Question

I don’t know if anyone could kindly explain this code for me?

unsigned int x = 0;
(x ^= x ) || x++ || ++x || x++;

printf("%d\n", x);

when I compile this on my computer using gcc 4.2, the output is 2.

Originally i thought maybe this behavior is unspecified but then i figure || will have lower precedence over other operators, so shouldn’t the answer be 3? Since there are three “++”.

Can someone explain? Thanks

Answer 1

• (x ^= x) is evaluated and it yields 0, therefore:
• (x++) is evaluated and it yields 0, therefore:
• (++x) is evaluated and it yields 2, therefore it stops

It all boils down to one rule: || only evaluates its right side if its left side is false.