Java – why is the output different in the case of \ u0026 \ u0026, &, |?

This is a code snippet

Can you explain why the output changes?

1)

public static ShortCkt {
    public static void main(String args[]) {
        int i = 0;
        boolean t = true;
        boolean f = false,b;
        b = (t && ((i++) == 0));
        b = (f && ((i+=2) > 0));
        System.out.println(i);      
    }
}

The output in this case is 1

2)

public static ShortCkt {
    public static void main(String args[]) {
        int i = 0;
        boolean t = true;
        boolean f = false,b;
        b = (t & ((i++) == 0));
        b = (f & ((i+=2) > 0));
        System.out.println(i);      
    }
}

The output in this case is 3

3)

public static ShortCkt {
    public static void main(String args[]) {
        int i = 0;
        boolean t = true;
        boolean f = false,b;
        b = (t || ((i++) == 0));
        b = (f || ((i+=2) > 0));
        System.out.println(i);      
    }
}

The output in this case is 2

4)

public static ShortCkt {
    public static void main(String args[]) {
        int i = 0;
        boolean t = true;
        boolean f = false,b;
        b = (t | ((i++) == 0));
        b = (f | ((i+=2) > 0));
        System.out.println(i);      
    }
}

The output in this case is 3

Solution

As in C / C + + & & is evaluated as "lazy" rather than & not

If a is false then & & B will return false without even evaluating B

The same applies to | B: if the first operand A is true, the entire expression is true, and the second operand B is never evaluated For | however, both a and B will be evaluated

If there are side effects when using & & and / or (or |), as shown in your example

Sidenote: few Java programmers know that ^ (XOR) also applies to Boolean values (a ^ ^ version does not exist just because it is redundant.)