Java &= 运算符是应用 & 还是 && ？
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Does the Java &= operator apply & or &&?
提问by pakore
Assuming
假设
boolean a = false;
I was wondering if doing:
我想知道是否这样做：
a &= b;
is equivalent to
相当于
a = a && b; //logical AND, a is false hence b is not evaluated.
or on the other hand it means
或者另一方面这意味着
a = a & b; //Bitwise AND. Both a and b are evaluated.
采纳答案by Stephen C
From the Java Language Specification - 15.26.2 Compound Assignment Operators.
来自 Java 语言规范 - 15.26.2 复合赋值运算符。
A compound assignment expression of the form
E1 op= E2is equivalent toE1 = (T)((E1) op (E2)), whereTis the type ofE1, except thatE1is evaluated only once.
形式的复合赋值表达式
E1 op= E2等价于E1 = (T)((E1) op (E2))，其中T是 的类型E1，只是E1只计算一次。
So a &= b;is equivalent to a = a & b;.
所以a &= b;相当于a = a & b;.
(In some usages, the type-casting makes a difference to the result, but in this one bhas to be booleanand the type-cast does nothing.)
（在某些用法中，类型转换对结果有影响，但在这种b情况下必须如此boolean，类型转换什么也不做。）
And, for the record, a &&= b;is not valid Java. There is no &&=operator.
而且，为了记录，a &&= b;Java 是无效的。没有&&=运营商。
In practice, there is little semantic difference between a = a & b;and a = a && b;. (If bis a variable or a constant, the result is going to be the same for both versions. There is only a semantic difference when bis 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.)
实际上，a = a & b;和之间的语义差异很小a = a && b;。（如果b是变量或常量，则两个版本的结果将相同。只有当b是具有副作用的子表达式时的语义差异。在这种&情况下，副作用总是发生。在&&如果它发生取决于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 bis non-trivial, the performance difference between the two versions is too small to be worth considering.
在性能方面，权衡是在评估b成本、测试和分支的成本以及a避免对 进行不必要分配的潜在节省之间进行权衡a。分析并不直截了当，但除非计算成本b非同寻常，否则两个版本之间的性能差异太小，不值得考虑。
回答by Philippe Leybaert
It's the last one:
这是最后一个：
a = a & b;
回答by stew
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).
参见JLS 的 15.22.2。对于布尔操作数，&运算符是布尔值，而不是按位。布尔操作数&&和&布尔操作数的唯一区别在于&&它是短路的（这意味着如果第一个操作数的计算结果为假，则不计算第二个操作数）。
So in your case, if bis a primitive, a = a && b, a = a & b, and a &= ball do the same thing.
所以在你的情况下， ifb是一个原始的，a = a && b, a = a & b，并且a &= b都做同样的事情。
回答by Daniel Testa
i came across a similar situation using booleans where I wanted to avoid calling b() if a was already false.
我在使用布尔值时遇到了类似的情况，如果 a 已经是假的，我想避免调用 b()。
This worked for me:
这对我有用：
a &= a && b()
回答by user7291698
Here's a simple way to test it:
这是一个简单的测试方法：
public class OperatorTest {
public static void main(String[] args) {
boolean a = false;
a &= b();
}
private static boolean b() {
System.out.println("b() was called");
return true;
}
}
The output is b() was called, therefore the right-hand operand is evaluated.
输出为b() was called，因此评估右侧操作数。
So, as already mentioned by others, a &= bis the same as a = a & b.
所以，正如其他人已经提到的， a &= b与a = a & b.
