将静态方法作为参数传递，不需要地址运算符？
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，我注意到在将静态成员函数作为参数传递时是否使用运算符 (&) 的地址似乎并不重要。没关系吗？如果没有，为什么？一种方法比另一种方法更正确吗？
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);
}
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?
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实际上并不重要。函数（自由函数和静态成员函数，而不是非静态成员函数）衰减为函数指针。没有一种方法比另一种更正确，但我碰巧更喜欢明确的方法。
C++11 标准，4.3/1：
C++11 标准，5.2.2/1 - 函数调用：
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 Standard, 4.3/1:
C++11 Standard, 5.2.2/1 - Function call: