Bitwise Operators

[sudha siva]

Hai everyone,
class Test{
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; // line 1
System.out.println(x);
}
}

The output of the above code is 0.And in the answer the line 1 is evaluated as (b1 | (b2 & (b3 ^ b4)).But the AND operator should be evaluated first in this type of expression.(I think i,m correct).I'm confused of this code.Can anyone clear my doubt.

Thanks
sudha

[Simeon Shi]

hi,sudha...
u r right about the operators' priority--&->^->|,
then the result of (b1 | b2 & b3 ^b4) is true.
okay, the next step is x = x++; yes,it is where u confused.
such code as x=x--;x=x++; won't change the the value of x.
so x still is 0;
the output should be 0

[Philip Hung]

The Operator Precedence would be & then ^ then |. So b2 & b3 gets evaluated first (results to true), then ^ b4 (results to false) lastly | b1 (results to true). Evaluate x++ which returns 0.

[vishal avad]

hi sudha siva
i dont think it has any relation with bitwise operator.it is has problem with increment operator
for ref.
see post increment operator posted by ravish kumar
i m also waiting 4 its answer. its really complicated.
rgds
vishal

[R K Singh]

Hi Sudha
operators' priority is & -> ^ -> |
it is evaluated as (b1 | ((b2 & b3) ^b4))
so
b2 & b3 = true
true ^ b4 = false
b1 | false = true
hence x = x++;
but now I am confused ??

------------------
Regards
Ravish

[Uma Viswanathan]

x=(b1 | b2 & b3 ^b4) ? x++ : --x;
First let us evaluate
true | ( (true & true ) ^ true )
i.e., true | (true ^ true )
i.e., true | false
i.e., true
Now, the expression is equivalent to
x = x++;
The steps are
1) Return the value of x
2) Increment the value of x
3) Assign the value returned by step 1 to L.H.S variable
So,
The value 0 is assigned to the variable x
Hope this helps...
Uma