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Operator Precedence & Associativity

Chad Patrick
Greenhorn

Joined: Feb 01, 2003
Posts: 11

I had a question about the result of this code from Mughal's Java Cert book (problem 3.17 p74)
public class Logic{
public static void main( String args[]){
int i = 0;
int j = 0;
boolean t = true;
boolean r;
r= (t & 0<(i+=1));
r= (t && 0<(i+=2));
r= (t | 0<(j+=1));
r= (t || 0<(j+=2));
System.out.println(i + " "+j);
}
}
Select all valid answers.
(a) The first digit printed is 1.
(b) The first digit printed is 2.
(c) The first digit printed is 3.
(d) The second digit printed is 1.
(e) The second digit printed is 2.
(f) The second digit printed is 3.
According to the answer in the back of the book, the correct answers are (c) and (d). I understand how the shortcircuited and logical operators work, but since "i+=" and "j+=" are in parenthesis, wouldnt they be executed first in each expression/line, making the correct answers (c) and (f)... I guess im getting the left to right associativity confused with the precedence.
Can anyone clarify this?
Thanks
Chad
Anonymous
Ranch Hand

Joined: Nov 22, 2008
Posts: 18944
This seems to be the problem line: r= (t || 0<(j+=2));
For any expression 'X || Y', subexpression Y isn't evaluated at all if subexpression X happens to be true, no matter whether or not subexpression Y is parenthesized. So, if 't' happens to be true (which it is), expression '0 < '(j+=2) will not be evaluated.
For completeness, for any epxression 'X && Y', subexpression Y isn't evaluated at all if subexpression X happens to be false.
kind regards
 
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