Why does the “or” go before the “and”?

Question

int it=9, at=9;

if(it>4 || ++at>10 && it>0)  
{
    System.out.print(\"stuff\");    
}

System.out.print(at);

prints out stuff9 a

Answer 1

Operator precedence only controls argument grouping; it has no effect on execution order. In almost all cases, the rules of Java say that statements are executed from left to right. The precedence of || and && causes the if control expression to be evaluated as

it>4 || (++at>10 && it>0)

but the higher precedence of && does not mean that the && gets evaluated first. Instead,

it>4

is evaluated, and since it's true, the short-circuit behavior of || means the right-hand side isn't evaluated at all.