在构造函数初始化程序中使用map的初始化程序列表

Question

尝试创建int成员函数指针映射并在构造函数初始化程序中初始化它。 像这样：

class X 
{ 
    using STATEFUNC = void(X::*)(int); 
public: 
    X() : m{ { 1, &setState1 } } {} 

    void setState1(int x) { cout << "state1" << endl; } 

    void setState2(int x) { cout << "state2" << endl; } 


    std::map<int, STATEFUNC> m; 
}; 

我会说这是正确的，但Visual Studio的2017年说：

Error C2664 'std::map,std::allocator>>::map(std::initializer_list>)': cannot convert argument 1 from 'initializer list' to 'std::initializer_list>'

Error C2276 '&': illegal operation on bound member function expression

当您从成员函数删除操作的地址的第一个错误信息保持不变，但第二个更改为：

Error C3867 'X::setState1': non-standard syntax; use '&' to create a pointer to member

如何将int映射初始化为构造函数初始值设定项列表中的成员函数指针？

Answer 1

与

X() : m{ { 1, &X::setState1 } } {} 

只使用&setState1我得到的，从G ++，下面的消息错误

error: ISO C++ forbids taking the address of an unqualified or parenthesized non-static member function to form a pointer to member function. Say ‘&X::setState1’ [-fpermissive]

从铛++错误尝试仅仅是

error: must explicitly qualify name of member function when taking its address

- 编辑 -

我的答案解释了如何解决问题。

要理解为什么&X::setState1作品和&setState1没有，请参阅说书人的答案（+1）

Answer 2

的answer通过max66是修复。至于为什么这是修复：原因是你的代码不会创建一个指向成员的指针。引述n4659（最后一个C++ 17草案，但以前的标准版本说的一样）：

[expr.unary.op/4]

A pointer to member is only formed when an explicit & is used and its operand is a qualified-id not enclosed in parentheses. [ Note: That is, the expression &(qualified-id), where the qualified-id is enclosed in parentheses, does not form an expression of type “pointer to member”. Neither does qualified-id, because there is no implicit conversion from a qualified-id for a non-static member function to the type “pointer to member function” as there is from an lvalue of function type to the type “pointer to function” ([conv.func]). Nor is &unqualified-id a pointer to member, even within the scope of the unqualified-id's class.  — end note ]

X::setState1是一个合格的ID，但setState1不是。