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by stromgo 3816 days ago
The casts are needed for "dirty pointer math". Without the casts, &A[1] - &A[0] returns 1, not 4, for an int *A.
1 comments

The code is still wrong for another reason: if (uintptr_t)ptrOld < (uintptr_t)ptrNew, then the subtraction is calculated modulo some power of 2 yielding a large number. The result of converting this to a signed integral type like ptrdiff_t is implementation defined.

In practice it works only if sizeof(ptrdiff_t) == sizeof(uintptr_t) AND the system uses two's complement for signed integers. Neither property is guaranteed by the standard.

I think the correct way to find the byte offset is this:

  ptrdiff_t diff = (char*)ptrOld - (char*)ptrNew;
And it's wrong for yet another reason. You can convert a pointer to uintprt_t, but arithmetic on the result isn't necessarily meaningful.
This thread makes me eager to get back into the minimal C programming that I left behind in my college assembly class.

~1 beat~

Yikes!

~leaves self-shaped dust cloud behind as hurriedly runs toward higher [level] ground~