How to switch with pointer to member functions? - c++

Question

Well, all I want to do is a "switch" with a function pointer, but with methods pointers. The switch is that if I call the method Run(), it will either redirect to A::RunOn() or A::RunOff() according to Run ptr is pointing to these member functions.
I know it can be done. I did it in plain c but I have searched and googled to do the same thing in c++ but no luck.
class A
{
typedef (void)(A::*RunPtr)(int);
RunPtr RunMethod;
public:
RunMethod Run;
A()
{
Run = RunOff;
}
void SetOn(bool value)
{
if (value)
Run = RunOn;
else
Run = RunOff;
}
void RunOn(int)
{
// RunOn stuff here
}
void RunOff(int)
{
// RunOff stuff here
}
};
So I can call Run() and there will be a switch between the function calls, which I think is more efficient than just doing:
if (on)
RunOn();
else
RunOff();
Don't know how to do it!

Answer 1

Your member function pointer typedef is wrong (Despite the other issues in the shown code). You need
typedef void(A::*RunPtr)(int);
Or you can provide the alias for the member function pointer of class A with the help of using keyword as follows:
using RunPtr = void(A::*)(int);
RunPtr RunMethod;
Now in the SetOn you can do member pointer assignment as follows
void SetOn(bool value)
{
RunMethod = value ? &A::RunOn : &A::RunOff;
}
Now, in order to call the stored member function pointer, you may/ can provide a Run member function as follows:
void Run(int arg)
{
std::invoke(RunMethod, this, arg);
// do something...
}
The call to member function is a bit tricky.
However, this can be done using more generic std::invoke from <functional> header (Since c++17).
Here is the complete example:
#include <iostream>
#include <functional> // std::invoke
class A
{
using RunPtr = void(A::*)(int);
// or with typedef
// typedef void(A::*RunPtr)(int);
RunPtr RunMethod;
public:
void SetOn(bool value)
{
RunMethod = value ? &A::RunOn : &A::RunOff;
}
void Run(int arg)
{
std::invoke(RunMethod, this, arg);
// do something...
}
void RunOn(int arg) { std::cout << "RunOn: " << arg << "\n"; }
void RunOff(int arg) { std::cout << "RunOff: " << arg << "\n"; }
};
int main()
{
A obj;
obj.SetOn(true);
obj.Run(1); // prints: RunOn: 1
obj.SetOn(false);
obj.Run(0); // prints: RunOff: 0
}
(See a Demo)

Answer 2

Your code works fine once you fix the syntax mistakes in it, namely:
Class needs to be class.
in RunMethod Run;, RunMethod is not a type, it is a member variable. You meant to use RunPtr Run; instead (and get rid of RunMethod), but that won't work so well for you (see further below for why).
in Run = RunOn; and Run = RunOff;, you need to fully qualify the method name, and prefix it with the & operator, eg Run = &A::RunOn;.
Try the following:
class A {
public:
typedef void (A::*RunPtr)(int);
RunPtr Run;
A()
{
Run = &A::RunOff;
}
void SetOn(bool value)
{
if (value)
Run = &A::RunOn;
else
Run = &A::RunOff;
}
void RunOn(int param)
{
//RunOn stuff here
}
void RunOff(int param)
{
//RunOff stuff here
}
};
Note, however, that even though you can use Run as a public method pointer like this, the caller will still need to use operator.* or operator->* to actually call it, and that would not look so nice, eg:
A a;
(a.*a.Run)(...);
Online Demo
If you want to be able to call Run() like a.Run(...) then you have to make Run() be a standard method, and have it use a method pointer internally, eg:
class A {
typedef void (A::*RunPtr)(int);
RunPtr RunMethod;
public:
A()
{
RunMethod = &A::RunOff;
}
void SetOn(bool value)
{
if (value)
RunMethod = &A::RunOn;
else
RunMethod = &A::RunOff;
}
void RunOn(int param)
{
//RunOn stuff here
cout << "RunOn: " << param << endl;
}
void RunOff(int param)
{
//RunOff stuff here
cout << "RunOff: " << param << endl;
}
void Run(int param)
{
(this->*RunMethod)(param);
}
};
A a;
a.Run(...);
Online Demo