C++ class function pointer

Question

I have a request for function pointer by C++. below is the sample what I need:

in API file:

class MyClass {
    public:
        void function1();
        void function2();
        void function3();
        void function4();
};

in main file:

MyClass globalglass;

void global_function_call(???)// <---- how to do declaration of argument???
{
    //Do *function
}

int main()
{
    global_function_call(&globalglass.function1()); // <----  pseudocode, I need to use global class 
    global_function_call(&globalglass.function2());
    global_function_call(&globalglass.function3());
    global_function_call(&globalglass.function4());

    return 1;   
}

I have no idea to do declaration...

What is global_function_call expected to do? Should it call functionX on globalglass specifically or do you want to pass the object on to it as well? — user17732522
– user17732522, Commented Jan 10, 2022 at 3:47
Only having a function pointer to a member &MyClass::function1 is not enough to be able to call it, you also need to provide a class instance to call. Have a look at : learncpp.com/cpp-tutorial/…. I find that syntax the most useful/reusable : auto lambda1 = [&globalclass]{global_class.function1();}; global_function_call(lambda1); — Pepijn Kramer
– Pepijn Kramer, Commented Jan 10, 2022 at 3:48
Oh and do try to come up with a design where you will not need a global instance. It is not good design for maintainability/testability. You might want to learn about dependency injection : en.wikipedia.org/wiki/Dependency_injection_ — Pepijn Kramer
– Pepijn Kramer, Commented Jan 10, 2022 at 3:53

Remy Lebeau · Accepted Answer · 2022-01-10 04:47:11Z

2

To do what you are asking for, you can use a pointer-to-member-method, eg:

MyClass globalglass;

void global_function_call(void (MyClass::*method)())
{
    (globalglass.*method)();
}

int main()
{
    global_function_call(&MyClass::function1); 
    global_function_call(&MyClass::function2);
    global_function_call(&MyClass::function3);
    global_function_call(&MyClass::function4);

    return 1;   
}

Online Demo

edited Jan 10, 2022 at 4:47

answered Jan 10, 2022 at 4:41

Remy Lebeau

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