CODEWITHSUNDEEP
c++templatesc++11decltype
Adrian
Adrian

Reputation: 10911

How do I alias a template specialization?

I have a type T that has a member function fn with a return type RT. I have a template that takes T and RT as parameters. I'd like to alias this template class so that my code isn't so ugly and hard to read. How would I accomplish that?

template <typename X, typename Y>
struct I
{};

struct T
{
  int& fn(int);
};

So I want to alias this or something like this type so I can write a function like this:

template <typename C>
I< typename std::remove_reference<std::decltype(std::declval(C).fn(0))>::type, C> fn(C& c)
{
  return I< typename std::remove_reference<std::decltype(std::declval(C).fn(0))>::type, C>();
}

But with less mess. I've not used std::decltype before, so I'm not even sure if I'm using it right as I was getting errors.

I was thinking about using a function like that and doing a decltype on it, but I was having some difficulty with it and I'd like it to look cleaner too.

Upvotes: 0

Views: 545

Answers (1)

Piotr Skotnicki
Piotr Skotnicki

Reputation: 48457

decltype() is a built-in operator, not a function from the std namespace like std::declval<T>().

If you want to shorten the syntax with an alias template, you can declare one as shown below:

#include <type_traits>    

template <typename C>
using IT = I<C, typename std::remove_reference<decltype(std::declval<C&>().fn(0))>::type>;

template <typename C>
IT<C> fn(C& c)
{
  return IT<C>(&c, c.fn(1));
}

DEMO

Since the above code fails to compile successfully in VC++ for an unknown reason, you can instead create an alias template that queries the result type of a helper function that you already have:

template <typename C>
I<C, typename std::remove_reference<decltype(std::declval<C&>().fn(0))>::type> test();

template <typename C>
using IT = decltype(test<C>());

template <typename C>
IT<C> fn(C& c)
{
  return IT<C>(&c, c.fn(1));
}

DEMO 2

Upvotes: 1

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