Question

I have simple code,

#include "stdafx.h"
#include <malloc.h>
int main()
{
  char *p = (char*) malloc(10);
  p = "Hello";

  free(p);

  return 0;
}

This code is giving run time exception while terminating. Below is error snippiest,

---

Microsoft Visual C++ Debug Library

Debug Assertion Failed!

Program: …\my documents\visual studio 2010\Projects\samC\Debug\samC.exe
File: f:\dd\vctools\crt_bld\self_x86\crt\src\dbgheap.c
Line: 1322

Expression: _CrtIsValidHeapPointer(pUserData)

For information on how your program can cause an assertion
failure, see the Visual C++ documentation on asserts.

(Press Retry to debug the application)

Abort Retry Ignore

Answer 1

This is how you write a string in the memory allocated by malloc to a char pointer.

strcpy(p, "Hello");

Replace the line

p = "Hello";

with the strcpy one & your program will work fine.

You also need to

#include <string.h>

malloc returns a pointer to allocated memory. Say the address is 95000 (just a random number I pulled out).

So after the malloc – p will hold the address 95000
The p containing 95000 is the memory address which needs to be passed to free when you are done with the memory.

However, the next line p = "Hello"; puts the address of the string literal “Hello” (which say exists at address 25000) into p.

So when you execute free(p) you are trying to free 25000 which wasn’t not allocated by malloc.

OTOH, when you strcpy, you copy the string “Hello” into the address starting at p (i.e. 95000). p remains 95000 after the strcpy.

And free(p) frees the right memory.

You can also avoid the malloc and use

char *p = "Hello";

However, in this method, you cannot modify the string.

i.e. if after this you do *p = 'B' to change the string to Bello, it becomes an undefined operation. This is not the case in the malloc way.

If instead, you use

char p[] = "Hello";

or

char p[10] = "Hello";

you get a modifiable string which need not be freed.