C++ Reference to member function work around (computation speed) - c++

Question

It is well known that you cannot create a reference to a member function in C++ [source].
For those that don't know. The issue come when you want to do something similar to
class A
{
public:
void Add(int a, int b)
{
std::cout << "Sum is " << a + b << std::endl;
}
void CallAdd(int a, int b, void (*func)(int, int))
{
func(a, b);
}
};
and then call Add through CallAdd:
A a;
a.Add(3, 7); // This works fine
a.CallAdd(3, 7, &a.CallAdd); //This does not compile
The error is
error: cannot create a non-constant pointer to member function
a.CallAdd(3, 7, &a.CallAdd);
Which would not occur if it wiere outside a class.
There is a work around using std::function/lambda. Like this:
class A
{
public:
function<void(int, int)> AddFunc = [](int a, int b)
{
std::cout << "Sum is " << a + b << std::endl;
};
void CallAdd(int a, int b, std::function<void(int, int)> &func)
{
func(a, b);
};
};
int main()
{
A a;
a.CallAdd(3, 7, a.AddFunc);
}
This works fine but the problem is that computation time increases a lot (obviously this is just a minimal reproducible example) compared to simply calling the function.
Is there a way to increase computation speed or is this the best one can do?
For context, I have an algorithm to integrate functions, and I want to change integrands at will so the integrand must be a function parameter.

Answer 1

Pass the function object (or lambda) to a templated member function as follows:
#include <concepts>
#include <iostream>
inline auto myAddFunc = [](int a, int b) {
std::cout << "Sum is " << a + b << std::endl;
};
class A
{
public:
template <std::regular_invocable<int,int> Func>
void CallFunc(int a, int b, Func func)
{
func(a, b);
};
};
int main()
{
A a;
auto mySubFunc = [](int a, int b) {
std::cout << "Difference is " << a - b << std::endl;
};
a.CallFunc(3, 7, myAddFunc);
a.CallFunc(7, 3, mySubFunc);
}