线程安全的单例在C ++ 11 [英] Thread-safe singleton in C++11

问题描述

我知道以下是在C ++ 11中实现单例的线程安全方法：

I know the following is a thread-safe way to implement a singleton in C++11:

Foo* getInst()
{
    static Foo* inst = new Foo(...);
    return inst;
}

我读过这个回答下面也是线程安全的：

I read in this answer that the following is also thread-safe:

Foo& getInst()
{
    static Foo inst(...);
    return inst;
}

它是真线程安全的吗？

Isn'那么一个问题是Foo 的实例将被分配在单个堆栈帧中不会被分配在堆上？

Is it really thread-safe?
Isn't it a problem that the instance of Foo will be allocated in a single stack frame will not be allocated on the heap?

如果是

推荐答案

静态变量是

Both版本使用内部编译器保护（即 __ cxa_guard_acquire（）和 __ cxa_guard_release（），其功能等同于 mutex :: lock（）和 mutex :: unlock（）），以确保序列化访问特殊变量，

Both versions use internal compiler guards (namely __cxa_guard_acquire() and __cxa_guard_release(), which are functionally equivalent to mutex::lock() and mutex::unlock()) to ensure serialized access to a special variable that tells whether or not your global instances are already initialized or not.

您的代码：

Foo& getInst()
{
    static Foo inst(...);
    return inst;
}

在编译之后实际上会是这样：

will actually look like that after the compilation:

Foo& getInst()
{
    static Foo inst; // uninitialized - zero
    static guard instGuard; // zero

    if (is_initialized(instGuard) == false)
    {
        __cxa_guard_acquire(instGuard);
        // do the initialization here - calls Foo constructor
        set_initialized(instGuard);
        __cxa_guard_release(instGuard);
    }

    return inst;
}

所以你的两个exampes都是线程安全的。

So both of your exampes are thread safe.

这篇关于线程安全的单例在C ++ 11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！