How to pass const member function as non-const member function

Question

How to pass a const member function as a non-const member function to the template?

class TestA 
{
public:
    void A() {
    }

    void B() const {
    }
};

template<typename T, typename R, typename... Args>
void regFunc(R(T::*func)(Args...)) 
{}

void test() 
{
    regFunc(&TestA::A); // OK
    regFunc(&TestA::B); // ambiguous
}

Don't want to add something like:

void regFunc(R(T::*func)(Args...) const)

Is there a better way?

Do you need the function to accept a function pointer? Just taking a generic function type or a std::function should work. — NathanOliver
– NathanOliver, Commented Dec 3, 2018 at 14:46

JeJo · Accepted Answer · 2018-12-03 15:21:58Z

5

Why not simply pass it to a generic template function:

see live

#include <iostream>
#include <utility>

class TestA
{
public:
    void A() { std::cout << "non-cost\n"; }
    void B() const { std::cout << "cost with no args\n"; }
    void B2(int a) const { std::cout << "cost with one arg\n"; }
    const void B3(int a, float f) const { std::cout << "cost with args\n"; }
};
template<class Class, typename fType, typename... Args>
void regFunc(fType member_fun, Args&&... args)
{
    Class Obj{};
    (Obj.*member_fun)(std::forward<Args>(args)...);
}

void test()
{
    regFunc<TestA>(&TestA::A); // OK
    regFunc<TestA>(&TestA::B); // OK
    regFunc<TestA>(&TestA::B2, 1); // OK
    regFunc<TestA>(&TestA::B3, 1, 2.02f); // OK
}

output:

non-cost
cost with no args
cost with one arg: 1
cost with args: 1 2.02

answered Dec 3, 2018 at 15:21

JeJo

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1 Comment

ejkoy Over a year ago

This approach doesn't fit my situation. I need to handle the arguments(get it from something like lua_toxxxx) and the return value ( push it into something like lua_pushxxxx). I finally use function overloading to do this. so ugly but work.. :(

Caleth · Accepted Answer · 2018-12-03 15:14:26Z

2

No, you have to specify the cv and ref qualifiers to match. R(T::*func)(Args...) is a separate type to R(T::*func)(Args...) const for any given R, T, Args....

As a terminology note, it isn't ambiguous. There is exactly one candidate, it doesn't match. Ambiguity requires multiple matching candidates.

answered Dec 3, 2018 at 15:14

Caleth

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