Do &= and
In other words, do the following two statements behave the same way?
isFoobared = isFoobared && methodWithSideEffects();
isFoobared &= methodWithSideEffects();
I realize I could just write up a test, but someone might know this offhand, and others might find the answer useful.
No, |= and &= do not shortcircuit, because they are the compound assignment version of & and | , which do not shortcircuit.
JLS 15.26.2 Compound Assignment Operators
A compound assignment expression of the form E1 op= E2 is equivalent to E1 = (T)((E1) op (E2)) , where T is the type of E1 , except that E1 is evaluated only once.
Thus, assuming boolean & , the equivalence for isFoobared &= methodWithSideEffects() is:
isFoobared = isFoobared & methodWithSideEffects(); // no shortcircuit
On the other hand && and || do shortcircuit, but inexplicably Java does not have compound assignment version for them. That is, Java has neither &&= nor ||= .
See also
What is this shortcircuiting business anyway?
The difference between the boolean logical operators ( & and | ) compared to their boolean conditional counterparts ( && and || ) is that the former do not "shortcircuit"; the latter do. That is, assuming no exception etc:
& and | always evaluate both operands && and || evaluate the right operand conditionally; the right operand is evaluated only if its value could affect the result of the binary operation. That means that the right operand is NOT evaluated when: && evaluates to false false ) || evaluates to true true ) References
& , ^ , and | && || No, they do not, because x &= y is short for x = x & y and x |= y is short for x = x | y x = x | y . Java has no &&= or ||= operators which would do what you want.
The & and | operators (along with ~ , ^ , << , >> , and >>> ) are the bitwise operators. The expression x & y will, for any integral type, perform a bitwise and operation. Similarly, | performs a bitwise or. To perform a bitwise operation, each bit in the number is treated like a boolean, with 1 indicating true and 0 indicating false . Thus, 3 & 2 == 2 , since 3 is 0...011 in binary and 2 is 0...010 . Similarly, 3 | 2 == 3 3 | 2 == 3 . Wikipedia has a good complete explanation of the different operators. Now, for a boolean, I think you can get away with using & and | as non-short-circuiting equivalents of && and || , but I can't imagine why you'd want to anyway.
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