browsing the Internet, or if I wake up the computer and open Firefox (sometimes I have connection and sometimes I don't) . Please note that when the internet connection is on, it is not slow (as speedtest.net results show) In the beginning, I thought it was a problem with the driver r8169 for my RTL8111/8168B Ethernet card, so I downloaded the r8168 from Realtek website, followed the detailed instructions (blacklisted r8169, changed the file to '.bsh' ...), but still the same problem persisted. So I switched to a wireless connection, and I got the same problem with internet connection dropping randomly. Any ideas? Thanks in advance Output from 'lspci -v' Code: 00:00.0 Host bridge: Intel Corporation 2nd Generation Core Processor Family DRAM Controller (rev 09) Subsystem: Dell Device 04a7 Flags: bus master, fast devsel, latency 0 Capabilities: [e0] Vendor Specific Information: Len=0c <?> 00:01.0 PCI bridge: Intel Corporation Xeon E3-1200/2nd Generation Core Processor Family PCI Express Root Port (rev 09) (prog-if 00 [Normal decode]) Flags: bus master, fast devsel, latency 0 Bus: primary=00, secondary=01, subordinate=01, sec-latency=0 I/O behind bridge: 0000e000-0000efff Memory behind bridge: f8000000-fa0fffff Prefetchable memory behind bridge: 00000000d0000000-00000000dbffffff Capabilities: [88] Subsystem: Dell Device 04a7 Capabilities: [80] Power Management version 3 Capabilities: [90] MSI: Enable+ Count=1/1 Maskable- 64bit- Capabilities: [a0] Express Root Port (Slot+), MSI 00 Capabilities: [100] Virtual Channel Capabilities: [140] Root Complex Link Kernel driver in use: pcieport Kernel modules: shpchp 00:01.1 PCI bridge: Intel Corporation Xeon E3-1200/2nd Generation Core Processor Family PCI Express Root Port (rev 09) (prog-if 00 [Normal decode]) Flags: bus master, fast devsel, latency 0 Bus: primary=00, secondary=02, subordinate=02, sec-latency=0 I/O behind bridge: 0000d000-0000dfff Memory behind bridge: f4000000-f60fffff Prefetchable memory behind bridge: 00000000c0000000-00000000cbffffff Capabilities: [88] Subsystem: Dell Device 04a7 Capabilities: [80] Power Management version 3 Capabilities: [90] MSI: Enable+ Count=1/1 Maskable- 64bit- Capabilities: [a0] Express Root Port (Slot+), MSI 00 Capabilities: [100] Virtual Channel Capabilities: [140] Root Complex Link Kernel driver in use: pcieport Kernel modules: shpchp 00:16.0 Communication controller: Intel Corporation 6 Series/C200 Series Chipset Family MEI Controller #1 (rev 04) Subsystem: Dell Device 04a7 Flags: bus master, fast devsel, latency 0, IRQ 52 Memory at f6108000 (64-bit, non-prefetchable) [size=16] Capabilities: [50] Power Management version 3 Capabilities: [8c] MSI: Enable+ Count=1/1 Maskable- 64bit+ Kernel driver in use: mei Kernel modules: mei 00:1a.0 USB controller: Intel Corporation 6 Series/C200 Series Chipset Family USB Enhanced Host Controller #2 (rev 05) (prog-if 20 [EHCI]) Subsystem: Dell Device 04a7 Flags: bus master, medium devsel, latency 0, IRQ 16 Memory at f6107000 (32-bit, non-prefetchable) [size=1K] Capabilities: [50] Power Management version 2 Capabilities: [58] Debug port: BAR=1 offset=00a0 Capabilities: [98] PCI Advanced Features Kernel driver in use: ehci_hcd 00:1b.0 Audio device: Intel Corporation 