我可以在没有互斥量的线程中读取bool变量吗? [英] Can I read a bool variable in a thread without mutex?

问题描述

如果我不使用互斥锁，那么以下源代码有什么错误吗?

Is there anything wrong in following source code if I don't use mutex?

bool bStop = false;

void thread1_fun()
{
    while (!bStop)
    {
        doSomething();
    }
}

void thread2_fun()
{
    bStop = true;
}

推荐答案

在一个线程中写入对象而另一个线程完全访问对象时，这是未定义的行为.

It is undefined behaviour to write to an object in one thread while another thread accesses the object at all.

除非您明确告知编译器应该有围栏，例如使用std::atomic，std::mutex等，否则所有投注都将关闭.

Unless you specifically inform the compiler that there should be a fence, such as the use of std::atomic, std::mutex et al, all bets are off.

编译器有权按以下方式重新编写代码:

The compiler is within its rights to re-write the code as this:

bool bStop = false;

void thread1_fun()
{
    const bool stopped = bStop;
    // compiler reasons: "because there is no fence, the variable clearly cannot
    // have changed to true, since no-other thread will modify it, since
    // to modify it without a fence would be UB." 
    while (!stopped)  
    {  
        doSomething();
    }
}

void thread2_fun()
{
    // this happens in my thread's view of the world, 
    // but no other thread need see it.
    bStop = true;  
} 

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