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  • C++ Sentinel/Count Controlled Loop beginning programming

    - by Bryan Hendricks
    Hello all this is my first post. I'm working on a homework assignment with the following parameters. Piecework Workers are paid by the piece. Often worker who produce a greater quantity of output are paid at a higher rate. 1 - 199 pieces completed $0.50 each 200 - 399 $0.55 each (for all pieces) 400 - 599 $0.60 each 600 or more $0.65 each Input: For each worker, input the name and number of pieces completed. Name Pieces Johnny Begood 265 Sally Great 650 Sam Klutz 177 Pete Precise 400 Fannie Fantastic 399 Morrie Mellow 200 Output: Print an appropriate title and column headings. There should be one detail line for each worker, which shows the name, number of pieces, and the amount earned. Compute and print totals of the number of pieces and the dollar amount earned. Processing: For each person, compute the pay earned by multiplying the number of pieces by the appropriate price. Accumulate the total number of pieces and the total dollar amount paid. Sample Program Output: Piecework Weekly Report Name Pieces Pay Johnny Begood 265 145.75 Sally Great 650 422.50 Sam Klutz 177 88.5 Pete Precise 400 240.00 Fannie Fantastic 399 219.45 Morrie Mellow 200 110.00 Totals 2091 1226.20 You are required to code, compile, link, and run a sentinel-controlled loop program that transforms the input to the output specifications as shown in the above attachment. The input items should be entered into a text file named piecework1.dat and the ouput file stored in piecework1.out . The program filename is piecework1.cpp. Copies of these three files should be e-mailed to me in their original form. Read the name using a single variable as opposed to two different variables. To accomplish this, you must use the getline(stream, variable) function as discussed in class, except that you will replace the cin with your textfile stream variable name. Do not forget to code the compiler directive #include < string at the top of your program to acknowledge the utilization of the string variable, name . Your nested if-else statement, accumulators, count-controlled loop, should be properly designed to process the data correctly. The code below will run, but does not produce any output. I think it needs something around line 57 like a count control to stop the loop. something like (and this is just an example....which is why it is not in the code.) count = 1; while (count <=4) Can someone review the code and tell me what kind of count I need to introduce, and if there are any other changes that need to be made. Thanks. [code] //COS 502-90 //November 2, 2012 //This program uses a sentinel-controlled loop that transforms input to output. #include <iostream> #include <fstream> #include <iomanip> //output formatting #include <string> //string variables using namespace std; int main() { double pieces; //number of pieces made double rate; //amout paid per amount produced double pay; //amount earned string name; //name of worker ifstream inFile; ofstream outFile; //***********input statements**************************** inFile.open("Piecework1.txt"); //opens the input text file outFile.open("piecework1.out"); //opens the output text file outFile << setprecision(2) << showpoint; outFile << name << setw(6) << "Pieces" << setw(12) << "Pay" << endl; outFile << "_____" << setw(6) << "_____" << setw(12) << "_____" << endl; getline(inFile, name, '*'); //priming read inFile >> pieces >> pay >> rate; // ,, while (name != "End of File") //while condition test { //begining of loop pay = pieces * rate; getline(inFile, name, '*'); //get next name inFile >> pieces; //get next pieces } //end of loop inFile.close(); outFile.close(); return 0; }[/code]

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  • Images to video - converting to IplImage makes video blue

    - by user891908
    I want to create a video from images using opencv. The strange problem is that if i will write image (bmp) to disk and then load (cv.LoadImage) it it renders fine, but when i try to load image from StringIO and convert it to IplImage, it turns video to blue. Heres the code: import StringIO output = StringIO.StringIO() FOREGROUND = (0, 0, 0) TEXT = 'MY TEXT' font_path = 'arial.ttf' font = ImageFont.truetype(font_path, 18, encoding='unic') text = TEXT.decode('utf-8') (width, height) = font.getsize(text) # Create with background with place for text w,h=(600,600) contentimage=Image.open('0.jpg') background=Image.open('background.bmp') x, y = contentimage.size # put content onto background background.paste(contentimage,(((w-x)/2),0)) draw = ImageDraw.Draw(background) draw.text((0,0), text, font=font, fill=FOREGROUND) pi = background pi.save(output, "bmp") #pi.show() #shows image in full color output.seek(0) pi= Image.open(output) print pi, pi.format, "%dx%d" % pi.size, pi.mode cv_im = cv.CreateImageHeader(pi.size, cv.IPL_DEPTH_8U, 3) cv.SetData(cv_im, pi.tostring()) print pi.size, cv.GetSize(cv_im) w = cv.CreateVideoWriter("2.avi", cv.CV_FOURCC('M','J','P','G'), 1,(cv.GetSize(cv_im)[0],cv.GetSize(cv_im)[1]), is_color=1) for i in range(1,5): cv.WriteFrame(w, cv_im) del w

