关于指向非静态类成员的指针的澄清
当我需要指向类成员的指针时，我会执行以下操作
struct MyStruct
{
int foo();
};
int (MyStruct::*p)() = &MyStruct::foo;
我的问题是为什么我需要使用 &运算符获取地址，如果它是静态函数，则可以忽略该地址。另外，我听说指向成员的指针并不是真正的指针，有人可以澄清一下吗？
When I need a pointer to member of class, I do as following
struct MyStruct
{
int foo();
};
int (MyStruct::*p)() = &MyStruct::foo;
My question is why do I need to use & operator to take the address which may be ignored if it were a static function. Also, I've heard pointer to members aren't really a pointer, Can someone clarify that?
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如果它是静态函数，它的工作方式就像常规非成员函数指针：函数名称本身可以隐式转换为函数指针。
如果它是一个非静态成员函数，它就不再与非成员函数是一回事了：
this参数；this指针。这已经使得不可能使用指针来存储指向成员函数的指针。雷蒙德·陈 就此写了一篇有趣的文章，其中包含更多详细信息和示例。
If it's a static function, it works just as a regular non-member function pointer: the function name itself can be implicitly converted to a function pointer.
If it's a non-static member function, it's no longer the same thing as a non-member function:
thisparameter;thispointer may need to be adjusted before the call. This already makes it impossible to use a pointer to store a pointer-to-member-function.Raymond Chen wrote an interesting article about this with more details and examples.
您是对的，在指向成员函数语法的指针的情况下，理想情况下可以省略
&。我认为，由于历史惯例，&语法的存在可能。这是不正确的。唯一的区别是它们是指向成员函数的指针。由于
class非静态成员包含一个隐式this指针作为其参数，因此它们具有特殊的签名。此外，它们不能与具有相同签名的普通函数指针相互转换。在您的代码示例中，理论上
p指向：You are right, in the case of pointer to member function syntax ideally
&can be omitted. I think,&syntax is there may be due to historical convention.That's not correct. The only difference is that they are pointer to member function. Since,
classnon-static member contain an implicitthispointer as their argument, they have a special signature. Also, they are not inter-convertible with normal function pointer with same signature.In your code example, theoritically
pis pointing to: