Java &= 运算符是否适用 &或&&? [英] Does the Java &= operator apply & or &&?

问题描述

假设

boolean a = false;

我想知道是否要这样做:

I was wondering if doing:

a &= b; 

相当于

a = a && b; //logical AND, a is false hence b is not evaluated.

或者另一方面它意味着

a = a & b; //Bitwise AND. Both a and b are evaluated.

推荐答案

来自 Java 语言规范 - 15.26.2 复合赋值运算符.

From the Java Language Specification - 15.26.2 Compound Assignment Operators.

E1 op= E2 形式的复合赋值表达式等价于 E1 = (T)((E1) op (E2))，其中T 是 E1 的类型，除了 E1 只计算一次.

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.

所以 a &= b; 等价于 a = a &b;.

(在某些用法中，类型转换会对结果产生影响，但在这种情况下，b 必须是 boolean 并且类型转换什么都不做.)

(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.)

而且，根据记录，a &&= b; 不是有效的 Java.没有 &&= 运算符.

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

在实践中，a = a & 之间几乎没有语义差异.b; 和 a = a &&b;.(如果 b 是变量或常量，则两个版本的结果将相同.当 b 是具有边的子表达式时，只有语义差异-effects.在&的情况下，副作用总是发生.在&&的情况下，它的发生取决于a.)

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.)

在性能方面，权衡是在评估 b 的成本与 a 的值的测试和分支的成本之间，以及避免对 a 进行不必要的分配的潜在节省.分析不是直截了当的，但除非计算b的代价不小，否则两个版本之间的性能差异太小，不值得考虑.

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.

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