Does the Java &= operator apply & or &&?

Solution 1:

From the Java Language Specification - 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.

So a &= b; is equivalent to a = a & b;.

(In some usages, the type-casting makes a difference to the result, but in this one b has to be boolean and the type-cast does nothing.)

And, for the record, a &&= b; is not valid Java. There is no &&= operator.

In practice, there is little semantic difference between a = a & b; and a = a && b;. (If b is a variable or a constant, the result is going to be the same for both versions. There is only a semantic difference when b is a subexpression that has side-effects. In the & case, the side-effect always occurs. In the && case it occurs depending on the value of a.)

On the performance side, the trade-off is between the cost of evaluating b, and the cost of a test and branch of the value of a, and the potential saving of avoiding an unnecessary assignment to a. The analysis is not straight-forward, but unless the cost of calculating b is non-trivial, the performance difference between the two versions is too small to be worth considering.

Solution 2:

see 15.22.2 of the JLS. For boolean operands, the & operator is boolean, not bitwise. The only difference between && and & for boolean operands is that for && it is short circuited (meaning that the second operand isn't evaluated if the first operand evaluates to false).

So in your case, if b is a primitive, a = a && b, a = a & b, and a &= b all do the same thing.