Unary Operators

Am I correct
class Question {
public static void main(String[] args) {
int x = 0;
boolean b1,b2,b3,b4;
b1=b2=b3=b4=true;
x=(b1 | b2 & b3 ^ b4) ? x++ : --x;//here value of i =0
System.out.println(x); //here it is 1
}
}
A)1--true
b)2
c)3
D)0

What happens when you compile it?

Answer is D ie 0 on compilation, but how??

Sounds weird, right? Even the expression,
int x =0;
x = x++;
results in assigning 0 to x;
I think we need to see how this is done step by step.
First the expression is evaluvated. In this case, we get x = 0.
Then, x is incremented; x = x + 1;
Finally the evaluvated expression is executed, so x = 0 is executed resulting in assigning 0 to x.
I hope i haven't missed anything.

it is simple.
it goes via order of precedence.
ie "XOR" "AND" and "OR".
first it would return false.ie "true" XOR "true" is "false".
then it would return false. ie "false" AND "true" is "false".
then it would return true. ie "false" OR "false" is "true".

Uday-
Actually "AND" is higher in precedence than "XOR". So it evaluates like this:
<code><pre>
(b1 | b2 & b3 ^ b4)
(b1 | ((b2 & b3) ^ b4))
(true | ((true & true) ^ true))
(true | ((true) ^ true))
(true | (false))
(true)
</pre></code>
The rest of the question is, how is this evaluated?
<code><pre>int x = 0;
x = (true) ? x++ : --x;</pre></code>
This is equivalent to:
<code><pre>int x = 0;
x = x++;</pre></code>
...which is what Thandapani correctly explained.

[This message has been edited by Jim Yingst (edited February 10, 2000).]

hi jim,
thanx for the suggestion.
i thought it was xor first.
and thandapani was right.

hmmm, most interesting question, not because of the boolean tests but because of what results from the following expressions:
int x = 0;
x = x++;
System.out.println("x = " + x); // prints 0

int y = 0;
y = ++y;
System.out.println("y = " + y); // prints 1
Since BOTH the pre and postfix ++ operators have higher precedence than the assignment operator, one would think the result should be 1 in both cases. Go figure....