can realloc move pointer if new size smaller?

Question

I am wondering whether the C or C++ standard guarantees that a pointer is not changed when realloc is called with a smaller (nonzero) size:

size_t n=1000;
T*ptr=(T*)malloc(n*sizeof(T));
//<--do something useful (that won't touch/reallocate ptr of course)
size_t n2=100;//or any value in [1,n-1]
T*ptr2=(T*)realloc(ptr,n2*sizeof(T));
//<-- are we guaranteed that ptr2==ptr ?

Basically, can the OS decide on its own that since we freed a large memory block, he wants to take advantage of all reallocs to defragment the memory, and somehow move ptr2 ?

Answer 1

With realloc, you get absolutely no guarantees about where the memory will live afterwords. I believe that libc's default malloc will only begrudgingly copy memory around, so practically speaking you may be OK. But don't count on it.

Answer 2

http://opengroup.org/onlinepubs/007908775/xsh/realloc.html

Upon successful completion with a size not equal to 0, realloc() returns a pointer to the (possibly moved) allocated space.

Nope, no guarantee

Answer 3

On Windows, the C-Runtime grabs a heap, and then allocates memory from that heap. So the OS won't know about individual memory allocations, and thus won't move things around.

Answer 4

There's no guarantee realloc will return the same location, period.

Answer 5

realloc is not required to leave the block in place even if it would fit, and in fact the simplest stub implementation is an example where it might not:

• malloc: call sbrk.
• realloc: call malloc and memcpy.
• free: no-op.

This may sound ridiculous, but sometimes for embedded systems an implementation like I've just described is actually the optimal one.