C programming ++ operator

Question

Why does this code always produce x=2?

unsigned int x = 0;
x++ || x++ || x++ || x++ || ........;
printf("%d\n",x);

Answer 1

the 1st x++ changes x to 1 and returns 0
the 2nd x++ changes x to 2 and returns 1

at which point the or short circuits, returns true, and leaves x at 2.