CODEWITHSUNDEEP
javaconditional-statementsoperator-keyword
user2388736
user2388736

Reputation: 23

Conditional operator behaviour

Can anybody explain the logic behind the output of below code?

    if(true||false&&false){
        System.out.println("right to left ");
    }
    if (false && true||true){
        System.out.println("left to right");
    }

Upvotes: 2

Views: 95

Answers (4)

Mikhail
Mikhail

Reputation: 4223

First of all such if cases are eliminated at compile time, as long as the result is constant.

||(or) - addition

&&(and) - multiplication

that's why && has a higher precedence than ||

So by precedence expressions are equal to:

true + (false * false) and (false * true) + true

Upvotes: 0

Daniel Kaplan
Daniel Kaplan

Reputation: 67320

if(true||false&&false) can be simplified to true so it prints the line. It can be simplified to true because you have a true on the left hand side. Since there's an or/|| on the right of it, and things are already true, it means it can short circuit and not evaluate the rest. You can prove this to yourself by taking the false&&false, putting it in its own method, and then printing a line before it returns. You'll see that it doesn't print. That's because the (false && false) is not even evaluated.

false && true||true can be simplified to true so it prints the line. The false && true evaluates to false, but since there's an or left over, it evaluates the other side of that. That turns out to be true. Since false||true == true, that evaluates to true. There is no short circuiting here though. You can prove this by extracting the false && true into its own method and printing something on the first line. You'll see that it prints:

public static void main(String args[]) {
    if(true|| doSomething()){
        System.out.println("right to left ");
    }
    if (doSomethingElse() ||true){
        System.out.println("left to right");
    }

}

private static boolean doSomething() {
    System.out.println("do something"); //this does NOT print
    return false&&false;
}

private static boolean doSomethingElse() {
    System.out.println("do something else");    //this DOES print
    return false && true;
}

Upvotes: 1

Keppil
Keppil

Reputation: 46209

If you take a look at this Operator precedence table, you can see that && has a higher precedence than ||.

Therefore, the two expressions are evaluated like this:

true || false && false == true || (false && false) == true

false && true || true == (false && true) || true == true

Upvotes: 5

Noctua
Noctua

Reputation: 5208

Java gives the different operator precedence to && and ||. That means your expressions can be written as true || (false && false) and (false && true) || true. When you see an "or true" in a boolean expression, it's always true. Which is why both strings are printed.

If you want your expressions to be clear, you can always put brackets around the x && y parts.

Upvotes: 1

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