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  • Basic PHP question for Drupal Views theming

    - by oalo
    My PHP and programming knowledge is extremely basic, so this is probably a dumb question: I am theming a View in Drupal 6, and I want to add an id with a consecutive number to each item in the view (first item would have the id #item1, the second #item2, etc). I am customizing the style output (views-view-unformatted--MYVIEWNAME.tpl.php) and the row style output (views-view-fields--MYVIEWNAME.tpl.php), and I want to add a counter variable in the foreach loop in the style output tpl, and then use that variable in the row style output tpl, but the last one is not recognizing the variable. It does not give me any errors, but doesnt print the number. I understand this has probably something to do with variables visibility, how can I declare the counter variable in the style .tpl so I can the use it in the row style .tpl? Thank you

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  • Cannot work for 2nd iteration because of writing delay.

    - by karikari
    My code's IF-THEN does not work for 2nd iteration. This is due to, the jar processing take some time to write it result inside the output.txt. Since the writing is a bit late, my code's 2nd iteration will always read the previous written value inside the output.txt in order to pass it to the IF-THEN. For example, in 1st iteration: output.txt -- 0.9888 twrite.txt -- msg: ok 2nd iteration: output.txt -- 0.5555 twrite.txt -- msg: ok //the IF-THEN still gives this result which is based on previous iteration. it should be msg: not ok . since it is < 0.7 I need help, how to solve this 'delay' problem? HRESULT CButtonDemoBHO::onDocumentComplete(IDispatch *pDisp, VARIANT *vUrl){ ATLTRACE("CButtonDemoBHO::onDocumentComplete %S\n", vUrl->bstrVal); WinHttpClient client(vUrl->bstrVal); client.SendHttpRequest(); wstring httpResponseHeader = client.GetHttpResponseHeader(); wstring httpResponse = client.GetHttpResponse(); writeToLog(httpResponse.c_str()); if (isMainFrame(pDisp)){ m_normalPageLoad=false; FILE *child = _popen("javaw -jar c:\\simmetrics.jar c:\\chtml.txt c:\\thtml.txt > c:\\output.txt", "r"); fclose(child); char readnumber[10]; float f = 0; FILE *file11 = fopen("c:\\output.txt","r"); char* p = fgets(readnumber,10,file11); std::istringstream iss(p); iss >> f; if (f > 0.7) { wfstream file12 ("c:\\twrite.txt", ios_base::out); file12 << "Msg: ok"; file12.close(); } else { wfstream file12 ("c:\\twrite.txt", ios_base::out); file12 << "Msg: not ok"; file12.close(); } iss.clear(); fclose(file11); return S_OK; } return S_OK; }

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  • Why does order matter for ImageMagick's -colorspace operation?

    - by Mark Trapp
    Starting with ImageMagick 6, the command style changed to solve a bunch of problems outlined in the Basic Usage document. That document does imply that for simple operations, one should only need to move the options from before the source file to between the source and output files to convert from the "old" style to the "new" style. However, this doesn't seem to work for the -colorspace operation. When I use the following command, I get an output file with the correct colors: convert -colorspace rgb input.pdf output.png But when I try to use the new command style, the -colorspace operation is never applied: convert input.pdf -colorspace rgb output.png Samples: Why does this occur?

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  • cURL + HTTP_POST, keep getting 500 error. Has no idea?

