c++ : variadic template and function overloading - c++

Question

see example below live : https://onlinegdb.com/Hkg6iQ3ZNI
#include <iostream>
#include <utility>
#include <type_traits>
class A
{
public:
A(int v=-10):v_(v){}
void print()
{
std::cout << "called A: " << v_ << std::endl;
}
private:
int v_;
};
void f(int v)
{
std::cout << "called f: " << v << std::endl;
}
template<typename T,typename ... Args>
void run(A&& a,
T&& t,
Args&& ... args)
{
a.print();
t(std::forward<Args>(args)...);
}
template<typename T,typename ... Args>
void run(T&& t,
Args&& ... args)
{
run(A(),
std::forward<T>(t),
std::forward<Args>(args)...);
}
int main()
{
int v_function=1;
int v_a = 2;
run(f,v_function);
return 0;
}
The code above compiles, runs and print (as expected):
called A: -10
called f: 1
but if the main function is modified to:
int main()
{
int v_function=1;
int v_a = 2;
run(f,v_function);
// !! added lines !!
A a(v_a);
run(a,f,v_function);
return 0;
}
then compilation fails with error:
main.cpp:30:6: error: no match for call to ‘(A) (void (&)(int), int&)’
t(std::forward(args)...);
~^~~~~~~~~~~~~~~~~~~~~~~~~~~~~
which seems to indicate that even when an instance of A is passed as first argument, the overload function
void(*)(T&&,Args&&...)
is called, and not
void(*)(A&&,T&&,Args&&...)

Answer 1

With
template<typename T,typename ... Args>
void run(A&& a,
T&& t,
Args&& ... args)
a is not a forwarding reference, but an rvalue reference. That means when you do run(a,f,v_function); that function will not be selected because a is an lvalue and those can't be bound to rvalue references. There are two quick ways to fix this. First, use std::move on a like
run(std::move(a),f,v_function);
but this isn't great. a isn't actually moved in the function so you are kind of violating the principle of least surprise.
The second option is to make A in the function a template type so it becomes a forwarding reference and then you can constrain it to be of type A like
template<typename A_, typename T,typename ... Args, std::enable_if_t<std::is_same_v<std::decay_t<A_>, A>, bool> = true>
void run(A_&& a,
T&& t,
Args&& ... args)
{
a.print();
t(std::forward<Args>(args)...);
}

Answer 2

Your code works, if you are calling run with an rvalue.
Playable example here.
As NathanOliver already sad: void run(A&& a, T&& t, Args&& ... args) expects an rvalue reference.
Basic idea of an rvalue reference: You are passing an rvalue to a function (e.g. a string literal). That value will be copied to the function. This work is unnecessary. Instead, you are just "moving" the reference to that value, so that it is "owned" by a different part of your program. Move constructors are a good starting point for understanding this problem.