'Use of class template requires template arguments' error with default specified non-type template parameter value

Question

Compiling this:

#include <iostream>
#include <memory>

template <auto val = 42>
struct A
{
    A()
    {
        std::cerr << val << "\n";
    }
};


int main(int argc, const char* argv[])
{
    std::shared_ptr<A> a_ptr {new A {}};

    return 0;
}

gives error: use of class template 'A' requires template arguments . Although I provide a default value for non-type template parameter and expect it to be seen and used by compiler somehow. What do I miss here?

Thanks.

Answer 1

It's still a template, and it still requires < and > :

std::shared_ptr<A<>> a_ptr {new A<> {}};

Think about this. If you have a function with a default parameter:

int foo(int baz=42);

Do you think you can simply call it without the parenthesis?

int foobar=foo;

Of course not, this won't work. You still need the parenthesis:

int foobar=foo();

Same thing with templates.

Answer 2

A doesn't name a class , it names a class template.

You need to use A<> to name the instantiation of the template that uses the default value:

std::shared_ptr<A<>> a_ptr {new A<> {}};
//              ^~~ here and    ^~~ here

If you are in c++17 or above, as is implied by the auto template parameter, you may remove the <> from A 's construction since this qualifies as CTAD, but it's still needed in the type for shared_ptr

std::shared_ptr<A<>> a_ptr {new A {}};
//              ^~~ only here

Note: This only applies if there is no ambiguity to the type. If a constructor qualifies for CTAD, this will deduce a (possibly different) type than A<>

For example, if A were defined as:

template <typename T = int>
struct A
{
    A(){}
    A(T x){}
};

Then:

std::shared_ptr<A<>> a_ptr {new A {}};

would succeed, but

std::shared_ptr<A<>> a_ptr {new A {0.1}};

would fail, since new A {0.1} deduces A<double> , but A<> names A<int>