Need an explanation about evaluating a given expression

Question

The expression is as follows

(ux-uy) == -(unsigned) (y-x)

where x and y are random integers and ux and uy are declared and defined as follows

unsigned ux = (unsigned) x;
unsigned uy = (unsigned) y;

I tested the expression in c with various numbers and it was correct but I cant prove why it is correct. please explain.

Whether `signed` or not is `ux - uy == -(uy - ux)` or not? You think of it as a mathematical problem, and you shall see what is going on. — Iharob Al Asimi, Oct 05 '15 at 04:05
do a case by case analysis, i.e., when x greater than y by n, y greater than x by n, equal. — perreal, Oct 05 '15 at 04:06
I am confused about the - sign in front of RHS. what exactly does it do ? — user1335175, Oct 05 '15 at 04:16
@user1335175: The `-` sign in front of the cast means that the result is negated, and then converted back to an `(unsigned int)` again (because the LHS of the equality is an `unsigned int`) before the comparison is done. — Jonathan Leffler, Oct 05 '15 at 04:28

score 2 · Answer 1 · answered Oct 05 '15 at 04:17

2

y-x --> undefined behavior should the int subtraction overflow.

So the equality (ux-uy) == -(unsigned) (y-x) fails, in general.

answered Oct 05 '15 at 04:17

1 Answers1