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将ThreadLocal与包含静态成员的现有类一起使用

multithreading java

CPerkins · 技术社区 · 15 年前

我想用 ThreadLocal 为预先存在的非线程安全类提供线程安全,但遇到问题。似乎没有执行隔离——线程仍然共享静态,而不是每个线程的本地。

我相信我的用法几乎与 SimpleDateFormatter 描述于 this StackOverflow question 但这并不是我希望的那样。

我所希望的是,有人使用了这个工具,他会指出我必须犯的极其愚蠢的错误…所以我想我的问题是:你能发现我在这里做错了什么吗?

这是我的简单课程:

public class SimpleClassWithStaticMembers {
    private static String theStaticString =
        "StaticStringInClassWithStaticMember";
    public void setTheStaticString (String val) {
        SimpleClassWithStaticMembers.theStaticString = val; 
    }
    public String getTheStaticString () {
        return SimpleClassWithStaticMembers.theStaticString;
    }
}

这是线程类,它创建 SimpleClassWithStaticMembers :

public class SimpleTesterThread extends Thread {
    private void showMsg (String msg) {  
        System.out.println (msg); 
        System.out.flush(); 
    }
    public SimpleTesterThread (String threadId) {
        super(threadId);
    }
    public void run() {
        try { Thread.sleep(2000); } catch (InterruptedException ex) { }
        ThreadLocal<SimpleClassWithStaticMembers> localizedClass =
            new ThreadLocal<SimpleClassWithStaticMembers>();
        localizedClass.set(new SimpleClassWithStaticMembers());
        // repeating here to be sure we overlap all with all
        for (int ii=0; ii < 3; ii++) { 
            localizedClass.get().setTheStaticString ("Setby_" + this.getName());
            try { Thread.sleep(2000); } catch (InterruptedException ex) { }
            showMsg("            Thread [" + this.getName() + "] - " 
                + localizedClass.get().getTheStaticString() + "'.");
        }
        showMsg ("Thread [" + this.getName() 
            + "] complete. Our 'threadlocal' string is now - " 
            + localizedClass.get().getTheStaticString() + "'.");
        localizedClass.remove();
    }
}

当 SimpleTesterThread 创建(为它们提供不同的线程名称,如“aaaaaaa”、“bbbbbbb”等),然后启动,输出清楚地显示它们共享实例。日志输出包括:

...
            Thread [JJJJJJJJJJ] - Setby_CCCCCCCCCC'.
            Thread [DDDDDDDDDD] - Setby_JJJJJJJJJJ'.
            Thread [IIIIIIIIII] - Setby_DDDDDDDDDD'.
            Thread [GGGGGGGGGG] - Setby_IIIIIIIIII'.
            Thread [EEEEEEEEEE] - Setby_GGGGGGGGGG'.
Thread [EEEEEEEEEE] complete. Our 'threadlocal' string is now - Setby_GGGGGGGGGG'.
            Thread [HHHHHHHHHH] - Setby_GGGGGGGGGG'.
            Thread [BBBBBBBBBB] - Setby_GGGGGGGGGG'.
            Thread [FFFFFFFFFF] - Setby_GGGGGGGGGG'.
Thread [FFFFFFFFFF] complete. Our 'threadlocal' string is now - Setby_GGGGGGGGGG'.
...
            Thread [JJJJJJJJJJ] - Setby_GGGGGGGGGG'.
Thread [JJJJJJJJJJ] complete. Our 'threadlocal' string is now - Setby_GGGGGGGGGG'.
Thread [HHHHHHHHHH] complete. Our 'threadlocal' string is now - Setby_GGGGGGGGGG'.
            Thread [GGGGGGGGGG] - Setby_GGGGGGGGGG'.
Thread [GGGGGGGGGG] complete. Our 'threadlocal' string is now - Setby_GGGGGGGGGG'.
            Thread [IIIIIIIIII] - Setby_GGGGGGGGGG'.
            Thread [CCCCCCCCCC] - Setby_GGGGGGGGGG'.
Thread [CCCCCCCCCC] complete. Our 'threadlocal' string is now - Setby_GGGGGGGGGG'.
Thread [AAAAAAAAAA] complete. Our 'threadlocal' string is now - Setby_GGGGGGGGGG'.
            Thread [DDDDDDDDDD] - Setby_GGGGGGGGGG'.
Thread [DDDDDDDDDD] complete. Our 'threadlocal' string is now - Setby_GGGGGGGGGG'.
Thread [IIIIIIIIII] complete. Our 'threadlocal' string is now - Setby_GGGGGGGGGG'.
============== all threads complete.

我还没有包括创建、启动和连接线程的类——如果觉得这有帮助,我很乐意编辑来添加它。

2 回复 | 直到 15 年前

Tom Hawtin - tackline 15 年前

拥有单独的实例没有帮助,因为所有实例都是相同的静态字段。可变静态是邪恶的。

如果您真的不能更改类,那么您可能只想使用一个锁,这样每个客户机就可以一次使用一个静态字段。如果您想拥有静态字段的不同实例,那么您可能需要使用类加载器(另一个明显的解决方案是重写字节码,这样做更不容易)。

Vineet Reynolds 15 年前

ThreadLocal类不会更改静态字段修饰符的行为。静态字段将继续在类的多个实例中有一个体现,因为它们是在类初始化时创建和初始化的。

为了更清楚地说明这一点,线程本地成员在内部使用每个线程基础上存在的映射进行管理。get()和set()调用在此映射上操作。在threadlocal中没有“魔力”,即静态成员失去了拥有单一化身的属性。将静态成员设置为线程局部变量时,只需将其添加到映射中,以供线程执行的代码的另一部分引用。这不会阻止第二个线程获取对静态成员的引用,并对成员字段执行操作。

正是因为这个原因,汤姆·霍丁的陈述应该被认真对待——可变的静力学根本不构成好的设计。

PS:看看线程的实现,threadlocal和threadlocal.threadlocalmap类将有助于消除对threadlocal对象行为的任何误解。