并发代码中赋值运算符的返回值 [英] Return value of assignment operator in concurrent code

问题描述

给定以下类：

class Foo {
  public volatile int number;

  public int method1() {
    int ret = number = 1;
    return ret;
  }

  public int method2() {
    int ret = number = 2;
    return ret;
  }
}

在同一个 Foo 实例上同时使用方法1（）和 method2（）

and given multiple threads calling method1() and method2() concurrently on the same Foo instance, can a call to method1() ever return anything other than 1?

推荐答案

JLS 15.26规定：

The JLS 15.26 specifies:

有12个赋值运算符;所有都是语法右相关的（他们从右到左）。因此，a = b = c表示a =（b = c），它将c的值赋给b，然后将b的值赋给a。

There are 12 assignment operators; all are syntactically right-associative (they group right-to-left). Thus, a=b=c means a=(b=c), which assigns the value of c to b and then assigns the value of b to a.

Ted Hopp的回答表明Sun的javac不会遵循这个行为，可能是一个优化。

Ted Hopp's answer shows that Sun's javac doesn't follow this behaviour, possibly as an optimisation.

由于这里的线程，行为method1将是未定义的。如果Sun的编译器使得行为常数，那么它不会从未定义的行为中断。

Due to the threading here, the behaviour of method1 would be undefined. If Sun's compiler makes the behaviour constant then it doesn't break from the undefined behaviour.

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