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Heap corruption while freeing memory

I have a class as follows

 struct CliHandler {
     CliHandler(int argc, char** argv);
     ~CliHandler();

     int doWork();

     int argc_; 
     char** argv_;  
     private:
     CliHandler(const CliHandler&){}
     CliHandler& operator=(const CliHandler&){} 
 };

//Constructor

 CliHandler::CliHandler(int argc,
 char** argv) {
     //set command line parameters
     argc_ = argc; 

     argv_ = (char**) malloc(argc_ * sizeof(char*));

     for(int i=0; i<argc_; ++i)
     {
         std::cout<<sizeof(argv[i]); 
         argv_[i] = (char*) malloc(strlen(argv[i]) *
 sizeof(char));
         StrCpy(argv_[i], argv[i]);
     } }

// destructor

 CliHandler::~CliHandler() {
     for(int i=0; i<argc_; ++i)
         free(argv_[i]); 
     free(argv_);  }

While I debug, I get an error " Heap corruption detected. CRT detected that application wrote to memory after end of heap buffer. " My question id "Where exactly am i making a mistake ? How do I fix it". I am using visual stdio 2008.

Edit:I did something like this to add 1

argv_[i] = (char*) malloc(strlen(argv[i] + 1) * sizeof(char));

Which is terrible as it increments the pointer argv[i] by one. My co-worker pointed out that subtle issue. It should be

argv_[i] = (char*) malloc( (strlen(argv[i]) + 1) * sizeof(char));

like image 447
Eternal Learner Avatar asked May 09 '11 19:05

Eternal Learner


1 Answers

Change the code to:

 argv_[i] = (char*) malloc(strlen(argv[i]) + 1) ; 
 strcpy(argv_[i], argv[i]); 

It's because your StrCpy likely trashes your memory. You have to account for the terminating nul byte as well, provided your StrCpy works like the standard strcpy (which it has to in order to be useful, better just use the standard strcpy() unless you have a good reason not to).

sizeof(char) is by definition 1, so that can be omitted too.

like image 112
nos Avatar answered Sep 28 '22 05:09

nos