6 Series/C200 Series Chipset Family High Definition Audio Controller (rev 05) Subsystem: Dell Device 04a7 Flags: bus master, fast devsel, latency 0, IRQ 53 Memory at f6100000 (64-bit, non-prefetchable) [size=16K] Capabilities: [50] Power Management version 2 Capabilities: [60] MSI: Enable+ Count=1/1 Maskable- 64bit+ Capabilities: [70] Express Root Complex Integrated Endpoint, MSI 00 Capabilities: [100] Virtual Channel Capabilities: [130] Root Complex Link Kernel driver in use: snd_hda_intel Kernel modules: snd-hda-intel 00:1c.0 PCI bridge: Intel Corporation 6 Series/C200 Series Chipset Family PCI Express Root Port 1 (rev b5) (prog-if 00 [Normal decode]) Flags: bus master, fast devsel, latency 0 Bus: primary=00, secondary=03, subordinate=03, sec-latency=0 Memory behind bridge: fa400000-fa4fffff Capabilities: [40] Express Root Port (Slot+), MSI 00 Capabilities: [80] MSI: Enable- Count=1/1 Maskable- 64bit- Capabilities: [90] Subsystem: Dell Device 04a7 Capabilities: [a0] Power Management version 2 Kernel driver in use: pcieport Kernel modules: shpchp 00:1c.1 PCI bridge: Intel Corporation 6 Series/C200 Series Chipset Family PCI Express Root Port 2 (rev b5) (prog-if 00 [Normal decode]) Flags: bus master, fast devsel, latency 0 Bus: primary=00, secondary=04, subordinate=04, sec-latency=0 I/O behind bridge: 0000c000-0000cfff Prefetchable memory behind bridge: 00000000dc100000-00000000dc1fffff Capabilities: [40] Express Root Port (Slot+), MSI 00 Capabilities: [80] MSI: Enable- Count=1/1 Maskable- 64bit- Capabilities: [90] Subsystem: Dell Device 04a7 Capabilities: [a0] Power Management version 2 Kernel driver in use: pcieport Kernel modules: shpchp 00:1c.2 PCI bridge: Intel Corporation 6 Series/C200 Series Chipset Family PCI Express Root Port 3 (rev b5) (prog-if 00 [Normal decode]) Flags: bus master, fast devsel, latency 0 Bus: primary=00, secondary=05, subordinate=05, sec-latency=0 I/O behind bridge: 0000b000-0000bfff Memory behind bridge: fa300000-fa3fffff Capabilities: [40] Express Root Port (Slot+), MSI 00 Capabilities: [80] MSI: Enable- Count=1/1 Maskable- 64bit- Capabilities: [90] Subsystem: Dell Device 04a7 Capabilities: [a0] Power Management version 2 Kernel driver in use: pcieport Kernel modules: shpchp 00:1c.3 PCI bridge: Intel Corporation 6 Series/C200 Series Chipset Family PCI Express Root Port 4 (rev b5) (prog-if 00 [Normal decode]) Flags: bus master, fast devsel, latency 0 Bus: primary=00, secondary=06, subordinate=06, sec-latency=0 I/O behind bridge: 0000a000-0000afff Memory behind bridge: fa200000-fa2fffff Capabilities: [40] Express Root Port (Slot+), MSI 00 Capabilities: [80] MSI: Enable- Count=1/1 Maskable- 64bit- Capabilities: [90] Subsystem: Dell Device 04a7 Capabilities: [a0] Power Management version 2 Kernel driver in use: pcieport Kernel modules: shpchp 00:1d.0 USB controller: Intel Corporation 6 Series/C200 Series Chipset Family USB Enhanced Host Controller #1 (rev 05) (prog-if 20 [EHCI]) Subsystem: Dell Device 04a7 Flags: bus master, medium devsel, latency 0, IRQ 23 Memory at f6106000 (32-bit, non-prefetchable) [size=1K] Capabilities: [50] Power Management version 2 Capabilities: [58] Debug port: BAR=1 offset=00a0 Capabilities: [98] PCI Advanced Features Kernel driver in use: ehci_hcd 00:1f.0 ISA bridge: Intel Corporation P67 Express Chipset Family LPC Controller (rev 05) Subsystem: Dell Device 04a7 Flags: bus master, medium devsel, latency 0 Capabilities: [e0] Vendor Specific Information: Len=0c <?