Why variadic template constructor matches better than copy constructor?

Question

The following code does not compile:

#include 
#include 

struct Foo
{
    Foo() { std::cout << \"Foo()\" << std::         


        

Answer 1

Another way to avoid the variadic constructor being selected is to supply all forms of the Bar constructor.

It's a little more work, but avoids the complexity of enable_if, if that's important to you:

#include 
#include 

struct Foo
{
    Foo() { std::cout << "Foo()" << std::endl; }
    Foo(int) { std::cout << "Foo(int)" << std::endl; }
};

template 
struct Bar
{
    Foo foo;

    Bar(const Bar&) { std::cout << "Bar(const Bar&)" << std::endl; }
    Bar(Bar&) { std::cout << "Bar(Bar&)" << std::endl; }
    Bar(Bar&&) { std::cout << "Bar(Bar&&)" << std::endl; }

    template 
    Bar(Args&&... args) : foo(std::forward(args)...)
    {
        std::cout << "Bar(Args&&... args)" << std::endl;
    }
};

int main()
{
    Bar bar1{};
    Bar bar2{bar1};
}