    - by mysqllearner
    Okay, I want to make a HTTP_POST using cURL to a SSL site. I already imported the certificate to my server. This is my code: $url = "https://www.xxx.xxx"; $post = "";# all data that going to send $ch = curl_init(); curl_setopt($ch, CURLOPT_URL, $url); curl_setopt($ch, CURLOPT_POST, true); curl_setopt($ch, CURLOPT_POSTFIELDS, $post); curl_setopt($ch, CURLOPT_RETURNTRANSFER, true); curl_setopt($ch, CURLOPT_SSL_VERIFYPEER, FALSE); curl_setopt($ch, CURLOPT_SSL_VERIFYHOST, FALSE); curl_setopt($ch, CURLOPT_USERAGENT, 'Mozilla/4.0 (compatible; MSIE 5.01; Windows NT 5.0'); $exe = curl_exec($ch); $getInfo = curl_getinfo($ch); if ($exe === false) { $output = "Error in sending"; if (curl_error($ch)){ $output .= "\n". curl_error($ch); } } else if($geInfo['http_code'] != 777){ $output = "No data returned. Error: " . $geInfo['http_code']; if (curl_error($ch)){ $output .= "\n". curl_error($ch); } } curl_close($c); echo $output; It keep returned "500". Based on w3schools, 500 means Internal Server Error. Is my server having problem? How to solve/troubleshoot this?

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  • issue with $.ParseJSON which converts json string to null

    - by Aby
    I am using spring mvc in which i convert the arraylist into json string. I have one object 1) results. My output from spring looks like this: { "data":"[{\"userName\":\"test1\",\"firstName\":\"test\",\"lastName\":\"user\"}, {\"userName\":\"test2\",\"firstName\":\"test1\",\"lastName\":\"user1\"}]", } I get output as null when i do '$.parseJSON' with this output. When i tried testing only with data object it works fine Any help would be great.

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  • Executing system command in php, differs in using broswer and in using command line

    - by Amit
    Hi, I have to execute a Linux "more" command in php from a particular offset, format the result and display the result in Browser. My Code for the above is : <html> <head> <META HTTP-EQUIV=REFRESH CONTENT=10> <META HTTP-EQUIV=PRAGMA CONTENT=NO-CACHE> <title>Runtime Access log</title> </head> <body> <?php $moreCommand = "more +3693 /var/log/apache2/access_log | grep -v -e '.jpg' -e '.jpeg' -e '.css' -e '.js' -e '.bmp' -e '.ico'| wc -l"; exec($moreCommand, $accessDisplay); echo "<br/>No of lines are : $accessDisplay[0] <br/>"; ?> The output at the browser is :: No of lines are : 3428 (This is wrong) While executing the same command using command line gives a different output. My code snippet for the same is : <?php $moreCommand = "more +3693 /var/log/apache2/access_log | grep -v -e '.jpg' -e '.jpeg' -e '.css' -e '.js' -e '.bmp' -e '.ico'| wc -l"; exec($moreCommand, $accessDisplay); echo "No of lines are : $accessDisplay[0] \n"; ? The output at the command line is :: No of lines are : 279 (This is correct) While executing the same command directly in command line, gives me output as 279. I am unable to understand why the output of the same command is wrong in the browser. Its actually giving the word count of lines, ignoring the offset parameter. Please help !! Thanks, Amit

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  • Function behaviour on shell(ksh) script

    - by footy
    Here are 2 different versions of a program: this Program: #!/usr/bin/ksh printmsg() { i=1 print "hello function :)"; } i=0; echo I printed `printmsg`; printmsg echo $i Output: # ksh e I printed hello function :) hello function :) 1 and Program: #!/usr/bin/ksh printmsg() { i=1 print "hello function :)"; } i=0; echo I printed `printmsg`; echo $i Output: # ksh e I printed hello function :) 0 The only difference between the above 2 programs is that printmsg is 2times in the above program while printmsg is called once in the below program. My Doubt arises here: To quote Be warned: Functions act almost just like external scripts... except that by default, all variables are SHARED between the same ksh process! If you change a variable name inside a function.... that variable's value will still be changed after you have left the function!! But we can clearly see in the 2nd program's output that the value of i remains unchanged. But we are sure that the function is called as the print statement gets the the output of the function and prints it. So why is the output different in both?

