java conditional operator,why following code is giving output as true?

Question

class test {
    public static void main (String[] args) {
        boolean a = false;
        boolean b = true;

        if (a && a || b) {
            System.out.println(true);
        }
    }
} //--why it always true????o/p is true but why??

Answer 1

Order of operations.

&& has higher precedence than || and therefore is evaluated first . Your if condition can be rewritten as follows:

(a && a) || b
(false && false) || true
false || true
true

This condition will always be false || true false || true which is always true for the conditions you listed.

Check here for an official table from Oracle which lists the precedence of all operators.

Answer 2

Your code has the equivalence to this statement ::

If A is true or B is true the statement is true

Since B is set to true your statement is true. Also, there is no need to test A twice so instead of doing

(a && a || b) // old 
(a || b) //new

&& has a higher order of operation then || so it is evaluated first. To work around this you can use braces

if( a && ( a || b )) //tests as i believe you wanted although its redundent

Answer 3

a && a || b :
    a && a => false && false => false
    false || b => false || true => true

If there had been parens to change the operator precedence:

a && (a||b) :
    a || b => false || true => true
    a && true => false && true => false

then the result would be different.

Moral: if you are unsure of the precedence use parens so you are sure.

Answer 4

A = False
B = True

A && A (False && False) = False
( A && A(False) ) || B (True) = True

Your last expression is False || True = True False || True = True

Answer 5

This is because the AND(&&) operator is solved first then the OR(||) operator. run

  if (b||a&&a) {

        System.out.println(true);
    }

it will give you true again because && is calculated first.