将（部分）模板化模板函数作为std :: function（或函数指针）传递

Question

#include <vector>
#include <functional>

template<class F>
class Foo
{
public:
    template <class T>
    void std_function(std::function<F(std::vector<T>)> functor)
    {
        /* something */
    }

    template <class T>
    void func_ptr(F (*funtor)(std::vector<T>))
    {
        /* something else */
    }
};

template<class T, class F>
F bar(std::vector<T>)
{
    return F();
}

int main()
{
    Foo<double> test;
    std::function<double(std::vector<int>)> barz = bar<int, double>;

    test.std_function(bar<int, double>); //error 1
    test.std_function(barz); //OK 1
    test.func_ptr(bar<int, double>); //OK 2

    test.std_function(bar<int>); //error 2::1
    test.func_ptr(bar<int>); //error 2::2

    return 0;
}

Question 1.

Line error 1 : I am trying to pass explicitly instantiated template function( bar<int, double> ) as std::function , but It is not legal.

Line OK 1 : If I wrap bar<int, double> into std::function<double(std::vector<int>)> and pass wrapped functor, it is legal now.

Line OK 2 : If I pass bar<int, double> through Foo::func_ptr , which gets function pointer as argument instead of std::function , it is also legal.

I want to make the Line error 1 legal. As in the Line OK 2 , it is possible to pass bar<int, double> without any wrapper(unlike Line OK 1 ) and keep same form. But, the parameter type is different. I want to pass as std::function , not function pointer.

Question 2.

Line error 2::1 and 2::2 : What I am trying to achieve here is, I want class Foo to deduce return type of bar as its class template type F (for the code above, F is double ). So I can just pass as bar<int> , not bar<int, double> .

But It seems to fail deduction, because even if I pass bar<int> through Foo::func_ptr , it still produces error. How can I make this code work as my intention?

Answer 1

For error 1, what's happening is that the compiler is attempting to substitute for T in std::function , but it cannot because ultimately a function pointer and a std::function are different types, and there's no conversion defined for a function pointer to std::function

This worked:

std::function<double(std::vector<int>)> barz = bar<int, double>

Because std::function was cleverly written with type-erasure to have a constructor that can accept any callable that is convertible to the type it needs. Note that this isn't the same as type-deduction in the above error because here we are already specifying the template arguments for std::function .

Note that we can do a little work to get Foo::std_function to work properly. First change its signature to take a forwarding reference:

template <class T>
void std_function(T&& functor){/*I'll talk about this in a bit*\}

Then we can construct our std::function internally (What you're looking to pass into it, I don't know) by using some helper structs to determine its type. For a function pointer we can do the following:

// base
template<class... T>
struct function_type_impl;

// specialization for function ptrs and static class fns
template<class Ret, class... Args>
struct function_type_impl<Ret(*)(Args...)>
{
   using type = std::function<Ret(Args...)>;
};

// type alias so we don't need to keep typing typename ... ::type
template<class... T>
using function_type = typename function_type_impl<std::decay_t<T>...>::type;

and then we can modify our std_function signature:

template <class T>
void std_function(T&& functor)
{
    function_type<T> myFunction = std::forward<T>(functor);
    // do something with our std::function
}

Then you can call it as

test.std_function(&::bar<int, double>);

However, if we want to be more complete, and accept functors, lambdas, and even other std::functions , we can add more specializations:

namespace detail
{
template<class... T>
struct function_type_impl;

template<class Callable>
struct function_type_impl<Callable>
{
    using type = typename function_type_impl<decltype(&Callable::operator())>::type;
};

template<class C, class Ret, class... Args>
struct function_type_impl<Ret(C::*)(Args...) const>
{
    using type = std::function<Ret(Args...)>;
};

template<class Ret, class... Args>
struct function_type_impl<Ret(*)(Args...)>
{
    using type = std::function<Ret(Args...)>;
};

template<class... T>
using function_type = typename function_type_impl<std::decay_t<T>...>::type;
}// detail namespace

And now the following will work, too:

struct MyFunctor
{
    double operator()(std::vector<int>) const
    {
        return 42;
    }
};

struct MyFunctor2
{
    static double foo(std::vector<int>)
    {
        return 42;
    }
};

int main()
{
    Foo<double> test;
    std::function<double(std::vector<int>)> barz = bar<int, double>;
    test.std_function(&::bar<int, double>);
    test.std_function(barz);
    test.std_function([](std::vector<int>)->double{return 42;});
    test.std_function(MyFunctor{});
    test.std_function(MyFunctor2::foo);
}

Live Demo

For errors 2::1 and 2::2, the issue is simpler; functions do not exist at all until they're completely instantiated. That is, you cannot create a function pointer to a partially templated function. You have to specify all the template arguments when trying to get a function pointer. Since you have already specified the return type, you can allow the compiler to instantiate the rest of the pointer for you if you explicitly tell func_ptr what to deduce for T :

test.func_ptr<int>(bar);