Search Results

Search found 7755 results on 311 pages for 'word wrap'.

Page 61/311 | < Previous Page | 57 58 59 60 61 62 63 64 65 66 67 68  | Next Page >

  • Is there a way to synchronize sections of different Word documents? Alternatives?

    - by David Hodgson
    Hi, I'm working on how my company does documentation (especially programming documentation). I'd like to be able to synchronize sections of different Word documents, such that if a section in one document changes, the change is reflected in the other document, and vice versa. Is there a way to do this with Word, and if not, is there some word processing program that is good at this?

    Read the article

  • How to count the Chinese word in a file using regex in perl?

    - by Ivan
    I tried following perl code to count the Chinese word of a file, it seems working but not get the right thing. Any help is greatly appreciated. The Error message is Use of uninitialized value $valid in concatenation (.) or string at word_counting.pl line 21, <FILE> line 21. Total things = 125, valid words = which seems to me the problem is the file format. The "total thing" is 125 that is the string number (125 lines). The strangest part is my console displayed all the individual Chinese words correctly without any problem. The utf-8 pragma is installed. #!/usr/bin/perl -w use strict; use utf8; use Encode qw(encode); use Encode::HanExtra; my $input_file = "sample_file.txt"; my ($total, $valid); my %count; open (FILE, "< $input_file") or die "Can't open $input_file: $!"; while (<FILE>) { foreach (split) { #break $_ into words, assign each to $_ in turn $total++; next if /\W|^\d+/; #strange words skip the remainder of the loop $valid++; $count{$_}++; # count each separate word stored in a hash ## next comes here ## } } print "Total things = $total, valid words = $valid\n"; foreach my $word (sort keys %count) { print "$word \t was seen \t $count{$word} \t times.\n"; } ##---Data---- sample_file.txt ??????,???????,????.??????.????:"?????????????,??????,????????.????????,?????????, ???????????.????????,???????????,??????,??????.???:`??,???????????.'?????, ??????????."??????,??????.????.???, ????????????,????,??????,?????????,??????????????. ????????,??????,???????????,????????,????????.????,????,???????, ??????????,??????,????????.??????.

    Read the article

  • All applications quit when printing on Mac OS X 10.5.8

    - by Tamany
    I recently ran a software update. I'm not sure if my problems are associated with this but I'm pretty sure they are as I printed successfully before update. I checked the log at time of printing: 03/05/2010 22:03:15 Microsoft Word[697] *** -[NSCFString _getValue:forType:]: unrecognized selector sent to instance 0x17a82b50 03/05/2010 22:03:15 [0x0-0x51051].com.microsoft.Word[697] Ignoring Quickdraw drawing between QDBeginCGContext and QDEndCGContext 03/05/2010 22:03:16 [0x0-0x51051].com.microsoft.Word[697] Ignoring Quickdraw drawing between QDBeginCGContext and QDEndCGContext 03/05/2010 22:03:16 [0x0-0x51051].com.microsoft.Word[697] Ignoring Quickdraw drawing between QDBeginCGContext and QDEndCGContext 03/05/2010 22:03:16 [0x0-0x51051].com.microsoft.Word[697] Ignoring Quickdraw drawing between QDBeginCGContext and QDEndCGContext 03/05/2010 22:03:16 [0x0-0x51051].com.microsoft.Word[697] Ignoring Quickdraw drawing between QDBeginCGContext and QDEndCGContext 03/05/2010 22:03:16 [0x0-0x51051].com.microsoft.Word[697] Ignoring Quickdraw drawing between QDBeginCGContext and QDEndCGContext 03/05/2010 22:03:17 [0x0-0x51051].com.microsoft.Word[697] Mon May 3 22:03:17 leopards-imac-2.local Word[697] <Error>: The function `CGPDFDocumentGetMediaBox' is obsolete and will be removed in an upcoming update. Unfortunately, this application, or a library it uses, is using this obsolete function, and is thereby contributing to an overall degradation of system performance. Please use `CGPDFPageGetBoxRect' instead. 03/05/2010 22:22:09 Microsoft Word[697] *** -[NSCFString _getValue:forType:]: unrecognized selector sent to instance 0x1b036500 Any thoughts on how to fix this?

    Read the article

  • In Java, how do you parse through a single word string?