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  • How would you go about tackling this problem? [SOLVED in C++]

    - by incrediman
    Intro: EDIT: See solution at the bottom of this question (c++) I have a programming contest coming up in about half a week, and I've been prepping :) I found a bunch of questions from this canadian competition, they're great practice: http://cemc.math.uwaterloo.ca/contests/computing/2009/stage2/day1.pdf I'm looking at problem B ("Dinner"). Any idea where to start? I can't really think of anything besides the naive approach (ie. trying all permutations) which would take too long to be a valid answer. Btw, the language there says c++ and pascal I think, but i don't care what language you use - I mean really all I want is a hint as to the direction I should proceed in, and perhpas a short explanation to go along with it. It feels like I'm missing something obvious... Of course extended speculation is more than welcome, but I just wanted to clarify that I'm not looking for a full solution here :) Short version of the question: You have a binary string N of length 1-100 (in the question they use H's and G's instead of one's and 0's). You must remove all of the digits from it, in the least number of steps possible. In each step you may remove any number of adjacent digits so long as they are the same. That is, in each step you can remove any number of adjacent G's, or any number of adjacent H's, but you can't remove H's and G's in one step. Example: HHHGHHGHH Solution to the example: 1. HHGGHH (remove middle Hs) 2. HHHH (remove middle Gs) 3. Done (remove Hs) -->Would return '3' as the answer. Note that there can also be a limit placed on how large adjacent groups have to be when you remove them. For example it might say '2', and then you can't remove single digits (you'd have to remove pairs or larger groups at a time). Solution I took Mark Harrison's main algorithm, and Paradigm's grouping idea and used them to create the solution below. You can try it out on the official test cases if you want. //B.cpp //include debug messages? #define DEBUG false #include <iostream> #include <stdio.h> #include <vector> using namespace std; #define FOR(i,n) for (int i=0;i<n;i++) #define FROM(i,s,n) for (int i=s;i<n;i++) #define H 'H' #define G 'G' class String{ public: int num; char type; String(){ type=H; num=0; } String(char type){ this->type=type; num=1; } }; //n is the number of bits originally in the line //k is the minimum number of people you can remove at a time //moves is the counter used to determine how many moves we've made so far int n, k, moves; int main(){ /*Input from File*/ scanf("%d %d",&n,&k); char * buffer = new char[200]; scanf("%s",buffer); /*Process input into a vector*/ //the 'line' is a vector of 'String's (essentially contigious groups of identical 'bits') vector<String> line; line.push_back(String()); FOR(i,n){ //if the last String is of the correct type, simply increment its count if (line.back().type==buffer[i]) line.back().num++; //if the last String is of the wrong type but has a 0 count, correct its type and set its count to 1 else if (line.back().num==0){ line.back().type=buffer[i]; line.back().num=1; } //otherwise this is the beginning of a new group, so create the new group at the back with the correct type, and a count of 1 else{ line.push_back(String(buffer[i])); } } /*Geedily remove groups until there are at most two groups left*/ moves=0; int I;//the position of the best group to remove int bestNum;//the size of the newly connected group the removal of group I will create while (line.size()>2){ /*START DEBUG*/ if (DEBUG){ cout<<"\n"<<moves<<"\n----\n"; FOR(i,line.size()) printf("%d %c \n",line[i].num,line[i].type); cout<<"----\n"; } /*END DEBUG*/ I=1; bestNum=-1; FROM(i,1,line.size()-1){ if (line[i-1].num+line[i+1].num>bestNum && line[i].num>=k){ bestNum=line[i-1].num+line[i+1].num; I=i; } } //remove the chosen group, thus merging the two adjacent groups line[I-1].num+=line[I+1].num; line.erase(line.begin()+I);line.erase(line.begin()+I); moves++; } /*START DEBUG*/ if (DEBUG){ cout<<"\n"<<moves<<"\n----\n"; FOR(i,line.size()) printf("%d %c \n",line[i].num,line[i].type); cout<<"----\n"; cout<<"\n\nFinal Answer: "; } /*END DEBUG*/ /*Attempt the removal of the last two groups, and output the final result*/ if (line.size()==2 && line[0].num>=k && line[1].num>=k) cout<<moves+2;//success else if (line.size()==1 && line[0].num>=k) cout<<moves+1;//success else cout<<-1;//not everyone could dine. /*START DEBUG*/ if (DEBUG){ cout<<" moves."; } /*END DEBUG*/ }

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  • How can I save a directory tree to an array in PHP?

