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javashort-circuiting

Do &= and |= short-circuit in Java?

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.

Solution

  • 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:
      • The left operand of && evaluates to false
        • (because no matter what the right operand evaluates to, the entire expression is false)
      • The left operand of || evaluates to true
        • (because no matter what the right operand evaluates to, the entire expression is true)

    References