Can realloc fail (return NULL) when trimming?

Question

If do the next:

int* array = malloc(10 * sizeof(int));

and them I use realloc:

array = realloc(array, 5 * sizeof(int));

On the second line (and only it), can it return NULL?

Answer 1

Yes, it can. There are no implementation guarantees on realloc(), and it can return a different pointer even when shrinking.

For example, if a particular implementation uses different pools for different object sizes, realloc() may actually allocate a new block in the pool for smaller objects and free the block in the pool for larger objects. Thus, if the pool for smaller objects is full, it will fail and return NULL.

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Or it may simply decide it's better to move the block

I just used the following program to get size of actually allocated memory with glibc:

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

int main()                                                                   
{                                                                            
    int n;                                                                   

    for (n = 0; n <= 10; ++n)                                                
    {                                                                        
        void* array = malloc(n * sizeof(int));                               
        size_t* a2 = (size_t*) array;                                        

        printf("%d -> %zu\n", n, a2[-1]);                                    
    }                                                                        
}

and for n <= 6, it allocates 32 bytes, and for 7-10 it is 48.

So, if it shrank int[10] to int[5], the allocated size would shrink from 48 to 32, effectively giving 16 free bytes. Since (as it just has been noted) it won't allocate anything less than 32 bytes, those 16 bytes are lost.

If it moved the block elsewhere, the whole 48 bytes will be freed, and something could actually be put in there. Of course, that's just a science-fiction story and not a real implementation ;).

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The most relevant quote from the C99 standard (7.20.3.4 The realloc function):

Returns

4 The realloc function returns a pointer to the new object (which may have the same value as a pointer to the old object), or a null pointer if the new object could not be allocated.

'May' is the key-word here. It doesn't mention any specific circumstances when that can happen, so you can't rely on any of them, even if they sound obvious at a first glance.

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By the way, I think you could consider realloc() somewhat deprecated. If you'd take a look at C++, the newer memory allocation interfaces (new / delete and allocators) don't even support such a thing. They always expect you to allocate a new block. But that's just a loose comment.

Answer 2

The other answers have already nailed the question, but assuming you know the realloc call is a "trimming", you can wrap it with:

void *safe_trim(void *p, size_t n) {
    void *p2 = realloc(p, n);
    return p2 ? p2 : p;
}

and the return value will always point to an object of size n.

In any case, since the implementation of realloc knows the size of the object and can therefore determine that it's "trimming", it would be pathologically bad from a quality-of-implementation standpoint not to perform the above logic internally. But since realloc is not required to do this, you should do it yourself, either with the above wrapper or with analogous inline logic when you call realloc.

Answer 3

The language (and library) specification makes no such guarantee, just like it does not guarantee that a "trimming" realloc will preserve the pointer value.

An implementation might decide to implement realloc in the most "primitive" way: by doing an unconditional malloc for a new memory block, copying the data and free-ing the old block. Obviously, such implementation can fail in low-memory situations.