'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.

Yeah, it works fine with A<>, but I thought there is some way around A<>. — psb
– psb, Commented Nov 26, 2020 at 18:28
You can use A() :D. Jokes aside, you won't get around this for pure type inference (i.e. without an instance). That's why A a; will be fine, but std::shared_ptr<A<>> ptr; requires more. — Cedric
– Cedric, Commented Nov 26, 2020 at 18:30

Davis Herring · Accepted Answer · 2023-09-22 13:36:16Z

5

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 parentheses?

int bar=foo;

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

int bar=foo();

Same thing with templates.

edited Sep 22, 2023 at 13:36

Davis Herring

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answered Nov 26, 2020 at 18:27

Sam Varshavchik

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2 Comments

Bitwize Over a year ago

Huh, I've used templates for years and never made the mental connection between function call syntax with default arguments and template instantiation with default arguments. This is a good answer

Catskul Over a year ago

"Do you think you simply call it witout the parenthesis? ...Of course not..." Except that sometimes you can use templates without <>, so the "Of course not" is not appropriate here.

Bitwize · Accepted Answer · 2020-11-26 18:41:31Z

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>

edited Nov 26, 2020 at 18:41

answered Nov 26, 2020 at 18:27

Bitwize

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3 Comments

psb Over a year ago

looks like there is should be another A<>, like std::shared_ptr<A<>> a_ptr {new A<> {}};

Cedric Over a year ago

@Human-Compiler : I believe you need the A<> in the shared_ptr, not the new.

Bitwize Over a year ago

Whoops, thanks for the catch -- tried typing that quickly and was still the second answer. I've added some additional notes to indicate this. Also A {} only works since it qualifies for CTAD, but in more complex expressions can produce an ambiguity that is different from A<>

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