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Short Circuit operators and Bitwise operators

 
Vinal Kalyan
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Hi,

I am having trouble understanding what happens when short-circuit operators and bitwise operators are combined in an expression producing a boolean result. The following code has both a && operator and a | operator.

I thought that the following code would update the count variable to 1, because the ba.b[++ba.count] | true would be first evaluated because " | " holds higher order of precedence than " && ". However, I found out that it does not evaluate that all, and short-circuits, with the count variable not being updated.

public class BoolArray {
boolean [] b = new boolean[3];
int count = 0;

public static void main(String[] args) {
BoolArray ba = new BoolArray();
ba.b[0] = false;
ba.b[1] = false;
ba.b[2] = false;

if (false && ba.b[++ba.count] | true) {
System.out.println("We are here");

}
System.out.println(ba.count);

Thanks

Vinal.
 
Jeff Bosch
Ranch Hand
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By having "(false&&", the condition is automatically false and will go no further in evaluation because you've told the compiler to use the shortcut operator. Had you used "(false&" it would continue evaluating.

That's my story and I'm sticking to it...
 
Bert Bates
author
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Disclaimer: I didn;t read either the question OR the answer in any detail...

I would like to say however that if you're studying for the Tiger exam you don't need to worry about the bitwise operators.
 
It is sorta covered in the JavaRanch Style Guide.
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