> Kernel modules: iTCO_wdt 00:1f.2 RAID bus controller: Intel Corporation 82801 SATA Controller [RAID mode] (rev 05) Subsystem: Dell Device 04a7 Flags: bus master, 66MHz, medium devsel, latency 0, IRQ 42 I/O ports at f070 [size=8] I/O ports at f060 [size=4] I/O ports at f050 [size=8] I/O ports at f040 [size=4] I/O ports at f020 [size=32] Memory at f6105000 (32-bit, non-prefetchable) [size=2K] Capabilities: [80] MSI: Enable+ Count=1/1 Maskable- 64bit- Capabilities: [70] Power Management version 3 Capabilities: [a8] SATA HBA v1.0 Capabilities: [b0] PCI Advanced Features Kernel driver in use: ahci 00:1f.3 SMBus: Intel Corporation 6 Series/C200 Series Chipset Family SMBus Controller (rev 05) Subsystem: Dell Device 04a7 Flags: medium devsel, IRQ 5 Memory at f6104000 (64-bit, non-prefetchable) [size=256] I/O ports at f000 [size=32] Kernel modules: i2c-i801 01:00.0 VGA compatible controller: NVIDIA Corporation Device 0dc5 (rev a1) (prog-if 00 [VGA controller]) Subsystem: NVIDIA Corporation Device 085b Flags: bus master, fast devsel, latency 0, IRQ 16 Memory at f8000000 (32-bit, non-prefetchable) [size=16M] Memory at d0000000 (64-bit, prefetchable) [size=128M] Memory at d8000000 (64-bit, prefetchable) [size=32M] I/O ports at e000 [size=128] Expansion ROM at fa000000 [disabled] [size=512K] Capabilities: [60] Power Management version 3 Capabilities: [68] MSI: Enable- Count=1/1 Maskable- 64bit+ Capabilities: [78] Express Endpoint, MSI 00 Capabilities: [b4] Vendor Specific Information: Len=14 <?> Capabilities: [100] Virtual Channel Capabilities: [128] Power Budgeting <?> Capabilities: [600] Vendor Specific Information: ID=0001 Rev=1 Len=024 <?> Kernel driver in use: nouveau Kernel modules: nouveau, nvidiafb 01:00.1 Audio device: NVIDIA Corporation GF106 High Definition Audio Controller (rev a1) Subsystem: NVIDIA Corporation Device 085b Flags: bus master, fast devsel, latency 0, IRQ 17 Memory at fa080000 (32-bit, non-prefetchable) [size=16K] Capabilities: [60] Power Management version 3 Capabilities: [68] MSI: Enable- Count=1/1 Maskable- 64bit+ Capabilities: [78] Express Endpoint, MSI 00 Kernel driver in use: snd_hda_intel Kernel modules: snd-hda-intel 02:00.0 VGA compatible controller: NVIDIA Corporation Device 0dc5 (rev a1) (prog-if 00 [VGA controller]) Subsystem: NVIDIA Corporation Device 085b Flags: bus master, fast devsel, latency 0, IRQ 17 Memory at f4000000 (32-bit, non-prefetchable) [size=32M] Memory at c0000000 (64-bit, prefetchable) [size=128M] Memory at c8000000 (64-bit, prefetchable) [size=64M] I/O ports at d000 [size=128] Expansion ROM at f6000000 [disabled] [size=512K] Capabilities: [60] Power Management version 3 Capabilities: [68] MSI: Enable- Count=1/1 Maskable- 64bit+ Capabilities: [78] Express Endpoint, MSI 00 Capabilities: [b4] Vendor Specific Information: Len=14 <?> Capabilities: [100] Virtual Channel Capabilities: [128] Power Budgeting <?> Capabilities: [600] Vendor Specific Information: ID=0001 Rev=1 Len=024 <?