    - by Greg
    I'm trying to take a directory with the structure: top folder1 file1 folder2 file1 file2 And save it into an array like: array ( 'folder1' => array('file1'), 'folder2' => array('file1', 'file2') ) This way, I can easily resuse the tree throughout my site. I've been playing around with this code but it's still not doing what I want: private function get_tree() { $uploads = __RELPATH__ . DS . 'public' . DS . 'uploads'; $iterator = new RecursiveIteratorIterator(new RecursiveDirectoryIterator($uploads), RecursiveIteratorIterator::SELF_FIRST); $output = array(); foreach($iterator as $file) { $relativePath = str_replace($uploads . DS, '', $file); if ($file->isDir()) { if (!in_array($relativePath, $output)) $output[$relativePath] = array(); } } return $output; }

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  • NSTask Tail -f Using objective C

    - by Bach
    I need to read the last added line to a log file, in realtime, and capture that line being added. Something similar to Tail -f. So my first attempt was to use Tail -f using NSTask. I can't see any output using the code below: NSTask *server = [[NSTask alloc] init]; [server setLaunchPath:@"/usr/bin/tail"]; [server setArguments:[NSArray arrayWithObjects:@"-f", @"/path/to/my/LogFile.txt",nil]]; NSPipe *outputPipe = [NSPipe pipe]; [server setStandardInput:[NSPipe pipe]]; [server setStandardOutput:outputPipe]; [server launch]; [server waitUntilExit]; [server release]; NSData *outputData = [[outputPipe fileHandleForReading] readDataToEndOfFile]; NSString *outputString = [[[NSString alloc] initWithData:outputData encoding:NSUTF8StringEncoding] autorelease]; NSLog (@"Output \n%@", outputString); I can see the output as expected when using: [server setLaunchPath:@"/bin/ls"]; How can i capture the output of that tail NSTask? Is there any alternative to this method, where I can open a stream to file and each time a line is added, output it on screen? (basic logging functionality)

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  • Is there "good" PRNG generating values without hidden state?

    - by actual
    I need some good pseudo random number generator that can be computed like a pure function from its previous output without any state hiding. Under "good" I mean: I must be able to parametrize generator in such way that running it for 2^n iterations with any parameters should cover all or almost all values between 0 and 2^n - 1, where n is the number of bits in output value. Combined generator output of n + p bits must cover all or almost all values between 0 and 2^(n + p) - 1 if I run it for 2^n iterations for every possible combination of its parameters, where p is the number of bits in parameters. For example, LCG can be computed like a pure function and it can meet first condition, but it can not meet second one. Say, we have 32-bit generator, m = 2^32 and it is constant, our p = 64 (two 32-bit parameters a and c), n + p = 96, so we must peek data by three ints from output to meet second condition. Unfortunately, condition can not be meet because of strictly alternating sequence of odd and even ints in output. To overcome this, hidden state must be introduced, but that makes function not pure and breaks first condition (period become much longer). Am I wanting too much?

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  • Unable to add FromName to e-mail using cdosys in SQL Server 2008

