How to call a function by its name. An approach using STL::map and Class

Question

Based on post How to call a function by its name (std::string) in C++?, tried to make a version using CLASS, but my approach does not work.

class A { public: int add(int i, int j) { return i+j; } int sub(int i, int j) { return i-j; } }; typedef int (*FnPtr)(int, int); int main(int argc, char* argv[]) { // initialization: std::map<std::string, FnPtr> myMap; A a; myMap["add"] = a.add; myMap["sub"] = a.sub;

Returns this erro:

main.cpp:31:22: error: cannot convert ‘A::add’ from type ‘int (A::)(int, int)’ to type ‘std::map<std::basic_string<char>, int (*)(int, int)>::mapped_type {aka int (*)(int, int)}’ main.cpp:32:22: error: cannot convert ‘A::sub’ from type ‘int (A::)(int, int)’ to type ‘std::map<std::basic_string<char>, int (*)(int, int)>::mapped_type {aka int (*)(int, int)}’

Does anyone know what is the error?

Member functions are not functions. You cannot call a member function (like "myMap["add"](1, 2)"), because that makes no sense. Instead, you need to combine the class member A::add and the instance object a in some way. — Kerrek SB
– Kerrek SB, Commented Oct 20, 2013 at 17:06
A pointer-to-member is not the same as a pointer-to-non-member. But you would have been able to find this yourself if you had googled the compiler error. Change typedef int (*FnPtr)(int, int); to typedef int (A::*FnPtr)(int, int); and call it like a.*(myMap["add"])(42, 1337);. Horrible, isn't it? — user529758
– user529758, Commented Oct 20, 2013 at 17:06
You have a more fundamental problem here: member functions are not the same as non-member functions. — juanchopanza
– juanchopanza, Commented Oct 20, 2013 at 17:07

Jerry Coffin · Accepted Answer · 2013-10-20 17:11:47Z

At least as you've shown things, your class A provides nothing but problems. If you turn it into a namespace, things will be a lot easier.

namespace A { int add(int i, int j) { return i+j; } int sub(int i, int j) { return i-j; } }; typedef int (*FnPtr)(int, int); int main(int argc, char* argv[]) { std::map<std::string, FnPtr> myMap; myMap["add"] = A::add; myMap["sub"] = A::sub; // ...

This way, add and sub aren't member functions, so you don't get the type mismatch. At least as shown, the instance of A provided no functionality beyond calling add and sub, so a namespace accomplishes just as much good while eliminating the problems.

Nice and straight-forward. Static class members would provide a similar alternative, with the benefit of being templatable.

Hannah Lenk · Accepted Answer · 2013-10-20 17:34:33Z

You could also solve the problem, as mentioned by Kal in the comments, using std::function and std::bind(this is C++11 so):

#include <iostream> #include <string> #include <map> #include <functional> class A { public: int add(int i, int j) { return i+j; } int sub(int i, int j) { return i-j; } }; int main(int argc, char* argv[]) { std::map<std::string, std::function<int(int,int)>> myMap; A a; myMap["add"] = std::bind(&A::add,&a,std::placeholders::_1,std::placeholders::_2); myMap["sub"] = std::bind(&A::sub,&a,std::placeholders::_1,std::placeholders::_2); std::cout<<myMap["sub"](7,2)<<std::endl; }

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