将静态方法作为参数传递，是否不需要地址运算符? [英] Passing static method as argument, no address-of operator required?

问题描述

class ThreadWorker
{
public:
    ThreadWorker(void);
    virtual ~ThreadWorker(void);

    static void DoSomething();
};


int main()
{
    boost::thread thread1(ThreadWorker::DoSomething);
    boost::thread thread2(ThreadWorker::DoSomething);
    boost::thread thread3(&ThreadWorker::DoSomething);
}

我正在使用Boost.Thread，我注意到在传递静态成员函数作为参数时，是否使用运算符(&)的地址并不重要.争论.没关系吗如果没有，为什么呢?一种方法比另一种更正确吗?

I'm playing around with Boost.Thread and I notice it doesn't seem to matter whether I use the address of operator (&) or not when passing a static member function as an argument. Does it not matter? And if not, why? Is one way more correct than the other?

推荐答案

实际上没关系.函数(自由函数和静态成员函数，而非非静态成员函数)衰减为函数指针.没有其他方法比其他方法更正确，不过我碰巧更喜欢显式方法.

It effectively does not matter. Functions (free functions and static member functions, not non-static member functions) decay to function pointers. No way is more correct than the other, I happen to prefer the explicit one though.

C ++ 11标准，4.3/1:

函数类型T的左值可以转换为指向T的指针"的前值.结果是指向该函数的指针.

An lvalue of function type T can be converted to a prvalue of type "pointer to T." The result is a pointer to the function.

C ++ 11 Standard，5.2.2/1-函数调用:

函数调用有两种:普通函数调用和成员函数调用.静态成员函数是普通函数.

There are two kinds of function call: ordinary function call and member function call. A static member function is an ordinary function.

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