为什么这不短路？

Question

public static void main(String args[]){
    String s = null;

    if(false && false || s.equalsIgnoreCase("x")) {    //throws an exception
        System.out.println("true");
    } else {
        System.out.println("false");
    }
}

I was expecting after this encounters false to short circuit but it doesn't. Why is that? I assumed if you have a false there is NO need to proceed to the next &&. Please help me to understand.

Answer 1

Because this:

a && b || c

is equivalent to this:

(a && b) || c

The lhs of the || is false, so c gets evaluated.

Answer 2

The && operator has higher precedence than the || operator . So, the condition is equivalent to

if( (false && false) || s.equalsIgnoreCase("x")) {    //throws an exception

which is then evaluated to leave:

if(false || s.equalsIgnoreCase("x")) {    //throws an exception

which doesn't short circuit. To make it short-circuit, use parentheses:

if(false && (false || s.equalsIgnoreCase("x"))) {    // fine

Answer 3

The and have higher precedence so it's bracketed like

((false && false) || s.equalsIgnoreCase("x"))

(false && false) reduces to false which is not short-circuit by || .

---

It's worth remembering that & is sort of like * ( 1 & 1 = 1 and 0 * 1 = 1 ) and + is sort of like | hence & have higher precedence. In maths sometimes + is used as or and nothing for both & and * . Logic operations have the same precedence as binary ones so && have higher then || .

Answer 4

Why should it short-circuit? || only short-circuits if the LH operand is true. && short-circuits if the LH operand is false. Here you have a short-circuing && and a non-short-circuiting ||.