Why do x&1==0 and !(x&1) not yield the same results in an if statement in C++?

Question

I had both of these conditions inside an if statement to check if x was even or odd, but the !(x&1) seemed to execute the body of the if in case of an even x, while x&1==0 didn't execute it.

I expected both to give 0 considering 1 & 0 is 0 and 1 in 32 or 64 bit representations will be 000..01 and if, say, x is something like 10010101100 (even), then their bit-wise and should yield 0. Hence, I'm still not sure why !(x&1) works. Please correct me if I am wrong in anywhere. Thank you.

Answer 1

It's just an operator precedence issue that you're facing.

In the expression x&1 == 0, the == operator has the highest precedence; it will be evaluated before the & operator. As a consequence, x&1 == 0 will evaluate to x & (1 == 0), which evaluates to x & 0 (always equal to 0).

You can find a reliable C++ Operator Precedence table on cppreference.com.

You can use parentheses to solve your issue: (x & 1) == 0