在多线程环境中读取的值 [英] Values read in Multithreading environment

问题描述

我正在阅读有关线程和锁的 JLS 文档http://docs.oracle.com/javase/specs/jls/se7/html/jls-17.html#jls-17.5.

I was going through the JLS documentation on Thread and Locks http://docs.oracle.com/javase/specs/jls/se7/html/jls-17.html#jls-17.5 .

class FinalFieldExample { 
final int x;
int y; 
static FinalFieldExample f;

public FinalFieldExample() {
    x = 3; 
    y = 4; 
} 

static void writer() {
    f = new FinalFieldExample();
} 

static void reader() {
    if (f != null) {
        int i = f.x;  // guaranteed to see 3  
        int j = f.y;  // could see 0
    } 
} 
}

我对本节中提到的关于如何将 f.y 视为零的上述示例(ex no 17.5-1)感到困惑.读取器线程要么将对象 f 读取为 null，在这种情况下它不会执行任何操作，要么将读取带有某个引用的对象 f.如果对象 f 有一个引用，那么即使多个写入线程正在运行，构造函数也必须完成它的执行，以便可以将引用分配给 f，如果构造函数已经执行，那么 f.y 应该被视为 4.

I am confused with above example (ex no 17.5-1) mentioned in the section as to how f.y could be seen as zero. The Reader Threads will either read the object f as null in which case it will not execute anything or it will read the object f with some reference. If the object f has a reference then the constructor must have completed its execution even though Multiple Writer Threads are running so that a reference could be assigned to f and if the constructor has executed then f.y should be seen as 4.

f.y =0 在什么条件下可能?

In what condition can f.y =0 be possible?

谢谢

推荐答案

f.y =0 在什么条件下可能?

In what condition can f.y =0 be possible?

Java 内存模型允许 JIT 编译器重新排列构造函数外部非最终字段的初始化顺序.x 字段是最终的，所以它必须由 JVM 初始化，但 y 不是最终的.因此，有可能 FinalFieldExample 已分配并设置在 static FinalFieldExample f 上，但 y 字段的初始化尚未完成.

The Java memory model allows the JIT compiler to reorder initialization of non final fields outside the constructor. The x field is final so it must be initialized by the JVM but y is not final. So there is the possibility that the FinalFieldExample is allocated and set on static FinalFieldExample f but the initialization of the y field has not been completed.

引用 17.5-1:

因为 writer 方法在对象的构造函数完成后写入 f，所以 reader 方法将保证看到 f.x 的正确初始化值:它将读取值 3.但是，f.y 不是 final；因此，读者方法不能保证看到它的值 4.

Because the writer method writes f after the object's constructor finishes, the reader method will be guaranteed to see the properly initialized value for f.x: it will read the value 3. However, f.y is not final; the reader method is therefore not guaranteed to see the value 4 for it.

因为 f.y 不是最终的，所以不能保证在构造函数完成和分配 static f 时它已经被设置.因此，创建了一个竞争条件，并且 reader 可能会看到 y 为 3 或 0，具体取决于这个竞争.

Because f.y is not final there is no guarantee that it has been set by the time the constructor finishes and the static f is assigned. So there is a race condition created and the reader may see y as 3 or 0 depending on this race.

这篇关于在多线程环境中读取的值的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！