Incorrect usage of free()

kobi_hikri · #1 08-Dec-2005, 02:12

Hey guys,

I wrote some code, and have a logical error. I wrote this code to demonstrate what I'm doing in my code :

This is the phase I'm getting to :

CPP / C++ / C Code:

#include <stdlib.h>
#include <stdio.h>

int main()
{
	void *p,*q;

	p = malloc(10000);
	printf("Original Allocation address Prior to <realloc>: 0x%.8X\n",p);
	q = realloc(p,20000);
	printf("Original Allocation address After <realloc>: 0x%.8X\n",p);
	printf("New Allocation address : 0x%.8X\n",q);
	return (0);
}

And here I do something very bad :

CPP / C++ / C Code:

#include <stdlib.h>
#include <stdio.h>

int main()
{
	void *p,*q,*dummy;

	p = malloc(10000);
	printf("Original Allocation address Prior to <realloc>: 0x%.8X\n",p);
	q = realloc(p,20000);
	printf("Original Allocation address After <realloc>: 0x%.8X\n",p);
	printf("New Allocation address : 0x%.8X\n",q);

	dummy = p;
	free(dummy);

	return (0);
}

Can someone point me to where I did something wrong ? Did I try to free something that doesn't belong to me ?

Best regards,
Kobi.

mikhail · #2 08-Dec-2005, 07:22

Quote:

Originally Posted by kobi_hikri

And here I do something very bad :

CPP / C++ / C Code:

#include <stdlib.h>
#include <stdio.h>

int main()
{
	void *p,*q,*dummy;

	p = malloc(10000);
	printf("Original Allocation address Prior to <realloc>: 0x%.8X\n",p);
	q = realloc(p,20000);
	printf("Original Allocation address After <realloc>: 0x%.8X\n",p);
	printf("New Allocation address : 0x%.8X\n",q);

	dummy = p;
	free(dummy);

	return (0);
}

Can someone point me to where I did something wrong ? Did I try to free something that doesn't belong to me ?

Best regards,
Kobi.

I just ran that second one in Dev-C++ and got no error. Executed fine too.

What compiler are you using?

alcoholic · #3 08-Dec-2005, 07:28

It may compile fine...but it should be

Code:

p = realloc(p,20000);

instead of

Code:

q = realloc(p,20000);

logically.
You can use malloc for q again

davekw7x · #4 08-Dec-2005, 08:00

Quote:

Originally Posted by kobi_hikri

Hey guys,

I wrote some code, and have a logical error. I wrote this code to demonstrate what I'm doing in my code :

This is the phase I'm getting to :

CPP / C++ / C Code:

#include <stdlib.h>
#include <stdio.h>

int main()
{
	void *p,*q;

	p = malloc(10000);
	printf("Original Allocation address Prior to <realloc>: 0x%.8X\n",p);
	q = realloc(p,20000);
	printf("Original Allocation address After <realloc>: 0x%.8X\n",p);
	printf("New Allocation address : 0x%.8X\n",q);
	return (0);
}

And here I do something very bad :

CPP / C++ / C Code:

#include <stdlib.h>
#include <stdio.h>

int main()
{
	void *p,*q,*dummy;

	p = malloc(10000);
	printf("Original Allocation address Prior to <realloc>: 0x%.8X\n",p);
	q = realloc(p,20000);
	printf("Original Allocation address After <realloc>: 0x%.8X\n",p);
	printf("New Allocation address : 0x%.8X\n",q);

	dummy = p;
	free(dummy);

	return (0);
}

Can someone point me to where I did something wrong ? Did I try to free something that doesn't belong to me ?

Best regards,
Kobi.

You can look it up on your favorite on-line language/library reference site. Such as http://www-ccs.ucsd.edu/c/.

You will find something like the following (where I have highlighted the significant part).

"realloc

void *realloc(void *ptr, size_t size);

The function allocates an object of size size, possibly obtaining initial stored values from the object whose address is ptr. It returns the address of the new object if successful; otherwise, it returns a null pointer. You can safely convert the return value to an object pointer of any type whose size is not greater than size.

If ptr is not a null pointer, it must be the address of an existing object that you first allocate by calling calloc, malloc, or realloc. If the existing object is not larger than the newly allocated object, realloc copies the entire existing object to the initial part of the allocated object. (The values stored in the remainder of the object are indeterminate.) Otherwise, the function copies only the initial part of the existing object that fits in the allocated object. If realloc succeeds in allocating a new object, it deallocates the existing object. Otherwise, the existing object is left unchanged.

If ptr is a null pointer, the function does not store initial values in the newly created object.
"

In other words, in your case realloc gives you a brand new space, pointed to by q and frees the space pointed to by p.

Regards,

Dave

davekw7x · #5 08-Dec-2005, 08:03

Quote:

Originally Posted by alcoholic

It may compile fine...but it should be

Code:

p = realloc(p,20000);

If realloc fails for some reason, then you can never free that previously allocated block. This would be a memory leak. (A Bad Thing.)

Quote:

Originally Posted by alcoholic

instead of

Code:

q = realloc(p,20000);

logically.
You can use malloc for q again

I don't see anything ironic about it; what I see is a memory leak. If you destroy the value returned by realloc (by setting q equal to some other value), then you can't free the memory. (Still A Bad Thing.)

Regards,

Dave

kobi_hikri · #6 08-Dec-2005, 15:44

Thank you all.
I'll re-think my approach.

Kobi.

davekw7x · #7 08-Dec-2005, 16:06

Quote:

Originally Posted by kobi_hikri

Thank you all.
I'll re-think my approach.

Kobi.

What's the problem?

If you have this:

CPP / C++ / C Code:


  p = malloc(size1);
  if (!p) {
    return 0;  /* abort or perform some appropriate recovery */
  }
  else {
    /* normal program execution continues */
  }
.
.
.
 temp = realloc(p, size2);
  if (!temp) {
    free(p);  /* or do whatever you need to do for the bad news case */
    return 0;
  }
  else {
    p = temp;
  }
.
.
.

Now the variable p still points to user's space, which has now been resized. Of course if this is done inside a function, and p is used outside of the function, the argument to the function will be a pointer to p, and the pointer will be dereferenced accordingly.

Everything is safe, and there are no memory leaks, assuming you free p before the program terminates.

Regards,

Dave

kobi_hikri · #8 08-Dec-2005, 23:27

Quote:

Originally Posted by davekw7x

What's the problem?

If realloc works fine, and the "old" allocation is deallocated, then any reference to the "old" allocation is illegal ...
Now I get it.

Kobi,

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