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Operators

 
Mathangi Shankar
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class EBH020 {
public static void main (String[] args) {
int a = 1 | 2 ^ 3 & 5;
int b = ((1 | 2) ^ 3) & 5;
int c = 1 | (2 ^ (3 & 5));
System.out.print(a + "," + b + "," + c);
}}

Please explain me the output of the above code.
 
Surasak Leenapongpanit
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int a = 1 | 2 ^ 3 & 5
= 0001 | 0010 ^ 0011 & 0101
-----------------------------------------
int b = ((1 | 2) ^ 3) & 5
= ((0001 | 0010) ^ 0011) & 0101
= (0011 ^ 0011) & 0101
= 0000 & 0101
= 0000
= 0
-----------------------------------------
int c = 1 | (2 ^ (3 & 5))
= 0001 | (0010 ^ (0011 & 0101))
= 0001 | (0010 ^ 0001)
= 0001 | 0011
= 0011
= 3
-----------------------------------------
 
Barry Gaunt
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a gives the same result as c (3) because the expressions are really the same due to the relative binding strength of the operators &, ^, and |.
 
Sumithab Baskaran
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so, for a), the order of precedence is & , ^ and then |. Is that right?
 
Barry Gaunt
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Don't take my word for it, do what I did; find a precedence table:
Roedy Green's for example.
[ December 03, 2004: Message edited by: Barry Gaunt ]
 
Mathangi Shankar
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Thanks All. I was struglling with the output of the 1st one then found that the order of precedence is |,^,& in the ascending order.

& has the highest order of precedence.


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

Mathangi.
 
Bruce Evans
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A word of advice if you're going to take the SCJP exam.

MEMORIZE the precedence rules! Just like you have to know the keywords,
all 49 of them, know these rules too.
-Bruce

Originally posted by Mathangi Shankar:
class EBH020 {
public static void main (String[] args) {
int a = 1 | 2 ^ 3 & 5;
int b = ((1 | 2) ^ 3) & 5;
int c = 1 | (2 ^ (3 & 5));
System.out.print(a + "," + b + "," + c);
}}

Please explain me the output of the above code.
 
With a little knowledge, a cast iron skillet is non-stick and lasts a lifetime.
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