> Kernel driver in use: nouveau Kernel modules: nouveau, nvidiafb 02:00.1 Audio device: NVIDIA Corporation GF106 High Definition Audio Controller (rev a1) Subsystem: NVIDIA Corporation Device 085b Flags: bus master, fast devsel, latency 0, IRQ 18 Memory at f6080000 (32-bit, non-prefetchable) [size=16K] Capabilities: [60] Power Management version 3 Capabilities: [68] MSI: Enable- Count=1/1 Maskable- 64bit+ Capabilities: [78] Express Endpoint, MSI 00 Kernel driver in use: snd_hda_intel Kernel modules: snd-hda-intel 03:00.0 USB controller: NEC Corporation uPD720200 USB 3.0 Host Controller (rev 03) (prog-if 30 [XHCI]) Subsystem: Dell Device 04a7 Flags: bus master, fast devsel, latency 0, IRQ 16 Memory at fa400000 (64-bit, non-prefetchable) [size=8K] Capabilities: [50] Power Management version 3 Capabilities: [70] MSI: Enable- Count=1/8 Maskable- 64bit+ Capabilities: [90] MSI-X: Enable+ Count=8 Masked- Capabilities: [a0] Express Endpoint, MSI 00 Capabilities: [100] Advanced Error Reporting Capabilities: [140] Device Serial Number ff-ff-ff-ff-ff-ff-ff-ff Capabilities: [150] Latency Tolerance Reporting Kernel driver in use: xhci_hcd 04:00.0 Ethernet controller: Realtek Semiconductor Co., Ltd. RTL8111/8168B PCI Express Gigabit Ethernet controller (rev 06) Subsystem: Dell Device 04a7 Flags: bus master, fast devsel, latency 0, IRQ 51 I/O ports at c000 [size=256] Memory at dc104000 (64-bit, prefetchable) [size=4K] Memory at dc100000 (64-bit, prefetchable) [size=16K] Capabilities: [40] Power Management version 3 Capabilities: [50] MSI: Enable+ Count=1/1 Maskable- 64bit+ Capabilities: [70] Express Endpoint, MSI 01 Capabilities: [b0] MSI-X: Enable- Count=4 Masked- Capabilities: [d0] Vital Product Data Capabilities: [100] Advanced Error Reporting Capabilities: [140] Virtual Channel Capabilities: [160] Device Serial Number 03-00-00-00-68-4c-e0-00 Kernel driver in use: r8168 Kernel modules: r8168 05:00.0 FireWire (IEEE 1394): VIA Technologies, Inc. VT6315 Series Firewire Controller (rev 01) (prog-if 10 [OHCI]) Subsystem: Dell Device 04a7 Flags: bus master, fast devsel, latency 0, IRQ 18 Memory at fa300000 (64-bit, non-prefetchable) [size=2K] I/O ports at b000 [size=256] Capabilities: [50] Power Management version 3 Capabilities: [80] MSI: Enable- Count=1/1 Maskable+ 64bit+ Capabilities: [98] Express Endpoint, MSI 00 Capabilities: [100] Advanced Error Reporting Capabilities: [130] Device Serial Number 00-10-dc-ff-ff-cf-56-1a Kernel driver in use: firewire_ohci Kernel modules: firewire-ohci 06:00.0 SATA controller: JMicron Technology Corp. JMB362 SATA Controller (rev 10) (prog-if 01 [AHCI 1.0]) Subsystem: Dell Device 04a7 Flags: bus master, fast devsel, latency 0, IRQ 19 I/O ports at a040 [size=8] I/O ports at a030 [size=4] I/O ports at a020 [size=8] I/O ports at a010 [size=4] I/O ports at a000 [size=16] Memory at fa210000 (32-bit, non-prefetchable) [size=512] Capabilities: [8c] Power Management version 3 Capabilities: [50] Express Legacy Endpoint, MSI 00 Kernel driver in use: ahci Note that my wireless card is not showing, I have the Ralink 3390 card (which apparently does not show up on Ubuntu for some reason), however I am able to connect to wireless network and connect to the internet (when it is working)

