Question

char *buf = malloc(bufsize)
char *ptr = buf;
…
while(condition) {
    ptrdiff_t offset = ptr - buf;    // <========== THIS LINE

    // offset will never be negative because we only ever *increase* ptr
    if ((size_t)offset > bufsize) {
        // we need more room
        bufsize += 128;
        buf = realloc(buf, bufsize);
        ptr = buf + offset;  // buf might be in a completely new location
    }
    *ptr++ = …  // write this byte
}

Is this valid or undefined?

I would have assumed that it’s valid, but I read something about it being undefined, so I googled it. These links seem to inescapably claim it’s undefined:

• Secure coding
• Is subtraction of pointers not pointing to different elements of same array valid in C?

However, no mention of it is made in these SO questions:

• Pointer subtraction confusion
• size_t return pointer subtraction
• Pointer Arithmetic In C

These all talk about not two pointers being in the same “array”. Does that actually mean a plain old C array on the stack?

If it is undefined, it seems very odd to me… Why force me to carry along a counter variable when I have access to one constant pointer and one moving pointer?

Answer 1

Pointers into a block of memory returned by malloc count as being into the same array:

c11

7.22.3 Memory management functions

1 – The
pointer returned [from malloc] if the allocation succeeds […] may be assigned to
a pointer to any type of object […] and then used
to access such an object or an array of such objects in the space allocated (until the space
is explicitly deallocated).