    - by Fraz
    I'm trying to parse though a string made up of a single word. How would you go about assigning the last letter of the word to a variable? I was thinking of using the Scanner class to parse the word and make each letter an element in an array but it seems Scanner.next() only goes through whole words and not the individual letters. Any help?

    Read the article

  • How to load MS Word document in C# (.NET)?

    - by Skuta
    Hi, How do I load MS Word document (.doc and .docx) to memory (variable) without doing this?: wordApp.Documents.Open I don't want to open MS Word, I just want that text inside. You gave me answer for DOCX, but what about DOC? I want free and high performance solution - not to open 12.000 instances of Word to process all of them. :( Aspose is commercial product, and 900$ is a way too much for what I do.

    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

  • How do I center my navigation bar and background?

    - by user2892958
    nav-wrap { background:url(nav-bg-blue.png) no-repeat top center; height:39px; padding-top:3px; } .no-header-page #nav-wrap { background:url(nav-bg-nobanner-blue.png) no-repeat top center; height:43px; padding-top:4px; margin-bottom:30px; } #nav-wrap .container { clear: both; overflow: hidden; position:center; width:100%; } #nav-wrap .container ul { list-style: none; float: center; } #nav-wrap .container ul li { list-style: none; float: left; background:url(nav-right-last.png) no-repeat top right; padding-right:20px; margin-left:-10px; position:auto; } #nav-wrap .container ul span li { background:url(nav-right-last.png) no-repeat top right; } #nav-wrap .container ul li a { float: center; display: block; font-family: 'News Cycle', sans-serif; color: #fff; text-decoration: none; padding: 5px 10px 8px 20px; border: 0; outline: 0; list-style-type: none; font-size: 14px; text-transform:uppercase; letter-spacing:2px; background:url(nav-left-first.png) no-repeat top left; line-height:25px; text-shadow:0 -1px 2px rgba(0,0,0,0.3); } #nav-wrap .container ul li#active, #nav-wrap .container ul li:hover{ background:url(nav-hover-right-last-brown-red.png) no-repeat topright; z-index:1; } #nav-wrap .container ul li:hover a, #nav-wrap .container ul li#active a, #nav-wrap .container ul li a:hover { border: 0; background:url(nav-hover-left-brown-red.png) no-repeat top left; } .wsite-nav-0 { margin-left:0 !important`` }

    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

  • That’s a wrap! Almost, there’s still one last chance to attend a SQL in the City event in 2012

    - by Red and the Community
    The communities team are back from the SQL in the City multi-city US Tour and we are delighted to have met so many happy SQL Server professionals and Red Gate customers. We set out to run a series of back-to-back events in order to meet, talk to and delight as many SQL Server and Red Gate enthusiasts as possible in 5 different cities in 11 days. We did it! The attendees had a good time too and 99% of them would attend another SQL in the City event in 2013 – so it seems we left an impression. There were a range of topics on the event agenda, ranging from ‘The Whys & Hows of Continuous Integration’, ‘Database Maintenance Essentials’, ‘Red Gate tools – The Complete Lifecycle’, ‘Automated Deployment: Application And Database Releases Without The Headache’, ‘The Ten Commandments of SQL Server Monitoring’ and many more. Videos and slides from the events will be posted to the event website in November, after our last event of 2012. SQL in the City Seattle – November 5 Join us for free and hear from some of the very best names in the SQL Server world. SQL Server MVPs such as; Steve Jones, Grant Fritchey, Brent Ozar, Gail Shaw and more will be presenting at the Bell Harbor conference center for one day only. We’re even taking on board some of the recent attendee-suggestions of how we can improve the events (feedback from the 65% of attendees who came to our US tour events), first off we’re extending the drinks celebration in the evening! Rather than just a 30 minute drink and run, attendees will have up to 2 hours to enjoy free drinks, relax and network in a fantastic environment amongst some really smart like-minded professionals. If you’re interested in expanding your SQL Server knowledge, would like to learn more about Red Gate tools, get yourself registered for the last SQL in the City event of 2012. It’s free, fun and we’re very friendly! I look forward to seeing you in Seattle on Monday November 5. Cheers, Annabel.

    Read the article

  • To 'seal' or to 'wrap': that is the question ...

