Array of Pointers on Member functions of current class

Question

I try to create an array of pointers of member function of my current class but without success for the moment...

I already tried many things, here is my code:

// Human.hpp

#include <iostream>

class Human
{
    private:
        void meleeAttack(std::string const & target);
        void rangedAttack(std::string const & target);
        void intimidatingShout(std::string const & target);
    public:
        void action(std::string const & action_name, std::string const & target);
};

// Human.cpp

#include "Human.hpp"

// ...

void    Human::action(std::string const & action_name, std::string const & target)
{
    std::string actionsStr[] = {"meleeAttack", "rangedAttack", "intimidatingShout"};

    typedef void (Human::*Actions)(std::string const & target);
    Actions actions[3] = {&Human::meleeAttack, &Human::rangedAttack, &Human::intimidatingShout};

    for (int i = 2; i >= 0; i--)
        if (actionsStr[i] == action_name)
        {
            actions[i](target); // there is an error HERE
            break;
        }
}

This code generate an error on compilation:

main.cpp:41:23: error: called object type 'Actions' (aka 'void (Human::*)(const std::string &)') is not a function or function pointer
            actions[i](target);
            ~~~~~~~~~~^
1 error generated.

What is the good way to do it ?

A pointer to a non-member function and a pointer to a member function is not the same. Pointers to member functions needs an object to be called on (what becomes the this pointer). You might want to read about std::function and std::bind. — Some programmer dude
– Some programmer dude, Commented Apr 4, 2017 at 11:29
You probably want to read about std::unordered_map as well. — Some programmer dude
– Some programmer dude, Commented Apr 4, 2017 at 11:31

Richard Hodges · Accepted Answer · 2017-04-04 11:46:27Z

7

for (int i = 2; i >= 0; i--)
    if (actionsStr[i] == action_name)
    {
        (this->*(actions[i]))(target);
        break;
    }

answered Apr 4, 2017 at 11:46

Richard Hodges

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2 Comments

Caio V. Over a year ago

Why this works?

Richard Hodges Over a year ago

Because language rules

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