  Read the article

• What word processor can manage left and right pages separately?

  - by chtfn

  I was after a word processor that can manage left and right pages separately so I can have my text on the right pages and my illustrations on the left pages. I know LibeOffice can set "left page" and "right page" formats, but there is no simple way to make the text jump from right page to right page when writing. I don't know what would be the best way to do that, but I guess I need an app that is designed for writing books and do that kind of thing, or an app that can associate a page format with an object format (so the "illustration" format can be exclusively associated with "left page" format, and "text body" exclusively associated with "right page", for example). Or is there an extension for LO or OOo out there that I didn't hear about? Cheers!

  Read the article

• Working with Timelines with LINQ to Twitter

  - by Joe Mayo

  When first working with the Twitter API, I thought that using SinceID would be an effective way to page through timelines. In practice it doesn’t work well for various reasons. To explain why, Twitter published an excellent document that is a must-read for anyone working with timelines: Twitter Documentation: Working with Timelines This post shows how to implement the recommended strategies in that document by using LINQ to Twitter. You should read the document in it’s entirety before moving on because my explanation will start at the bottom and work back up to the top in relation to the Twitter document. What follows is an explanation of SinceID, MaxID, and how they come together to help you efficiently work with Twitter timelines. The Role of SinceID Specifying SinceID says to Twitter, “Don’t return tweets earlier than this”. What you want to do is store this value after every timeline query set so that it can be reused on the next set of queries.  The next section will explain what I mean by query set, but a quick explanation is that it’s a loop that gets all new tweets. The SinceID is a backstop to avoid retrieving tweets that you already have. Here’s some initialization code that includes a variable named sinceID that will be used to populate the SinceID property in subsequent queries: // last tweet processed on previous query set ulong sinceID = 210024053698867204; ulong maxID; const int Count = 10; var statusList = new List<status>(); Here, I’ve hard-coded the sinceID variable, but this is where you would initialize sinceID from whatever storage you choose (i.e. a database). The first time you ever run this code, you won’t have a value from a previous query set. Initially setting it to 0 might sound like a good idea, but what if you’re querying a timeline with lots of tweets? Because of the number of tweets and rate limits, your query set might take a very long time to run. A caveat might be that Twitter won’t return an entire timeline back to Tweet #0, but rather only go back a certain period of time, the limits of which are documented for individual Twitter timeline API resources. So, to initialize SinceID at too low of a number can result in a lot of initial tweets, yet there is a limit to how far you can go back. What you’re trying to accomplish in your application should guide you in how to initially set SinceID. I have more to say about SinceID later in this post. The other variables initialized above include the declaration for MaxID, Count, and statusList. The statusList variable is a holder for all the timeline tweets collected during this query set. You can set Count to any value you want as the largest number of tweets to retrieve, as defined by individual Twitter timeline API resources. To effectively page results, you’ll use the maxID variable to set the MaxID property in queries, which I’ll discuss next. Initializing MaxID On your first query of a query set, MaxID will be whatever the most recent tweet is that you get back. Further, you don’t know what MaxID is until after the initial query. The technique used in this post is to do an initial query and then use the results to figure out what the next MaxID will be.  Here’s the code for the initial query: var userStatusResponse = (from tweet in twitterCtx.Status where tweet.Type == StatusType.User && tweet.ScreenName == "JoeMayo" && tweet.SinceID == sinceID && tweet.Count == Count select tweet) .ToList(); statusList.AddRange(userStatusResponse); // first tweet processed on current query maxID = userStatusResponse.Min( status => ulong.Parse(status.StatusID)) - 1; The query above sets both SinceID and Count properties. As explained earlier, Count is the largest number of tweets to return, but the number can be less. A couple reasons why the number of tweets that are returned could be less than Count include the fact that the user, specified by ScreenName, might not have tweeted Count times yet or might not have tweeted at least Count