Java operators precedence

Question

I have got confused due to what is true regarding the operator precedence in Java. I read in tutorials a long time ago that AND has a higher priority than OR, which is confirmed by the answers provided in the question. However, I am currently studying Java using the "Sun Certified Programmer for Java 6 Study Guide". This book contains the following example:

int y = 5;
int x = 2;
if ((x > 3) && (y < 2) | doStuff()) {
    System.out.println("true");
}

I am copying and citing the explanation of how the compiler handles the above code:

If (x > 3) is true, and either (y < 2) or the result of doStuff() is true, then print "true". Because of the short-circuit &&, the expression is evaluated as though there were parentheses around (y < 2) | doStuff(). In other words, it is evaluated as a single expression before the && and a single expression after the &&.

This implies though that | has higher precedence than &&. Is it that due to the use of the "non-short-circuit OR" and instead of the short circuit OR? What is true?

Probably a misprint in the book. A bitwise or b/w (y<2) and doStuff() doesn't make sense. Besides, the explanation talks about "either y<2 is true or doStuff() is true", which is logical OR. — Osiris
– Osiris, Commented Dec 27, 2012 at 17:25
| isn't the same as ||. One is bitwise. The other is logical. — Makoto
– Makoto, Commented Dec 27, 2012 at 17:28
Ok, false name, not "bitwise or" but "non-short-circuit OR". — arjacsoh
– arjacsoh, Commented Dec 27, 2012 at 17:30
Simply for easier understanding I prefer to make some more brackets even if not necessary. — MrSmith42
– MrSmith42, Commented Dec 27, 2012 at 17:49

Mike Samuel · Accepted Answer · 2012-12-27 17:37:57Z

8

That's because it is using the | operator instead of ||, which has a higher priority. Here's the table.

Use the || operator instead and it'll do what you think.

edited Dec 27, 2012 at 17:37

Mike Samuel

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answered Dec 27, 2012 at 17:28

mprivat

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5 Comments

JB Nizet Over a year ago

That said, the explanation of the book is completely wrong. The short-circuitness of the && operator doen't have anything to do with why (y < 2) | doStuff() is evaluated as if there were parenthese around it.

arjacsoh Over a year ago

@JB Nizet: The example was set on operators precedence not on how the && actually works. Surprisingly the book lacks a table presenting the whole precedence of the operators in this section, therefore got me confused according to my previous knowledge.

mprivat Over a year ago

Yeah I agree it's a goofy example. Bottom line is the operator precedence table tells all

JB Nizet Over a year ago

@arjacsoh: what I mean is that this sentence is wrong: Because of the short-circuit &&, the expression is evaluated as though there were parentheses around (y < 2) | doStuff(). It should in fact be: Because | has a higher precedence than &&, the expression is evaluated as though there were parentheses around (y < 2) | doStuff().

Alex Over a year ago

Having a hard time seeing how this even compiles - does the boolean (y < 2) get auto-promoted to int?

user207421 · Accepted Answer · 2012-12-28 09:13:59Z

0

Because of the short-circuit &&, the expression is evaluated as though there were parentheses around (y < 2) | doStuff() ...

This statement is incorrect, indeed meaningless. The fact that && is a short-circuit operator has nothing to do with the evaluation of (y < 2) | doStuff(), and indeed that could only compile if doStuff() returned a Boolean. What makes the difference (the implicit parentheses) is the precedence of && relative to |, which is defined in JLS ##15.22-23 as && being lower.

answered Dec 28, 2012 at 9:13

user207421

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