CODEWITHSUNDEEP
c++arraystemplatesstaticfunction-pointers
owagh
owagh

Reputation: 3528

How do I define a static array to member functions of a templated class?

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

Upvotes: 2

Views: 119

Answers (3)

Seth Carnegie
Seth Carnegie

Reputation: 75130

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.

Upvotes: 4

Stuart Golodetz
Stuart Golodetz

Reputation: 20616

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.

Upvotes: 2

Sarfaraz Nawaz
Sarfaraz Nawaz

Reputation: 361254

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.

Upvotes: 0

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