Why isn't the right side of && operator evaluated?

Question

Small doubt. Why is the output of the following code 1? Why not 3?

int i = 0;
boolean t = true, f = false, b;
b = (t && ((i++) == 0));
b = (f && ((i+=2) > 0));
System.out.println(i);

Answer 1

the Conditional-And operator - && - is short circuit. It doesn't evaluate the right operand if the left operand is false. That's why ((i+=2) > 0) is never evaluated, and i remains 1.

From the JLS 15.23:

The && operator is like & (§15.22.2), but evaluates its right-hand operand only if the value of its left-hand operand is true.

Answer 2

Here is the case

b = (f && ((i+=2) > 0)); // here f is false

Now false && anything is false. && is short circuit operator so it will not evaluate ((i+=2) part since left one is false. So i will remain 1

Just try to change

b = (f && ((i+=2) > 0));

To

b = (f & ((i+=2) > 0));// non short circuit 

Now you will get 3.

That is the two different behavior of short circuit and non short circuit AND.

For more info.

Answer 3

Agree with the posted answer as && is responsible for this.

You should note here that your statement b = (t && ((i++) == 0)); is equivalent to

   if(t){
        if(i++==0){
            b=true;
        }
    }

and second statement b = (f && ((i+=2) > 0)); is equivalent to,

  if(f==true){
        i=i+2;
        if(i>0){
            b=true;
        }
   }

Answer 4

For example:

if(p && q){
 // do something
}

if p is evaluated to be false, the rest of the statement need not be evaluated because of the very definition of && (p AND q must be true). If p is not true, then the statement cannot possibly be true.