operator precedence in C?

Question

int x=12;
int y=7;
int z=19;
int sum;

sum = ++x >= y * 2 || y % 2 && z++ % 2;

I am a little bit confused about the operator precedence? what is the first condition the compiler will read in the above code? Is it going to first evaluate y % 2 && z++ % 2 since && has precedence over || or is the compiler still going to go from left to right and short circuit if on the very left ++x >= y * 2 is true? i.e is the code going to be read in the following way by the compiler?

sum = (++x >= y * 2 || y % 2 )&& z++ % 2;

Answer 1

Your expression is grouped as

(++x >= (y * 2)) || ((y % 2) && (z++ % 2))

and this is assigned to sum. This is specified by the grammar of C.

Note also that the right hand side of || is not evaluated if the left hand side is 1: which will mean that z is not incremented in that case.

1. For avoidance of doubt, ++x is the new value of x, and z++ is the old value of z.

2. Note also that because || is a sequencing point, the expression would be well defined even had you written x++ on the right hand side, rather than z++.

3. Calling the result of this sum is an exercise in obfuscation.