Create a member variable using map cpp

Question

I am trying to call a function using a std::map key value pair. I found this stackoverflow article Calling a function depending on a variable? but the solution

std::map<std::string, std::function<void()>> le_mapo;

does not work and results in a error something like "error: lvalue required as unary ‘&’ operand" using it as so

std::map<std::string, std::function<void()>> le_mapo;
le_mapo["ftp"] = &ftp(); // ftp function is in the same class, this function is the constructor

I am trying to call a function in the same class with this method and it results in a lvalue error any idea what to do

I have also tried using at the top

#define BE_EVIL(map, function, x) map[ #x ] = & function ## x

NOTE: i have all the proper includes such as

#include <iostream>
#include <map>
#include <functional>

here is a Reproducible Example

#include <iostream>
#include <map>
#include <functional>


class stackoverflow
{
    private:
        void ftp();
    
    public:
        stackoverflow();
};

void ftp()
{
    std::cout << "Minimal reproduction" << std::endl;
}

stackoverflow::stackoverflow()
{
    std::map<std::string, std::function<void()>> le_mapo;
    le_mapo["ftp"] = &ftp();
}

Answer 1

Because ftp is a member function, you are going to have to use a lambda here (well, you could use std::bind, but lambdas have largely superseded that these days).

So you want:

le_mapo["ftp"] = [this](){ ftp(); };

This "captures" this, thus enabling you to call ftp() on the correct object.

&ftp() doesn't work for a number of reasons:

• appending () means that you are attempting to call the function (and then taking the address of whatever it returns), rather than taking the address of ftp directly.

• the syntax for taking the address of a member function is &stackoverflow::ftp, and not just &ftp.

• you are not passing this anywhere, so the caller won't have an object to call ftp on.

Answer 2

This:

le_mapo["ftp"] = &ftp();

is wrong because ftp() is calling the function and only then the address-of operator is applied. Though you cannot take the adress of void. The function pointer is &stackoverflow::ftp.

You need an object to call a member function. std::function has some magic to turn a member function into a callable that takes the object as parameter, hence this works:

stackoverflow::stackoverflow()
{
    std::map<std::string, std::function<void(stackoverflow&)>> le_mapo;
    le_mapo["ftp"] = &stackoverflow::ftp;
}

When calling the functions in the map you must pass a reference to an object of type stackoverflow that will be used to call the member function.

Alternatively you can wrap the object together with the member function in a lambda, as shown in the ohter answer.

---

PS: I suppose there is a typo in your code, because this:

void ftp()
{
    std::cout << "Minimal reproduction" << std::endl;
}

is a free function completely unrelated to stackoverflow::ftp. You don't need a lambda (other answer) or a std::function with an additional paramter when ftp is a free function. For free functions the answer in the Q&A you link is all you need.