    - by Alex Andronov
    I have a piece of cdosys code which runs correctly and generates e-mail with my SQL Server 2008 server talking to a MS Exchange 2003 Server. However the from name is not appearing on the e-mails when they arrive. Is there a fault in the code is it not possible this way? Thanks in advance usp_send_cdosysmail @from varchar(500), @to text, @bcc text , @subject varchar(1000), @body text , @smtpserver varchar(25), @bodytype varchar(10) as declare @imsg int declare @hr int declare @source varchar(255) declare @description varchar(500) declare @output varchar(8000) exec @hr = sp_oacreate 'cdo.message', @imsg out exec @hr = sp_oasetproperty @imsg, 'configuration.fields("http://schemas.microsoft.com/cdo/configuration/sendusing").value','2' exec @hr = sp_oasetproperty @imsg, 'configuration.fields("http://schemas.microsoft.com/cdo/configuration/smtpserver").value', @smtpserver exec @hr = sp_oamethod @imsg, 'configuration.fields.update', null exec @hr = sp_oasetproperty @imsg, 'to', @to exec @hr = sp_oasetproperty @imsg, 'bcc', @bcc exec @hr = sp_oasetproperty @imsg, 'from', @from exec @hr = sp_oasetproperty @imsg, 'fromname','A From Name' exec @hr = sp_oasetproperty @imsg, 'subject', @subject -- if you are using html e-mail, use 'htmlbody' instead of 'textbody'. exec @hr = sp_oasetproperty @imsg, @bodytype, @body exec @hr = sp_oamethod @imsg, 'send', null -- sample error handling. if @hr <>0 select @hr begin exec @hr = sp_oageterrorinfo null, @source out, @description out if @hr = 0 begin select @output = ' source: ' + @source print @output select @output = ' description: ' + @description print @output end else begin print ' sp_oageterrorinfo failed.' return end end exec @hr = sp_oadestroy @imsg

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  • Calculate the SUM of the Column which has Time DataType:

    - by thevan
    I want to calculate the Sum of the Field which has Time DataType. My Table is Below: TableA: TotalTime ------------- 12:18:00 12:18:00 Here I want to sum the two time fields. I tried the below Query SELECT CAST( DATEADD(MS, SUM(DATEDIFF(MS, '00:00:00.000', CONVERT(TIME, TotalTime))), '00:00:00.000' ) AS TOTALTIME) FROM [TableA] But it gives the Output as TOTALTIME ----------------- 00:36:00.0000000 But My Desired Output would be like below: TOTALTIME ----------------- 24:36:00 How to get this Output?

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  • use JSON variable in jQuery dynamically

    - by user1644123
    I have two DIVs, #placeholder AND #imageLoad. When the user clicks on a particular thumb its larger version (thumb2) should then appear in #imageLoad DIV. Here is the jQuery that needs to be fixed: $.getJSON('jsonFile.json', function(data) { var output="<ul>"; for (var i in data.items) { output+="<li><img src=images/items/" + data.items[i].thumb + ".jpg></li>"; } output+="</ul>"; document.getElementById("placeholder").innerHTML=output; }); //This is wrong!! Not working.. $('li').on({ mouseenter: function() { document.getElementById("imageLoad").innerHTML="<img src=images/items/" + data.items[i].thumb2 + ".jpg>"; } }); Here is the external JSON file below (jsonFile.json): {"items":[ { "id":"1", "thumb":"01_sm", "thumb2":"01_md" }, { "id":"2", "thumb":"02_sm", "thumb2":"02_md" } ]}

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  • How to proceed jpeg Image file size after read--rotate-write operations in Java?

    - by zamska
    Im trying to read a JPEG image as BufferedImage, rotate and save it as another jpeg image from file system. But there is a problem : after these operations I cannot proceed same file size. Here the code //read Image BufferedImage img = ImageIO.read(new File(path)); //rotate Image BufferedImage rotatedImage = new BufferedImage(image.getHeight(), image.getWidth(), BufferedImage.TYPE_3BYTE_BGR); Graphics2D g2d = (Graphics2D) rotatedImage.getGraphics(); g2d.rotate(Math.toRadians(PhotoConstants.ROTATE_LEFT)); int height=-rotatedImage.getHeight(null); g2d.drawImage(image, height, 0, null); g2d.dispose(); //Write Image Iterator iter = ImageIO.getImageWritersByFormatName("jpeg"); ImageWriter writer = (ImageWriter)iter.next(); // instantiate an ImageWriteParam object with default compression options ImageWriteParam iwp = writer.getDefaultWriteParam(); try { FileImageOutputStream output = null; iwp.setCompressionMode(ImageWriteParam.MODE_EXPLICIT); iwp.setCompressionQuality(0.98f); // an integer between 0 and 1 // 1 specifies minimum compression and maximum quality File file = new File(path); output = new FileImageOutputStream(file); writer.setOutput(output); IIOImage iioImage = new IIOImage(image, null, null); writer.write(null, iioImage, iwp); output.flush(); output.close(); writer.dispose(); Is it possible to access compressionQuality parameter of original jpeg image in the beginning. when I set 1 to compression quality, the image gets bigger size. Otherwise I set 0.9 or less the image gets smaller size. How can i proceed the image size after these operations? Thank you,