times within the maximum number of tweets that can be returned by the Twitter timeline API resource. Another reason could be because there aren’t Count tweets between now and the tweet ID specified by sinceID. Setting SinceID constrains the results to only those tweets that occurred after the specified Tweet ID, assigned via the sinceID variable in the query above. The statusList is an accumulator of all tweets receive during this query set. To simplify the code, I left out some logic to check whether there were no tweets returned. If  the query above doesn’t return any tweets, you’ll receive an exception when trying to perform operations on an empty list. Yeah, I cheated again. Besides querying initial tweets, what’s important about this code is the final line that sets maxID. It retrieves the lowest numbered status ID in the results. Since the lowest numbered status ID is for a tweet we already have, the code decrements the result by one to keep from asking for that tweet again. Remember, SinceID is not inclusive, but MaxID is. The maxID variable is now set to the highest possible tweet ID that can be returned in the next query. The next section explains how to use MaxID to help get the remaining tweets in the query set. Retrieving Remaining Tweets Earlier in this post, I defined a term that I called a query set. Essentially, this is a group of requests to Twitter that you perform to get all new tweets. A single query might not be enough to get all new tweets, so you’ll have to start at the top of the list that Twitter returns and keep making requests until you have all new tweets. The previous section showed the first query of the query set. The code below is a loop that completes the query set: do { // now add sinceID and maxID userStatusResponse = (from tweet in twitterCtx.Status where tweet.Type == StatusType.User && tweet.ScreenName == "JoeMayo" && tweet.Count == Count && tweet.SinceID == sinceID && tweet.MaxID == maxID select tweet) .ToList(); if (userStatusResponse.Count > 0) { // first tweet processed on current query maxID = userStatusResponse.Min( status => ulong.Parse(status.StatusID)) - 1; statusList.AddRange(userStatusResponse); } } while (userStatusResponse.Count != 0 && statusList.Count < 30); Here we have another query, but this time it includes the MaxID property. The SinceID property prevents reading tweets that we’ve already read and Count specifies the largest number of tweets to return. Earlier, I mentioned how it was important to check how many tweets were returned because failing to do so will result in an exception when subsequent code runs on an empty list. The code above protects against this problem by only working with the results if Twitter actually returns tweets. Reasons why there wouldn’t be results include: if the first query got all the new tweets there wouldn’t be more to get and there might not have been any new tweets between the SinceID and MaxID settings of the most recent query. The code for loading the returned tweets into statusList and getting the maxID are the same as previously explained. The important point here is that MaxID is being reset, not SinceID. As explained in the Twitter documentation, paging occurs from the newest tweets to oldest, so setting MaxID lets us move from the most recent tweets down to the oldest as specified by SinceID. The two loop conditions cause the loop to continue as long as tweets are being read or a max number of tweets have been read.  Logically, you want to stop reading when you’ve read all the tweets and that’s indicated by the fact that the most recent query did not return results. I put the check to stop after 30 tweets are reached to keep the demo from running too long – in the console the response scrolls past available buffer and I wanted you to be able to see the complete output. Yet, there’s another point to be made about constraining the number of items you return at one time. The Twitter API has rate limits and making too many queries per minute will result in an error from twitter that LINQ to Twitter raises as an exception. To use the API properly, you’ll have to ensure you don’t exceed this threshold. Looking at the statusList.Count as done above is rather primitive, but you can implement your own logic to properly manage your rate limit. Yeah, I cheated again. Summary Now you know how to use LINQ to Twitter to work with Twitter timelines. After reading this post, you have a better idea of the role of SinceID - the oldest tweet already received. You also know that MaxID is the largest tweet ID to retrieve in a query. Together, these settings allow you to page through results via one or more queries. You also understand what factors affect the number of tweets returned and considerations for potential error handling logic. The full example of the code for this post is included in the downloadable source code for LINQ to Twitter.   @JoeMayo

