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Unary Operators

josephine chen
Ranch Hand

Joined: Oct 29, 2002
Posts: 216
Am I correct
class Question {
public static void main(String[] args) {
int x = 0;
boolean b1,b2,b3,b4;
x=(b1 | b2 & b3 ^ b4) ? x++ : --x;//here value of i =0
System.out.println(x); //here it is 1
Jim Yingst

Joined: Jan 30, 2000
Posts: 18671
What happens when you compile it?

"I'm not back." - Bill Harding, Twister

Joined: Feb 02, 2000
Posts: 23
Answer is D ie 0 on compilation, but how??
Thandapani Saravanan
Ranch Hand

Joined: Oct 17, 1999
Posts: 117
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.

uday krishna

Joined: Feb 10, 2000
Posts: 5
it is simple.
it goes via order of precedence.
ie "XOR" "AND" and "OR".
first it would return "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".
Jim Yingst

Joined: Jan 30, 2000
Posts: 18671
Actually "AND" is higher in precedence than "XOR". So it evaluates like this:
(b1 | b2 & b3 ^ b4)
(b1 | ((b2 & b3) ^ b4))
(true | ((true & true) ^ true))
(true | ((true) ^ true))
(true | (false))
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).]
uday krishna

Joined: Feb 10, 2000
Posts: 5
hi jim,
thanx for the suggestion.
i thought it was xor first.
and thandapani was right.
Ranch Hand

Joined: Nov 22, 2008
Posts: 18944
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....
With a little knowledge, a cast iron skillet is non-stick and lasts a lifetime.
subject: Unary Operators