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

While I agree that it's counter-intuitive, the reason is that your copy constructor takes a const Bar& but bar1 is not const.

http://coliru.stacked-crooked.com/a/2622b4871d6407da

Since the universal reference can bind anything it is chosen over the more restrictive constructor with the const requirement.