How can I prevent a variadic constructor from being preferred to the copy constructor?

Question

I have a template \'Foo\', which owns a T, and I\'d like it to have a variadic constructor that forwards its arguments to T\'s constructor:

template

        

Answer 1

You can use some ugly SFINAE with std::enable_if, but I'm not sure it is better than your initial solution (in fact, I'm pretty sure it's worse!):

#include 
#include 

// helper that was not included in C++11
template using disable_if = std::enable_if;

template
struct Foo {

    Foo() = default;
    Foo(const Foo &) = default;

    template::type,
                Foo
            >::value
        >::type
    >
    Foo(Arg&& arg, Args&&... args)
        : t(std::forward(arg), std::forward(args)...) {}

    T t;
};

int main(int argc, char* argv[]) {
    Foo> x(new int(42));
    decltype(x) copy_of_x(x);
    decltype(x) copy_of_temp(Foo>(new int));
    return 0;
}