How to use string to call function in a class? - c++

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!!

Answer 1

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;
}

Answer 2

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(); };