What this code does

[Noorulain Khan]

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

------------------

[bami bal]

The first line evaluates t (true). Both operands must be true for b to become true, so the second operand will also be evaluated. The second operand adds 1 to i. i is not equal to 0, so the second operand = false. Therefore, b becomes false.
The second line first evaluates f. Because f = false, b cannot become true. The operator && prevents the second operand to be evaluated, so (i+=2) won't be executed, and the value of i remains 1.
[This message has been edited by bami bal (edited August 27, 2001).]

[Cindy Glass]

Actually the first line is the same as
b= true && (0 == 0); //the incrementing happens AFTER the compare
so b=true.
Now i is incremented to 1
b = false && (skipped) > 0);
b = false
Try it this way:

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

Prints out
true
1
false