  Read the article

• Does having a Google "stop word" in a domain name have less SEO benefit than not having it?

  - by Dan

  Let me explain. Let's say my keyword I want to optimize is "green giraffes". But the domain greengiraffes.com (singular, plural, no hyphen, hyphen, etc.) is not available. I know that the search results for "green giraffes" and "about green giraffes" are essentially the same because "about" is a "stop word". Does that therefore also mean that the domain name "aboutgreengiraffes.com" is as good as "greengiraffes.com" in terms of SEO value? Are all stop words equal in that regard, or a shorter one (such as "e" or "z") is better?

  Read the article

• How do you properly word a Google search when you don't even have a solution in mind? [closed]

  - by Bruno Romaszkiewicz

  So, I'm stuck on a problem, looking for a solution, my rubber duck can't help me, my co-workers can't help me. Next natural step is research, right? Google can help me, He always can. Or so I'm told. My problem is, I never found much use for Google when looking for a programming solution, it's very useful for finding how to implement one, but when you don't even know where to start, how do you properly word a Google search? Is there any other option?

  Read the article

• PHP Calculating Text to Content Ratio

  - by James

  I am using the following code to calculate text to code ratio. I think it is crazy that no one can agree on how to properly calculate the result. I am looking any suggestions or ideas to improve this code that may make it more accurate. <?php // Returns the size of the content in bytes function findKb($content){ $count=0; $order = array("\r\n", "\n", "\r", "chr(13)", "\t", "\0", "\x0B"); $content = str_replace($order, "12", $content); for ($index = 0; $index < strlen($content); $index ++){ $byte = ord($content[$index]); if ($byte <= 127) { $count++; } else if ($byte >= 194 && $byte <= 223) { $count=$count+2; } else if ($byte >= 224 && $byte <= 239) { $count=$count+3; } else if ($byte >= 240 && $byte <= 244) { $count=$count+4; } } return $count; } // Collect size of entire code $filesize = findKb($content); // Remove anything within script tags $code = preg_replace("@<script[^>]*>.+</script[^>]*>@i", "", $content); // Remove anything within style tags $code = preg_replace("@<style[^>]*>.+</style[^>]*>@i", "", $content); // Remove all tags from the system $code = strip_tags($code); // Remove Extra whitespace from the content $code = preg_replace( '/\s+/', ' ', $code ); // Find the size of the remaining code $codesize = findKb($code); // Calculate Percentage $percent = $codesize/$filesize; $percentage = $percent*100; echo $percentage; ?> I don't know the exact calculations that are used so this function is just my guess. Does anyone know what the proper calculations are or if my functions are close enough for a good judgement.

  Read the article

• What does the English word "for" exactly mean in "for" loops?

  - by kol

  English is not my first language, but since the keywords in programming languages are English words, I usually find it easy to read source code as English sentences: if (x > 10) f(); = "If variable x is greater than 10, then call function f." while (i < 10) ++i; = "While variable i is less than 10, increase i by 1." But how a for loop is supposed to be read? for (i = 0; i < 10; ++i) f(i); = ??? I mean, I know what a for loop is and how it works. My problem is only that I don't know what the English word "for" exactly means in for loops.

  Read the article

• Formatting an external HDD stuck at 70%

  - by mahmood

  My external HDD which is a 250GB WD (powered by USB) seems to have problem! Whenever i try to copy some files, it stuck while copying. I decided to format it. So I used windows tool and performed the format (not quickly) however at nearly 70% it stuck. Then I decided to perform a low level format with lowlevel. Again it stuck at 70%. I endup that the HDD has bad sector. So is there any tool that mark the bad sectors and bypass them? It is not very reasonable to through 250GB because of some bad sectors! P.S: I saw a similar topic but there were no conclusion there either. The smart data is Attribute, raw value, value, threshold, status Read Error Rate, 50, 200, 51, OK Spin-Up Time, 3275, 154, 21, OK Start/Stop Count, 2729, 98, 0, OK Reallocated Sectors Count,0, 200, 140, OK Seek Error Rate, 0, 100, 51, OK Power-On Hours (POH), 1057, 99, 0, OK Spin Retry Count, 0, 100, 51, OK Recalibration Retries ,0, 100, 51 , OK Power Cycle Count, 1385, 99, 0, OK Power-off Retract Count, 425, 200, 0, OK Load /Unload Cycle Count,12974, 196, 0, OK Temperature, 43, 43, 0, OK Reallocation Event Count,0, 200, 0, OK Current Pending Sector Count,23,200, 0, Degradation Uncorrectable Sector Count, 0, 100, 0, OK UltraDMA CRC Error Count,6, 200, 0, OK Write Error Rate/Multi-Zone Error Rate,0,100,51, OK It seems that the most important thing is this line Current Pending Sector Count,23,200, 0, Degradation Any idea on that?

  Read the article

< Previous Page | 83 84 85 86 87 88 89 90 91 92 93 94  | Next Page >