template template parameter of unknown type

Question

Trying to extract template parameter value in the following code:

template<std::size_t SIZE>
class Foo {};

template <template<std::size_t> class T, std::size_t K>
auto extractSize(const T<K>&) {
    return K;
}

int main() {
    Foo<6> f1;
    Foo<13> f2;
    std::cout << extractSize(f1) << std::endl;
    std::cout << extractSize(f2) << std::endl;
}

(As an answer for the question: Extract C++ template parameters).

However, is there a way to do it without knowing the type of the template parameter. Something like (code below doesn't compile...):

template <template<class SIZE_TYPE> class T, SIZE_TYPE K>
auto extractSize(const T<K>&) {
    return K;
}

The compilation error on the above, is:

error: unknown type name 'SIZE_TYPE'
template <template<class SIZE_TYPE> class T, SIZE_TYPE K>
                                             ^

Did you try to put '''typename SIZE_TYPE''' at the beggining of the list? — Robert Andrzejuk, Feb 04 '18 at 18:04
@RobertAndrzejuk Adding typename SIZE_TYPE [fails with C++14](http://coliru.stacked-crooked.com/a/8cfb21497d98525b). It does work [with C++17](http://coliru.stacked-crooked.com/a/368d3668d140ea71) - but if we are in C++17, the solution with `auto` proposed by Rakete1111 would be simpler. — Amir Kirsh, Feb 08 '22 at 13:41

score 8 · Accepted Answer · answered Feb 04 '18 at 14:03

8

auto to the rescue!

template <template<auto> class T, auto K>
auto extractSize(const T<K>&) {
    return K;
}

That way the type of the value you passed in as template parameter is automatically inferred.

answered Feb 04 '18 at 14:03

Rakete1111

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Doesn't that depend on `Foo` accepting an `auto` non-type parameter, and not a `std::size_t`? I'm not clear on the deduction rules here – StoryTeller - Unslander Monica Feb 04 '18 at 14:04
@StoryTeller Nope, it doesn't :) – Rakete1111 Feb 04 '18 at 14:05
1

Very nice indeed :) – StoryTeller - Unslander Monica Feb 04 '18 at 14:06
This is neat, didn't even think this would be possible! – Samaursa Feb 04 '18 at 14:07
3

Cool. Works well [for C++17](https://stackoverflow.com/questions/38026884/advantages-of-auto-in-template-parameters-in-c17), not for C++14. Any suggestion for C++11/14? – Amir Kirsh Feb 04 '18 at 14:08
3

@AmirKirsh `auto` template parameter is introduced in C++17, no such solution for C++14. – llllllllll Feb 04 '18 at 14:09
@AmirKirsh I can't find one. Maybe someone else can. I'll be still trying though. – Rakete1111 Feb 04 '18 at 14:19
1

There isn't a way in pre-C++17, assuming you are using a standard compliant compiler and not typing from memory (*smh*). My own attempt fails at https://timsong-cpp.github.io/cppwp/n4140/temp#deduct.type-5.3 – StoryTeller - Unslander Monica Feb 04 '18 at 14:28
@StoryTeller the compiler in one's head tends to be permissive :-) It seems that your answer closes it. Applause for C++17 auto template parameter! – Amir Kirsh Feb 04 '18 at 14:41

template template parameter of unknown type

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