free() Failing

[nickman]

Hi All

I have program:

#include<stdio.h>
void main()
{
    char *pd,*ps="I LOVE CAKE";
	int i = strlen(ps);
    pd=(char *)malloc(strlen(ps));
    strcpy(pd,ps);
    printf("%s",pd);
    free(pd);
	getchar();
}

In this free() is failing. I am not able to understand why>
Since pd has it know memory and I don't see any issue while deleting it.


Can any body help in this?

Thanks

[laserlight]

Observe this line:

pd=(char *)malloc(strlen(ps));

So, you allocated strlen(ps) number of bytes, but ps has strlen(ps) + 1 number of characters. Hence, when you do this:

strcpy(pd,ps);

You end up with a buffer overrun: the null character is copied, but no space was allocated for it. This leads to undefined behaviour. If you fix the malloc, your problem with free() might also be solved.

[nvoigt]

Your main function needs to be "int main".

Your string internally looks like this:

'I' blank 'L' 'O' 'V' 'E' blank 'C' 'A' 'K' 'E' binary zero

The binary zero at the end is added so all functions know where your string ends. So your string needs one byte of memory more than it's length.

As you are one byte short, your functions write to a byte that does not belong to you. That's bad in itself and it will probably make your compiler issue some kind of warning when you free the memory.

[AndiPersti]

Two problems:

1) you don't include "stdlib.h" and "string.h"
2) you allocate one byte too little (strlen returns the length of the string without the terminating '\0')

What is the exact error message you get?

Bye, Andreas

[nickman]

Oh Thanks.. I got it.
I changed it and now problem solved.

[inu11byte]

Just a few things to note:
What is the 'i' variable being used for?
Void main is bad. Read more about why, here:void main(void) - the Wrong Thing