c++ - How does size work with memcpy? -

-

मेरा प्रश्न निम्नलिखित कोड से सुरक्षित है: 

  #define ARRAY_SIZE 10 std :: सरणी & lt; BYTE, ARRAY_SIZE & gt; myArray; Char * string = "this_sentence_is_longer_than_10_bytes" मेम्क्पी (& amp; myArray, स्ट्रिंग, ARRAY_SIZE);

परिणाम: myArray को [0- 9] के साथ "this_sente"

से भरा है सुरक्षित होने के नाते, मुझे यह जानने की आवश्यकता है कि बाकी 'स्ट्रिंग' के साथ क्या हो रहा है क्या इसे पूरी तरह से दिए गए आकार की वजह से अनदेखा किया जा रहा है या क्या इसे सरणी के अंत में फेंक दिया जा रहा है?

संपादित करें: अब 

  #define ARRAY_SIZE 10 std: : सरणी & lt; BYTE, ARRAY_SIZE & gt; myArray; Char * string = "this_sentence_is_longer_than_10_bytes" अगर (स्ट्रेलन (स्ट्रिंग) & lt; ARRAY_SIZE) {BYTE स्पष्टअरेरे [ARRAY_SIZE] = {0}; Memcpy (& amp; myArray, साफ़अरे, ARRAY_SIZE); Memcpy (& amp; myArray, स्ट्रिंग, स्ट्रेलन (स्ट्रिंग)); } और {memcpy (& amp; myArray, स्ट्रिंग, ARRAY_SIZE); }

यह अब एसडीडी को बाहर कर देता है: शून्य के साथ सरणी यदि स्ट्रिंग 10 अक्षरों से कम है, अन्यथा यह प्रारंभिक विधि का उपयोग करता है।

< Div class = "post-text" itemprop = "text">

बाकी स्ट्रिंग स्थिर हो जाता है, जैसे कि यह वहां नहीं है। memcpy बंद हो जाता है जैसे ही यह आकार बाइट्स तक पहुंचता है।

ध्यान दें कि हालांकि यह स्मृति ब्लॉक को पास करना सुरक्षित है जो आकार से बड़ा है से memcpy , एक ब्लॉक से गुजर रहा है जो कम ट्रिगर होता है अनिर्धारित व्यवहार उदाहरण के लिए, अगर आप स्ट्रिंग पास करते हैं, जो नौ अक्षर से कम है, तो memcpy स्ट्रिंग के अंत में अमान्य वर्णों को myArray में कॉपी कर देगा। < / P>


Comments

Post a Comment

Popular posts from this blog

html - Trouble with image gallery on codepen -

-
I get my first project html & amp; CSS, but there is some problem. I am practicing using Codepean and my gallery was running well but I am trying to add the last line of images and it is all messing up and I do not know why For some reason, when I try to add that chess board it all gets worse. I am still very new so I hope to learn a lot. Thank you. [1]: http://codepen.io/zenturtle/pen/ezDGC Margins are causing issues when you use margins in the position of an element, then you push the elements around it too. Use a solution: Status: Relative and then use Positioning Properties: Above , Below , left , right example: #chess {status: relative; Left: 555px; } Another solution is to use status: absolute . Location Location: Relative on & lt; Div id = "perimeter" & gt; Status: Full on every Example: #chess {status: full; Bottom: 122px; Correct: 230px; } Similarly you are in the position of chess image...
Read more

java - org.apache.http.ProtocolException: Target host is not specified -

-
I have written a simple httprequest / response code and I am getting this error below: I used to type in my classpath httpclient, httpcore, common -Codecs and general logging. I am very new to Java and I do not know what's going on here. Please help me. code: import org.apache.http.client.HttpClient; Import org.apache.http.client.methods.HttpGet; Import org.apache.http.HttpResponse; Import org.apache.http.impl.client.HttpClientBuilder; Import org.apache.http.Header; Import org.apache.http.HttpHeaders; Public class UnshorteningUrl {public static zero main (string [] args throws exception {HttpGet request = zero; HTTP Client Client = HTTP ClientBuilder.credit (). Build (); Try {Request = New HTTP Gate ("trib.me/1lBFzSi"); HttpResponse httpResponse = client.execute (request); Header [] header = httpResponse.getHeaders (HttpHeaders.LOCATION); // Preconditions.checkState (headers.length == 1); String new url = header [0] .getValue (); System.out.println ("new url...
Read more

How to access user directory in lazarus? -

-
After I am trying to open file changelog.txt and it does not matter if I open it Need to have the user open it though it is always located in the ~ / directory. Enter the file here. Here is my code: Process TForm1.FormCreate (From: TObject); Var myFile: textfile; Line: String; Start AssignFile (myFile, '~ / changelog.txt'); Reset (myFile); While EOF (MyFile) does not readLn (myFile, Line); Label3.Caption: = (Label3.Capttion + Line + # 13 # 10); End; CloseFile (myFile); End; This does not work, however, if I replace ~ with the actual user name, then it works, however, I use the username of each user Can not know which will run my program. Any idea how can I get the username that started the program? ! Thanks Edit1: I have tried this, but this also includes a new line: RunCommand ('/ bin / bash', ['- - C ',' Whoami '], users); This is normal. The concept of "~" is an shell level and thus requires a separate shell elev...
Read more