C++ Map of string and member function pointer

Question

Hey so I am making a map with string as the key and a member function pointer as the value. I can't seem to figure out how to add to the map, this doesn't seem to be working.

#include <iostream>
#include <map>
using namespace std;

typedef string(Test::*myFunc)(string);
typedef map<string, myFunc> MyMap;


class Test
{
private:
    MyMap myMap;

public:
    Test(void);
    string TestFunc(string input);
};





#include "Test.h"

Test::Test(void)
{
    myMap.insert("test", &TestFunc);
    myMap["test"] = &TestFunc;
}

string Test::TestFunc(string input)
{
}

Seems to fix one error in the parameters, but I still get an error for insert — , Jan 19 '13 at 21:51
The insert said there no instance of the overloaded function and the [] assignment sais it must be a modifiable value — , Jan 19 '13 at 22:02
Be specific, quote the errors. Concerning insert(), you must convert it to the correct type, which is a pair, aka map::value_type. — Ulrich Eckhardt, Jan 19 '13 at 22:04
Does this answer your question? [std::map of member function pointers?](https://stackoverflow.com/questions/1924844/stdmap-of-member-function-pointers) — roschach, Aug 31 '21 at 17:08

Olaf Dietsche · Accepted Answer · 2018-08-02T17:07:48.540

9

See std::map::insert and std::map for value_type

myMap.insert(std::map<std::string, myFunc>::value_type("test", &Test::TestFunc));

and for operator[]

myMap["test"] = &Test::TestFunc;

You cannot use a pointer to member function without an object. You can use the pointer to member function with an object of type Test

Test t;
myFunc f = myMap["test"];
std::string s = (t.*f)("Hello, world!");

or with a pointer to type Test

Test *p = new Test();
myFunc f = myMap["test"];
std::string s = (p->*f)("Hello, world!");

See also C++ FAQ - Pointers to member functions

edited Aug 02 '18 at 17:07

answered Jan 19 '13 at 21:53

Olaf Dietsche

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+1, although because `std::map::value_type` is `pair` I prefer to insert `MyMap::value_type(a, b)` not `std::make_pair(a,b)` otherwise you get `pair ` which has to be converted to `pair` and that conversion cannot be elided. – Jonathan Wakely Jan 19 '13 at 22:02
I'm just wondering whether passing a string literal to make_pair is supposed to work? After all, the implied template type there is char[5], not std::string or somesuch. – Ulrich Eckhardt Jan 19 '13 at 22:05
@doomster, it's _const_ `char[5]`, and in C++03 `make_pair` takes its arguments by value, so the array decays to a pointer in the argument list, and in C++11 the return type uses std::decay::type` which also decays to a pointer. So it works perfectly. – Jonathan Wakely Jan 19 '13 at 22:08
@doomster You're right, this will be a `std::pair` and not `std::pair`. Although it works, Jonathan Wakely's version should be preferred. – Olaf Dietsche Jan 19 '13 at 22:10
Thanks got it, when trying to call the function with the key I'm having trouble, I tried: convertFunc fu = converters[input]; or map::const_iterator it = converters.find(converter); and return fu(input); or return (*fu)(input); – Jan 20 '13 at 03:11
Also, if you're trying to do this with a non-static function you can do this black magic `(this->*f)("Hello World")` – david May 17 '16 at 14:53

C++ Map of string and member function pointer

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