free() is killing my program?

[alwut]

So I'm writing a program and for some reason if I leave out the free part it doesn't work in random places so I was wondering if I did something wrong.

int* list = (int*)malloc(0);
int num = 0;

for(i=0; i < something; i++)
{
    if(something random happens)
    {
         num++;
         list = (int*) realloc(list, sizeof(int*) * num);
         list[num - 1] = something;
    }
}

free(list);

When I leave out the free part the program runs and finishes fine, but when I leave it in, it segfaults at a random and place. What actually happens is, another unrelated pointer is modified and when its accessed I get the error, am I doing something wrong?

[Elysia]

That usually means you've done something bad in the rest of your program.

[Dino]

Is this valid?

int* list = (int*)malloc(0);

If so, where does it point?

Todd

[MacGyver]

AFAIK, you can call realloc with a NULL pointer, and it will behave like malloc().... I think.

[Cactus_Hugger]

Calling realloc with NULL is valid:

If ptr is NULL, realloc() behaves like malloc() for the specified size.

This, however, is still questionable:

list = (int*) realloc(list, sizeof(int*) * num);

1) You should probably be allocating space for integers, and not integer pointers.
2) You're not checking the return value from realloc() - are you _sure_ it is succeeding at each call during your program's execution?

[brewbuck]

AFAIK, you can call realloc with a NULL pointer, and it will behave like malloc().... I think.

Supposed to, but can't be trusted to work everywhere.

[cpjust]

Is this valid?

int* list = (int*)malloc(0);

If so, where does it point?

Todd

Apparently it's valid and returns a pointer to a zero length array...

If size is 0, malloc allocates a zero-length item in the heap and returns a valid pointer to that item.

[cpjust]

So I'm writing a program and for some reason if I leave out the free part it doesn't work in random places so I was wondering if I did something wrong.

int* list = (int*)malloc(0);
int num = 0;

for(i=0; i < something; i++)
{
    if(something random happens)
    {
         num++;
         list = (int*) realloc(list, sizeof(int*) * num);
         list[num - 1] = something;
    }
}

free(list);

When I leave out the free part the program runs and finishes fine, but when I leave it in, it segfaults at a random and place. What actually happens is, another unrelated pointer is modified and when its accessed I get the error, am I doing something wrong?

Other than that fact you're casting malloc() & realloc() (see the FAQ), I don't see anything wrong in this piece of code.
However, I would strongly suggest setting list to NULL after you free it. That way you can easily check whether it's pointing to a valid pointer by comparing it to NULL.

[Elysia]

And the fact that you would actually malloc 0 bytes - I'm thinking that's kindof undefined behavior.
And if realloc fails, all the memory is lost and free will free an invalid pointer, probably NULL.

[MacGyver]

Supposed to, but can't be trusted to work everywhere.

Really? Under what conditions would it not work as specified?

[laserlight]

In fact, the behaviour is standard. From C99 section 7.20.3.4:
"If ptr is a null pointer, the realloc function behaves like the malloc function for the specified size."

So for those compilers for which it cannot be trusted to work, those compilers themselves cannot be trusted to conform to the standard.

[MacGyver]

BTW, with regard to malloc()ing a 0 byte block of memory, the results are implementation-defined according to this:

http://c-faq.com/ansi/malloc0.html

Apparently, it will either return a pointer that points to a block of 0 bytes, or return NULL. If it was my choice for my implementation, I believe I would return NULL, since I can't think of a real usage for a block of 0 bytes.

[robatino]

malloc(0) was recently discussed here:

http://cboard.cprogramming.com/showt...t=96232&page=2