    - by Simon Thorpe
    If you follow this blog you will already have a good idea of what Oracle Information Rights Management (IRM) does. By encrypting documents Oracle IRM secures and tracks all copies of those documents, everywhere they are shared, stored and used, inside and outside your firewall. Unlike earlier encryption products authorized end users can transparently use IRM-encrypted documents within standard desktop applications such as Microsoft Office, Adobe Reader, Internet Explorer, etc. without first having to manually decrypt the documents. Oracle refers to this encryption process as 'sealing', and it is thanks to the freely available Oracle IRM Desktop that end users can transparently open 'sealed' documents within desktop applications without needing to know they are encrypted and without being able to save them out in unencrypted form. So Oracle IRM provides an amazing, unprecedented capability to secure and track every copy of your most sensitive information - even enabling end user access to be revoked long after the documents have been copied to home computers or burnt to CD/DVDs. But what doesn't it do? The main limitation of Oracle IRM (and IRM products in general) is format and platform support. Oracle IRM supports by far the broadest range of desktop applications and the deepest range of application versions, compared to other IRM vendors. This is important because you don't want to exclude sensitive business processes from being 'sealed' just because either the file format is not supported or users cannot upgrade to the latest version of Microsoft Office or Adobe Reader. But even the Oracle IRM Desktop can only open 'sealed' documents on Windows and does not for example currently support CAD (although this is coming in a future release). IRM products from other vendors are much more restrictive. To address this limitation Oracle has just made available the Oracle IRM Wrapper all-format, any-platform encryption/decryption utility. It uses the same core Oracle IRM web services and classification-based rights model to manually encrypt and decrypt files of any format on any Java-capable operating system. The encryption envelope is the same, and it uses the same role- and classification-based rights as 'sealing', but before you can use 'wrapped' files you must manually decrypt them. Essentially it is old-school manual encryption/decryption using the modern classification-based rights model of Oracle IRM. So if you want to share sensitive CAD documents, ZIP archives, media files, etc. with a partner, and you already have Oracle IRM, it's time to get 'wrapping'! Please note that the Oracle IRM Wrapper is made available as a free sample application (with full source code) and is not formally supported by Oracle. However it is informally supported by its author, Martin Lambert, who also created the widely-used Oracle IRM Hot Folder automated sealing application.

    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

  • How do I get long command lines to wrap to the next line?

    - by BrianH
    Edit It was my .bashrc file. I've copied the same profile from machine to machine, and I used special characters in my $PS1 that are somehow throwing it off. I'm now sticking with the standard bash variables for my $PS1. Thanks to @ændrük for the tip on the .bashrc! ...End Edit... Something I have noticed in Ubuntu for a long time that has been frustrating to me is when I am typing a command at the command line that gets longer (wider) than the terminal width, instead of wrapping to a new line, it goes back to column 1 on the same line and starts over-writing the beginning of my command line. (It doesn't actually overwrite the actual command, but visually, it is overwriting the text that was displayed). It's hard to explain without seeing it, but let's say my terminal was 20 characters wide (Mine is more like 120 characters - but for the sake of an example), and I want to echo the English alphabet. What I type is this: echo abcdefghijklmnopqrstuvwxyz But what my terminal looks like before I hit the key is: pqrstuvwxyzghijklmno When I hit enter, it echos abcdefghijklmnopqrstuvwxyz so I know the command was received properly. It just wrapped my typing after the "o" and started over on the same line. What I would expect to happen, if I typed this command in on a terminal that was only 20 characters wide would be this: echo abcdefghijklmno pqrstuvwxyz Background: I am using bash as my shell, and I have this line in my ~/.bashrc: set -o vi to be able to navigate the command line with VI commands. I am currently using Ubuntu 10.10 server, and connecting to the server with Putty. In any other environment I have worked in, if I type a long command line, it will add a new line underneath the line I am working on when my command gets longer than the terminal width and when I keep typing I can see my command on 2 different lines. But for as long as I can remember using Ubuntu, my long commands only occupy 1 line. This also happens when I am going back to previous commands in the history (I hit Esc, then 'K' to go back to previous commands) - when I get to a previous command that was longer than the terminal width, the command line gets mangled and I cannot tell where I am at in the command. The only work-around I have found to see the entire long command is to hit "Esc-V", which opens up the current command in a VI editor. I don't think I have anything odd in my .bashrc file. I commented out the "set -o vi" line, and I still had the problem. I downloaded a fresh copy of Putty and didn't make any changes to the configuration - I just typed in my host name to connect, and I still have the problem, so I don't think it's anything with Putty (unless I need to make some config changes) Has anyone else had this problem, and can anyone think of how to fix it? Thanks in advance! Brian

    Read the article

  • How do I get long command lines to wrap to the next line?

