How to use string to call function in a class?

Question

I hope to use map library to call a function by a string with the function name, I've tested the following example and everything are working well.

#include <string>
#include <iostream>

using namespace std;

typedef void (*pFunc)();
map<string, pFunc> strFuncMap;

void func1()
{
    printf("this is func1!\n");
}

void buildMap()
{
    strFuncMap["func1"] = &func1;// 
}

void callFunc(const std::string& str)
{
    (*strFuncMap[str])();
}

int main()
{
    buildMap();

    callFunc("func1");
    return 0;
}

However, as I define all these things in a class, there is a compiler error occur:

#include <map>
#include <string>
#include <iostream>

using namespace std;

class theClass {
public:
    typedef void (*pFunc)();
    map<string, pFunc> strFuncMap;

    void func1()
    {
        printf("this is func1!\n");
    }

    void buildMap()
    {
        strFuncMap["func1"] = &func1;// a value of type cannot be assigned to an entity of type
    }

    void callFunc(const std::string& str)
    {
        (*strFuncMap[str])();
    }
};


int main()
{
    theClass a;
    a.buildMap();

    a.callFunc("func1");
    return 0;
}

I've tried to solve this problem for a couple of hours. Or is there any other way to use string to call function in a class? I will very appreciate if someone can help me. THANKS!!

Thank you so much! I found the mistake I did

Yichao Sun
– Yichao Sun

2021-07-10 01:33:58 +00:00
Commented Jul 10, 2021 at 1:33 — Yichao Sun
– Yichao Sun, Commented Jul 10, 2021 at 1:33

user15388024 · Accepted Answer · 2021-07-10 01:11:58Z

3

Your code doesn't work because func1 is a member function and the syntax for member functions is different.

You need a map of member function pointers (offsets)

typedef void (theClass::*pFunc)();
map<string, pFunc> strFuncMap;

Then you can store the pointer with

strFuncMap["func1"] = &theClass::func1;

And you need an object to call a member function

(this->*strFuncMap[str])();

The final code:

#include <map>
#include <string>
#include <iostream>

using namespace std;

class theClass {
public:
    typedef void (theClass::*pFunc)();
    map<string, pFunc> strFuncMap;

    void func1()
    {
        printf("this is func1!\n");
    }

    void buildMap()
    {
        strFuncMap["func1"] = &theClass::func1;
    }

    void callFunc(const std::string& str)
    {
        (this->*strFuncMap[str])();
    }
};


int main()
{
    theClass a;
    a.buildMap();

    a.callFunc("func1");
    return 0;
}

answered Jul 10, 2021 at 1:11

user15388024

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2 Comments

Yichao Sun Over a year ago

Thank you so much! I will use your final code!

Swift - Friday Pie Over a year ago

And if various objects and classes are involved, one would need do type erasure hacks in way similar to what some frameworks do for signal-slot architecture.

Silvio Mayolo · Accepted Answer · 2021-07-10 01:00:33Z

1

typedef void (*pFunc)();

This declares pFunc to be the type of function pointers. That is, the type of pointers to functions which exist at the top-level. This excludes member functions, lambda functions, and functors. Consider

using pFunc = std::function<void()>

Now your type will correctly accept anything that can reasonably be interpreted as a callable object. Note that member functions still need to be wrapped in a lambda, since you're closing around this.

strFuncMap["func1"] = [this]() { this->func1(); };

answered Jul 10, 2021 at 1:00

Silvio Mayolo

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1 Comment

Yichao Sun Over a year ago

Thank you so much! I found the mistake I did

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