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

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

  • Shortcut “or-assignment” (|=) operator in Java
  • What’s the difference between | and || in Java?
  • Why doesn’t Java have compound assignment versions of the conditional-and and conditional-or operators? (&&=, ||=)

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)


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