CODEWITHSUNDEEP
cmemorydynamicfreerealloc
Edward
Edward

Reputation: 47

I'm sure this is not supposed to work but it does

This is a function used to reallocate some memory on a dynamic array. I have typedef'd struct lottery to lot. I am passing by reference the array of structures (a) and the new size I want it to have (n). I am declaring a temporary array (b) so I can realloc to a.

My question is: I am creating some bytes on the heap with b = realloc() etc., but if I free(b) before I quit the function it doesn't work properly. If I call it again b == NULL becomes true, but if I remove the free(b) it works just fine, but I think that is not right thing because I am leaving garbage bytes on the heap. Can someone please explain to me the problem?

lot *Enterd(lot **a, int n) {
    lot *b = NULL;
    b = (lot *)realloc(*a, n * sizeof(lot));
    if (b == NULL) {
        printf("Memory could not be allocated for the new input.\n");
        return NULL;
   }
   *a = b;
   free(b);
   return *a;
}

Upvotes: 0

Views: 52

Answers (1)

melpomene
melpomene

Reputation: 85827

b = realloc(*a, X) frees the memory associated with *a and allocates a new allocation of size X, stored in b.

Now when you do *a = b, both *a and b reference this new allocation.

After free(b), that allocation is released and both b and *a become invalid pointers. At that point return *a has undefined behavior.

If you don't free b, all is fine. You don't leak memory because you still have a pointer to it: Both via *a (which references a variable in the caller) and the function's return value.

(Also, don't cast realloc().)

Upvotes: 1

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