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  • Ruby-Graphwiz does not render png

    - by auralbee
    I just tried the ruby-graphwiz gem (http://github.com/glejeune/Ruby-Graphviz). I followed the instructions (installed Graphwiz, gem and dependencies) and tried the example from the Github page. Unfortunately I am not able to render any output image (png,dot). # Create a new graph g = GraphViz.new( :G, :type => :digraph ) # Create two nodes hello = g.add_node( "Hello" ) world = g.add_node( "World" ) # Create an edge between the two nodes g.add_edge( hello, world ) # Generate output image g.output( :png => "hello_world.png" ) When I run the skript from the console I get no error message but also no output as expected. What could be the problem? Folders have read/write access for everybody. Thanks in advance. By the way, I´m working on a Mac (Leopard 10.6).

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  • Outputing UTF-8 string on Mac OS's Terminal

    - by SuperBloup
    I got a programm in haskell outputting utf-8 using the package utf8-string and using only the output functions of this package. I set the encoding of each file I write to this way : hSetEncoding myFile utf8 {- myFile may be stdout -} but when I try to output : alpha = [fromEnum 0x03B1] {- a -} instead of the nice alpha letter I got on Linux (or in a file on windows), I got the following : α The weird thing is even if I try to write the output on a file, I can't read it back with mvim as an utf-8 file. Is there any way to get the correct behaviour

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  • usage of 2 charectors in single qoutes in c

    - by user1632141
    #include<stdio.h> int main() { char ch = 'A'; printf("%d\n",'ag'); printf("%d\n",'a'); printf("%d, %d, %d, %d", sizeof(ch), sizeof('a'), sizeof('Ag'), sizeof(3.14f)); return 0; } I used to have many doubts on the output of this question while running on g++ and gcc. But I have cleared almost all the doubts by referring these links: Single and double quotes in C/C++ Single quotes vs. double quotes in C I still need to understand one thing about the output of this question. Can someone please explain the output of printf("%d\n",'ag'); mentioned above in the program. How is it actually stored in the memory? The output for the program on the Linux/GCC platform is: 24935 97 1, 4, 4, 4

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  • close file with fopen() but file still in use

    - by Marco
    Hi all, I've got a problem with deleting/overwriting a file using my program which is also being used(read) by my program. The problem seems to be that because of the fact my program is reading data from the file (output.txt) it puts the file in a 'in use' state which makes it impossible to delete or overwrite the file. I don't understand why the file stays 'in use' because I close the file after use with fclose(); this is my code: bool bBool = true while(bBool){ //Run myprogram.exe tot generate (a new) output.txt //Create file pointer and open file FILE* pInputFile = NULL; pInputFile = fopen("output.txt", "r"); // //then I do some reading using fscanf() // //And when I'm done reading I close the file using fclose() fclose(pInputFile); //The next step is deleting the output.txt if( remove( "output.txt" ) == -1 ){ //ERROR }else{ //Succesfull } } I use fclose() to close the file but the file remains in use by my program until my program is totally shut down. What is the solution to free the file so it can be deleted/overwrited? In reality my code isn't a loop without an end ; ) Thanks in advance! Marco

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  • cannot modify header information [closed]

    - by mohanraj
    Possible Duplicate: cannot modify header information Warning: Cannot modify header information - headers already sent by (output started at C:\xampp\htdocs\register&login\logout.php:1) in C:\xampp\htdocs\register&login\logout.php on line 4 Warning: Cannot modify header information - headers already sent by (output started at C:\xampp\htdocs\register&login\logout.php:1) in C:\xampp\htdocs\register&login\logout.php on line 5 Warning: Cannot modify header information - headers already sent by (output started at C:\xampp\htdocs\register&login\logout.php:1) in C:\xampp\htdocs\register&login\logout.php on line 6

