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, Dec 27 '12 at 17:25
`|` isn't the same as `||`. One is bitwise. The other is logical. — Makoto, Dec 27 '12 at 17:28
Ok, false name, not "bitwise or" but "non-short-circuit OR". — arjacsoh, Dec 27 '12 at 17:30
@owlstead: "Sun Certified Programmer for Java 6 Study Guide". Could you make a comment with regard to this book? — arjacsoh, Dec 27 '12 at 17:45
Simply for easier understanding I prefer to make some more brackets even if not necessary. — MrSmith42, Dec 27 '12 at 17:49
Tried to find errata on that book, but I didn't find any. Sun/Oracle hardly ever admit to mistakes anyway. — Maarten Bodewes, Dec 27 '12 at 17:57
I have formatted your question, maybe you could check out the resulting [markdown](http://stackoverflow.com/editing-help) formatting. Please use 4 spaces instead of tabs for code fragments. — Maarten Bodewes, Dec 27 '12 at 18:04

score 8 · Accepted Answer · edited Dec 27 '12 at 17:37

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 '12 at 17:37

Mike Samuel

118,113
30
216
245

answered Dec 27 '12 at 17:28

mprivat

21,582
4
54
64

1

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. – JB Nizet Dec 27 '12 at 17:32
@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. – arjacsoh Dec 27 '12 at 17:35
Yeah I agree it's a goofy example. Bottom line is the operator precedence table tells all – mprivat Dec 27 '12 at 17:36
2

@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()*. – JB Nizet Dec 27 '12 at 17:40
Having a hard time seeing how this even compiles - does the boolean `(y < 2)` get auto-promoted to int? – Alex Dec 27 '12 at 20:36

score 0 · Answer 2 · answered Dec 28 '12 at 09:13

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.

Java operators precedence

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