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# Operator Precedence & Associativity

Greenhorn
Posts: 11
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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);
}
}
(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