    - by BrianH
    Edit It was my .bashrc file. I've copied the same profile from machine to machine, and I used special characters in my $PS1 that are somehow throwing it off. I'm now sticking with the standard bash variables for my $PS1. Thanks to @ændrük for the tip on the .bashrc! ...End Edit... Something I have noticed in Ubuntu for a long time that has been frustrating to me is when I am typing a command at the command line that gets longer (wider) than the terminal width, instead of wrapping to a new line, it goes back to column 1 on the same line and starts over-writing the beginning of my command line. (It doesn't actually overwrite the actual command, but visually, it is overwriting the text that was displayed). It's hard to explain without seeing it, but let's say my terminal was 20 characters wide (Mine is more like 120 characters - but for the sake of an example), and I want to echo the English alphabet. What I type is this: echo abcdefghijklmnopqrstuvwxyz But what my terminal looks like before I hit the key is: pqrstuvwxyzghijklmno When I hit enter, it echos abcdefghijklmnopqrstuvwxyz so I know the command was received properly. It just wrapped my typing after the "o" and started over on the same line. What I would expect to happen, if I typed this command in on a terminal that was only 20 characters wide would be this: echo abcdefghijklmno pqrstuvwxyz Background: I am using bash as my shell, and I have this line in my ~/.bashrc: set -o vi to be able to navigate the command line with VI commands. I am currently using Ubuntu 10.10 server, and connecting to the server with Putty. In any other environment I have worked in, if I type a long command line, it will add a new line underneath the line I am working on when my command gets longer than the terminal width and when I keep typing I can see my command on 2 different lines. But for as long as I can remember using Ubuntu, my long commands only occupy 1 line. This also happens when I am going back to previous commands in the history (I hit Esc, then 'K' to go back to previous commands) - when I get to a previous command that was longer than the terminal width, the command line gets mangled and I cannot tell where I am at in the command. The only work-around I have found to see the entire long command is to hit "Esc-V", which opens up the current command in a VI editor. I don't think I have anything odd in my .bashrc file. I commented out the "set -o vi" line, and I still had the problem. I downloaded a fresh copy of Putty and didn't make any changes to the configuration - I just typed in my host name to connect, and I still have the problem, so I don't think it's anything with Putty (unless I need to make some config changes) Has anyone else had this problem, and can anyone think of how to fix it? Thanks in advance! Brian

    Read the article

  • Java dev learning Python: what concepts do I need to wrap my head around?

    - by LRE
    I've run through a few tutorials and written some small projects. I'm right in the middle of a small project now infact. All is going well enough thanks in no small part to Uncle Google (who usually points me to Stackoverflow ;-) Several times in the last few days I've found myself wondering "what am I missing?" - I feel that I'm still thinking in Java as I write in Python. This question over at StackOverflow is full of tips about what resources to read up on for learning Python, but I still feel that I'm a Java dev with a dictionary (pun unintended) to translate into Python. What I really want to do is refactor my head to be able to write Pythonic Python instead of Java disguised as Python (not that I want to loose my Java skills). So, the crux of my question is: what concepts does a Java dev really need to learn to think Pythonic? This includes anything that needs to be un-learnt. ps: I consider language syntax to not be particularly relevant to this question.

    Read the article

  • Java dev learning Python: what concepts do I need to wrap my head around?

    - by LRE
    I've run through a few tutorials and written some small projects. I'm right in the middle of a small project now infact. All is going well enough thanks in no small part to Uncle Google (who usually points me to Stackoverflow ;-) Several times in the last few days I've found myself wondering "what am I missing?" - I feel that I'm still thinking in Java as I write in Python. This question over at StackOverflow is full of tips about what resources to read up on for learning Python, but I still feel that I'm a Java dev with a dictionary (no pun intended) to translate into Python. What I really want to do is refactor my head to be able to write Pythonic Python instead of Java disguised as Python (not that I want to loose my Java skills). So, the crux of my question is: what concepts does a Java dev really need to learn to think Pythonic? This includes anything that needs to be un-learnt. ps: I consider language syntax to not be particularly relevant to this question.

    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

  • 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

  • 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

  • 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

  • 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

< Previous Page | 57 58 59 60 61 62 63 64 65 66 67 68  | Next Page >