Hi all,
I had a bug in my application that I THINK I fixed but I want to propose
my theory and see if it's correct.

What I was doing was this:

1) allocate memory via malloc() for var1 (char*)
2) allocate memory via malloc() for var2 (char*)
3) var1 points to a string "house"
4) var2 = strstr(var1, "s") so now var2 points to the "s" contained in var1
5) free (var1)
6) free (var2)

Step 6 was creating a SIGABRT from free().

My hypothesis:
Since var2 is pointing to var1 I never need to free var2 as long as I free var1,
since it is not a copy, but rather a pointer to the same memory allocated for var1.
Thus, when I free var1, var2's memory is also freed, since they're essentially the
same.

This means I also never need to malloc() for var2, since it will only point to
somewhere within var1, or NULL.

Assuming the above, my app seems to work fine, but I was hoping someone could
verify my thinking or correct it since there might be a bug lurking.

Thanks.

all right

Cool beans, thanks!

What do you mean exactly by this?
If "house" is a constant string (i.e. you are not going to change it), say, something like this:...then, that's it!. In this case you don't need any malloc, since the string "house" will already allocate 6 char's in the executable itself.

Or else you could do this and be able to free the memory:

char * var1 = malloc (6);
strcpy ( var1, "house");
char * var2 = strstr(var1,"s");
free(var1);
var2 = 0; //NULL

You r right. strstr() only returns the pointer to the location where 's' is present within the string. It does not copy anything to var2.

The scenario where u were allocating memory to var2 and then using var2 to store return value from strstr() will cause memory leaks because actual pointer to the memory allocated to the var2 will be lost.

#include <stdio.h>

main()
{
char *v1, *v2;

v1=(char*)malloc((strlen("house")+1)*sizeof(char));

v2=strstr(v1,"s");

free(v1);
}