CODEWITHSUNDEEP
c++c++11templatestypesstd-function
scravy
scravy

Reputation: 12283

How to deduce `std::function` parameters from actual function?

Given a class

class Foo {
 public:
  std::shared_ptr<const Bar> quux(const std::string&, std::uint32_t);
}

I can declare an std::function that has the same interface:

std::function<std::shared_ptr<const Bar>(const std::string&, std::uint32_t)> baz = ...

Is there a way of compressing that declaration such that the template arguments to std::function are derived from the declaration of that method, something like:

std::function<functype(X::quux)> baz = ...

where functype is an imaginary C++ operator similar to decltype. Is there a way to do this / does have such a capability?

I do see that the method has a slightly different signature actually as it would also take a reference/pointer to the this object; it would be fine for me to derive such a signature too.

Upvotes: 3

Views: 595

Answers (1)

aparpara
aparpara

Reputation: 2201

Yes, you can. Adapting How do I get the argument types of a function pointer in a variadic template class? to your request, we get:

template<typename T> 
struct function_traits;  

template<typename R, typename C, typename ...Args> 
struct function_traits<R(C::*)(Args...)>
{
    using type = std::function<R(Args...)>;
};

class Bar;

class Foo {
 public:
  std::shared_ptr<const Bar> quux(const std::string&, std::uint32_t);
};

int main()
{
   std::cout << std::is_same<
     std::function<std::shared_ptr<const Bar>(const std::string&, std::uint32_t)>,
     function_traits<decltype(&Foo::quux)>::type>::value << std::endl;
}

To make it work with constant methods you will need another specialization:

template<typename R, typename C, typename ...Args> 
struct function_traits<R(C::*)(Args...) const>
{
    using type = std::function<R(Args...)>;
};

But you will get problems with overloaded methods, because in order to resolve overloading you will need to specify the arguments anyway.

Upvotes: 6

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