Argthtjtr

Just curious: with C++17 and later we can use an auto placeholder for non-type template parameters:

template<typename A, auto B>

class C {

public:

    A foo() { return B; }

};

But can we pass instead of auto the template type parameter A?

example.cpp

template<typename A, A B>

class C {

public:

    A foo() { return B; }

};



int main()

{

    C<int, 5> c;

    std::cout << c.foo() << std::endl;



    return 0;

}

Well in practice we can, and clang with -std=c++11 allows to do that.

$ g++ -std=c++11 example.cpp

$ ./a.out

5

But what about the Standard? I didn't find any explicit rule for this.
Thanks!

But can we pass instead of "auto" template type parameter defined at left?

Sure. It's the old way.

But the auto way permit to avoid to pass the type A.

In C++17 you can write

template <auto B>

class C {

public:

    auto foo() { return B; }

};

so there is no needs to pass the type A.

I don't know a way to do the same in C++11/C++14.

I mean: if you want, in C++11/C++14 pass a value to a template... if the type is fixed, there is no problem

template <int I>

but the type itself can be different, in C++11/C++14 you have to pass first the type, then the value as in your example.cpp

template <typename T, T A>

The problem, in the old way, is that if you want to call that type of template, you have to be redundant and write something as

C<decltype(x), x>  some_C_variable;

and the only ways I know to avoid this redundancies is pass through a make function (something as make_tuple()) or a C-style macro.

In C++17 you can simply write

C<x>   some_C_variable

and, inside the C template class/struct, the type of x can be obtained from decltype(B).

For C++17 standard references...

First of all, the auto is defined a "placeolder"

From 10.1.7.4 (The auto specifier), point 1

The auto and decltype(auto) type-specifiers are used to designate a placeholder type that will be replaced later by deduction from an initializer.

This was true also in C++11/C++14

But C++17 standard add, for "Template parameters" (17.1), in point 4 enumeration, the new point 4.6

4. A non-type template-parameter shall have one of the following (optionally cv-qualified) types:

[...]

(4.6) a type that contains a placeholder type (10.1.7.4)

Template parameters can be divided in:

1. Type parameters

2. Non type parameters

3. Template template parameters

In your case, the statement A B is a non type template parameter whose type is the template type parameter A.

Keep in mind, in your example you do:

return B;

the above statement would fail on types A that are not copy constructible (This is easily expressed by a concept).

Keep in mind(and as was pointed in the comments), C++17 mandates copy elision.