Question

I came across this code:

    int main()
    {
        int i=1,j=2,k=0,m=0;
        m = ++i || ++j && ++k;
        printf("%d %d %d %d %d",i,j,k,m);
    }

The program returns 2 2 0 1.... Why?

&& has a higher priority than || so ++j && ++k should be evaluated first. Hence I would expect j=3 and k=1. It will return true hence || becomes true so ++i shouldn’t be evaluated. But it works other way around.

I would like others to explain to me.

Answer 1

Having higher precedence does not mean it gets evaluated first. It just means it binds tighter. In that example, that expression is equivalent to: ++i || (++j && ++k). What gets evaluated first is ++i because || evaluates left to right. Only if that evaluates to false will ++j && ++k be evaluated because || is short-circuiting.