Logical operator precedence in Java

Question

I'm not happy about this: http://docs.oracle.com/javase/specs/jls/se7/html/jls-15.html#jls-15.22 . It clearly states the following: "Each operator is commutative if the operand expressions have no side effects." and "These operators have different precedence, with & having the highest precedence and | the lowest precedence". How come the JVM doesn't comply to it's own rules. Take for instance the following example. ``` public static void main(String[] args){ boolean bool = isTrue1() | isFalse1() & isFalse2() ; System.out.println("Result : " + bool); } public static boolean isFalse1() { System.out.println("1 : " + false); return false ; } public static boolean isFalse2() { System.out.println("2 : " + false); return false ; } public static boolean isTrue1() { System.out.println("3 : " + true); return true ; } ``` The result is: ``` 3 : true 1 : false 2 : false Result : true ``` While the actual result should be, according to the fact that & operators are evaluated before | operators: ``…

Accepted Answer

"These operators have different precedence, with & having the highest precedence and | the lowest precedence".

Just because they have higher precedence, doesn't mean their operands will be evaluated first.

This

```
boolean bool = isTrue1() | isFalse1() & isFalse2() ;
```

becomes equivalent to

```
boolean bool = isTrue1() | ( isFalse1() & isFalse2() ) ;
```

That's all precedence is.

As the Java Language Specification says, operator operands are evaluated left to right.

The Java programming language guarantees that the operands of operators appear to be evaluated in a specific evaluation order, namely, from left to right.

Logical operator precedence in Java

Solution

Problem