Reputation: 53
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?
Upvotes: 4
Views: 238
Reputation: 32722
Why not simply pass it to a generic template function:
#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
Upvotes: 5
Reputation: 62636
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.
Upvotes: 2