如何为模板化类的成员函数定义静态数组？

Question

I want to do the following :-

#include <iostream>

template <typename I>
class A {
  public:
    I member_;
    void f() {}
    void g() {}
    typedef void (A::*fptr) ();
    static const fptr arr[];
};

template <typename I>
A<I>::fptr A<I>::arr[] = {
  &A<I>::f,
  &A<I>::g
};

How do I do this? I get the following errors :-

g++ memeber_func_ptr_array.cpp 
memeber_func_ptr_array.cpp:14:1: error: need ‘typename’ before ‘A<I>::fptr’ because ‘A<I>’ is a dependent scope
memeber_func_ptr_array.cpp:17:2: error: expected unqualified-id before ‘;’ token

Answer 1

Two things.

1. fptr is a dependent type so you need typename :

    template <typename I> const typename A<I>::fptr A<I>::arr[2] = { // also note the 2 and the const &A<I>::f, &A<I>::g }; 

   And as jrok noted in the comments, your declaration is const so the definition must be const as well.

2. Client code (files that just include the header) needs to know how big the array is so you need the actual size of the array in the declaration:

    static const fptr arr[2]; // include size 

   You can only use the automatic deduction of the size of the array when the array is declared and initialised in the same place.

Answer 2

You need to add const typename before A<I>::fptr . The typename is there to tell the compiler that fptr is a type within A<I> .

You might want to look at C++ Templates: The Complete Guide by Vandevoorde and Josuttis for more information.

Answer 3

Use typename as:

  template <typename I>
  typename A<I>::fptr A<I>::arr[] = { &A<I>::f, &A<I>::g };
//^^^^^^^^note this

It is because fptr is a dependent type.