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  • spoof mac address

    - by Cold-Blooded
    // macaddress.cpp : Defines the entry point for the console application. // #include "stdafx.h" #include <windows.h> #include <iostream> using namespace std; void readregistry(); void spoofmac(); void main(int argc, char* argv[]) { readregistry(); spoofmac(); } void spoofmac() { ////////////////////// ////////Write to Registry char buffer[60]; unsigned long size = sizeof(buffer); HKEY software; LPCTSTR location; char adapternum[10]=""; char numbers[11]="0123456789"; char editlocation[]="System\\CurrentControlSet\\Control\\Class\\{4D36E972-E325-11CE-BFC1-08002bE10318}\\0000"; char macaddress[60]; cout << "\n//////////////////////////////////////////////////////////////////\nPlease Enter Number of Network Adapter to Spoof or type 'E' to Exit.\nE.g. 18\n\nNumber: "; cin >> adapternum; if (adapternum[0]=='E') { exit(0); } if (strlen(adapternum)==2) { editlocation[strlen(editlocation)-2]=adapternum[0]; editlocation[strlen(editlocation)-1]=adapternum[1]; } if (strlen(adapternum)==1) { editlocation[strlen(editlocation)-1]=adapternum[0]; } if (strlen(adapternum)!=1 && strlen(adapternum)!=2) { cout << "Invaild Network Adapter Chosen\n\n"; exit(0); } cout << "Please Enter the Desired Spoofed Mac Address Without Dashes\nE.g. 00123F0F6D7F\n\nNew Mac: "; cin >> macaddress; location = editlocation; //error line strcpy(buffer,macaddress); size=sizeof(buffer); RegCreateKey(HKEY_LOCAL_MACHINE,location,&software); //RegSetValueEx(software,"NetworkAddress",NULL,REG_SZ,(LPBYTE)buffer,size); RegCloseKey(software); cout << "\nMac Address Successfully Spoofed.\n\nWritten by Lyth0s\n\n"; } void readregistry () { //////////////////////////////////// // Read From Registry char driver[60]=""; char mac[60]=""; char numbers[11]="0123456789"; char editlocation[]="System\\CurrentControlSet\\Control\\Class\\{4D36E972-E325-11CE-BFC1-08002bE10318}\\0000"; unsigned long driversize = sizeof(driver); unsigned long macsize = sizeof(mac); DWORD type; HKEY software; LPCTSTR location; int tenscount=0; int onescount=0; for (int x =0;x<=19; x+=1) { strcpy(driver,""); driversize=sizeof(driver); strcpy(mac,""); macsize=sizeof(mac); if (editlocation[strlen(editlocation)-1]=='9') { tenscount+=1; onescount=0; editlocation[strlen(editlocation)-2]=numbers[tenscount]; } editlocation[strlen(editlocation)-1]=numbers[onescount]; location=editlocation; //error line // cout << location << "\n"; // cout << "Checking 00" << location[strlen(location)-2] << location[strlen(location)-1] << "\n\n"; RegCreateKey(HKEY_LOCAL_MACHINE,location,&software); RegQueryValueEx(software,"DriverDesc",NULL,&type,(LPBYTE)driver,&driversize); //RegCloseKey(software); //RegCreateKey(HKEY_LOCAL_MACHINE,location,&software); RegQueryValueEx(software,"NetworkAddress",NULL,&type,(LPBYTE)mac,&macsize); RegCloseKey(software); cout << x << ": " << driver << "| Mac: " << mac << "\n"; onescount+=1; } } this program gives error as follows error C2440: '=' : cannot convert from 'char [83]' to 'LPCTSTR' why this error coming please explain

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  • What's the best way to write to more files than the kernel allows open at a time?

    - by Elpezmuerto
    I have a very large binary file and I need to create separate files based on the id within the input file. There are 146 output files and I am using cstdlib and fopen and fwrite. FOPEN_MAX is 20, so I can't keep all 146 output files open at the same time. I also want to minimize the number of times I open and close an output file. How can I write to the output files effectively? I also must use the cstdlib library due to legacy code.

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