Calling function in C++ without creating an object

Question

I've read of similar problems to this, but the solutions provided didn't work for me so.

I want to call a function that exists in another class located in a different .cpp file. I don't want to create an instance of the object, I just want to use the function.

My code that tries to call the function:

    switch (option)
        {
        case 1:
        cout << "\nDoing stuff\n\n" ;
        Controller::AlbumOps SayHey();
        //SayHey should have run but isn't working
        break;

And the function I'm trying to call:

#include "Menu.hpp"
#include "Album.hpp"
#include "stdio.h"
#include "AlbumOps.hpp"
#include <iostream>
using namespace std;


    namespace Controller
    {
        static void Controller::AlbumOps::SayHey ()
        { 
        cout << "Hey\n";
        }
    }

When I execute the code, the Hey is never printed. I thought the solution was to make the function static but that hasn't worked for me.

Answer 1

1. The call should be

     Controller::AlbumOps::SayHey(); // ^^ // double-colon 

2. You should put static on the in-class function declaration , not the out-of-class function definition (where it means something completely different, "internal linkage"). That is:

   in the header ( AlbumOps.hpp ):

    // ... namespace Controller { class AlbumOps { public: // ... static void SayHey(); // Note: 'static' here }; } // ... 

   and in the implementation file ( AlbumOps.cpp ): either:

    // ... void Controller::AlbumOps::SayHey() // Note: no 'static' { cout << "Hey\\n"; } // ... 

   or:

    // ... namespace Controller { // ... void AlbumOps::SayHey() // Note: no 'static', no repeated 'Controller::' { cout << "Hey\\n"; } // ... } // ... 

(For the record, what you current

        Controller::AlbumOps SayHey();
        //                  ^
        //                  space

does is locally declare a function named SayHey taking no parameter and returning a Controller::AlbumOps (search for "C++ most vexing parse").)