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 — user1670407
– user1670407, Commented Jan 19, 2013 at 21:51
Be specific, quote the errors. Concerning insert(), you must convert it to the correct type, which is a pair<T1 const, T2>, aka map::value_type. — Ulrich Eckhardt
– Ulrich Eckhardt, Commented Jan 19, 2013 at 22:04

Olaf Dietsche · Accepted Answer · 2018-08-02 17:07:48Z

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 2, 2018 at 17:07

answered Jan 19, 2013 at 21:53

Olaf Dietsche

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6 Comments

Jonathan Wakely Over a year ago

+1, although because std::map<A,B>::value_type is pair<const A,B> I prefer to insert MyMap::value_type(a, b) not std::make_pair(a,b) otherwise you get pair<A,B> which has to be converted to pair<const A,B> and that conversion cannot be elided.

Ulrich Eckhardt Over a year ago

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.

Jonathan Wakely Over a year ago

@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<const char[5]>::type` which also decays to a pointer. So it works perfectly.

Olaf Dietsche Over a year ago

@doomster You're right, this will be a std::pair<const char*, ...> and not std::pair<std::string, ...>. Although it works, Jonathan Wakely's version should be preferred.

user1670407 Over a year ago

Thanks got it, when trying to call the function with the key I'm having trouble, I tried: convertFunc fu = converters[input]; or map<string, convertFunc>::const_iterator it = converters.find(converter); and return fu(input); or return (*fu)(input);

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