std::function template with multiple template parameters

Question

When looking for documentation on std::function, I found several pages, that list two implementations of std::function for C++11:

https://en.cppreference.com/w/cpp/utility/functional/function

template< class > class function; /* undefined */ template< class R, class... Args > class function<R(Args...)>;

https://www.tutorialspoint.com/cpp_standard_library/functional.htm,

template <class T> function; // undefined template <class Ret, class... Args> class function<Ret(Args...)>;

https://cplusplus.com/reference/functional/function/

template<class > class function;

and

template< class R, class... Args > class function<R(Args...)>

I tried to use the multi-parameter version of the function<>-template, but my code does not compile, neither with visual c++ 2017, nor with XCode or g++. Here is my sample code:

#include <functional> int main(int argc, char*argv[]) { std::function<void, int> cb; }

All compilers complain on std::function taking only a single template parameter.

Can anybody explain this?

void, int doesn't look compatible with R(Args...). R - return type. Args... - pack of arguments. Specify it as void(int) — Aleksandr Medvedev
– Aleksandr Medvedev, Commented Sep 15, 2022 at 13:44
You basically want to specify the template type as ReturnType(Arg1Type, Arg2Type...). E.g.: godbolt.org/z/s4xoveqsE — rturrado
– rturrado, Commented Sep 15, 2022 at 14:05

Nikos Athanasiou · Accepted Answer · 2022-09-15 13:47:05Z

The template argument list of std::function has a single type, as seen in the base template:

template <class> class function;

What you see in the following:

template <class ReturnType, class... Arguments> class function<ReturnType(Arguments...)> { ... };

is a partial specialization, where the types mentioned are used for pattern matching. Essentially the specialization says:

"The type of std::function, that single type mentioned in the base template, is a callable type with such a ReturnType and such ArgumentTypes

Useless · Accepted Answer · 2022-09-15 13:54:33Z

that list two implementations of std::function for C++11:

No they don't. That isn't what they're showing at all.

template< class > class function; /* undefined */

is the base template, which (as it says), is never defined.

For example, std::function<int> would never make sense, so there is simply no template defined that could match that pattern.

template< class R, class... Args > class function<R(Args...)>;

is a partial specialization.

That is, std::function<T> is only defined at all when T has the form R(Args...), meaning T is the type of a function returning R and taking the arguments Args....

Hence your attempt should be std::function<void(int)> cb; ... exactly as shown in the extensive examples at the bottom of the cppreference page you linked.

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