short circuit operators

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);
}
}
answer 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);
}
}

0
1
2
3
answer is 3.
Please explain?

0
1
2
3

first answer

initially i=0;
in the sixth line it is incremented

so the final answer is 1....

second answer

same logic
i=0
6th line it is incremented by 1...so i=1
in the next line it is incremented by 2...so i=i+2>>>i=1+2

answer is 3

1.1.public static ShortCkt {
2.public static void main(String args[]) {
3.int i = 0;
4.boolean t = true;
5.boolean f = false, b;
6.b = (t && ((i++) == 0));
7.b = (f && ((i+=2) > 0));
8.System.out.println(i);
}
}
answer is 1.

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

Code 1 uses short circuit operator "&&". This evaluates the second part of the expression only if the first part is true. If the first part is false, the second part of the expression is ignored.
In line 6, in the first part, t is true. Hence, the second part is evaluated and i is incremented to 1.In line 7, first part, f is false so, the second half of the expression is ignored. Hence, i = 1.

Code 2 uses logical "&" operator. Hence both expressions are evaluated and i = 3.

Hope you are clear with the answer.

Ram...

this code has nothing to do with the short Circuit operators...they are asking the value of i....which you can evaluate just looking at the increment operators....its another way of confusing scjp aspirants....

Originally posted by suresh koutam:
this code has nothing to do with the short Circuit operators...they are asking the value of i....which you can evaluate just looking at the increment operators....its another way of confusing scjp aspirants....

If this code has nothing to do with the short circuit operators, then I ask you for the detailed explanation why the outputs are different.

Suresh,

The output of the first example does depend on the short circuit operation.

This line of code

does not change the value of i due to the += operator being on the right-hand side of the short circuit && operator when the left-hand side is false.

sorry guys..i was little excited seeing the answers...i agree with you guys..
thanks for correcting...

Suresh Koutam