Actually, with time, the sign of (iXX)((uXX)x-(uXX)y) is the "odometer comparison" of x and y, which handles overflow more gracefully than the typical < comparison.

You know what causes bugs in C (besides memory leaks)? Complex or clever expressions that don't reveal their semantics at first glance. As an example from elsewhere in the thread:

    return (x > y) - (x < y);

Quick. What's that do?

What will the next person who looks at the code think it does?

If it's in a function called "cmp", I think they will guess.

"I think they will guess."

I think tptacek must be ready to have his bib changed, with all the drooling he must be doing.

A C programmer should know that comparisons evaluate to 1 for true and 0 for false.

C programmers should also be familiar with carry propagation; I see no reason they shouldn't be expected to work out the correctness of

    int cmp(int x, int y) {
      const unsigned a = x;
      const unsigned b = y;
      const unsigned s = sizeof x * 8 - 1;
      return ((b^((b^(b-a))&(a^(b-a))))>>s)&~-((a^((a^(a-b))&(b^(a-b))))>>s)^-((a^((a^(a-b))&(b^(a-b))))>>s)&~-((b^((b^(b-a))&(a^(b-a))))>>s);
    }

The difference is that:

    return (x > y) - (x < y); 

Is a common expression which is very efficient (why else would you write anything in C if you don't care about code being fast/efficient?) and what you proposed is neither.

I write (and read) about 20,000 lines of C(++) a year, and have for the last 20 years. I've never seen that done.

That doesn't mean the expression is unreadable. It is still extremely straightforward in what it does.

The real difference is that it's simple. There are only 11 symbols in the expression, only two things that can vary, and only three interesting cases. As opposed to 129 symbols and some large number of interesting cases.

It returns 1 - 0 if x > y, 0 - 1 if x < y, and 0 - 0 if x == y.

Duh.