Behavior of realloc when the new size is the same as the old one

Question

I'm trying to make a code more efficient. I have something like this:

    typedef struct{
    ...
    }MAP;



    MAP* pPtr=NULL;
    MAP* pTemp=NULL;
    int iCount=0;
    while (!boolean){
    pTemp=(MAP*)realloc(pPtr,(iCount+1)*sizeof(MAP));
    if (pTemp==NULL){
    ...
    }
    pPtr=pTemp;
    ...
    iCount++;
    }

The memory is being allocated dynamically. I would like to reduce the realloc calls to make the code more efficient. I would like to know how realloc would behave if the new size is equal to the old one. Will the call be simply ignored?

Answer 1

It's not specified in standard C. All standard C guaranteed is: the contents of the new object shall be the same as that of the old object prior to deallocation, up to the lesser of the new and old sizes.

However, if you are using GNU libc, it says explicitely to return the same address, see here for detail.

If the new size you specify is the same as the old size, realloc is guaranteed to change nothing and return the same address that you gave.

Answer 2

I wouldn't count on realloc behaving as a no-op if the size doesn't change (although it seems a reasonable thing for it to do), but then you don't have to: if the value of iCount hasn't changed, don't make the call to realloc.

Answer 3

The C standard doesn't specify what will happen, so you're at the mercy of the implementation. I can't imagine any half-decent implementation that will not return the pointer passed in, but the safe option is to check whether the size of the allocation has changed in your own code. That will also certainly skip a function call.

(BTW, please don't cast the return value from realloc. It's not necessary and it may hide undefined behavior if you forget to #include <stdlib.h>.)