Template function specialization for template class - c++

Question

Is it possible to write something like this in C++11/14?
#include <iostream>
#include <vector>
template <typename T>
T Get();
template <typename T>
struct Data {
std::vector<T> data;
};
template <>
template <typename T>
Data<T> Get<Data<T>>() {
return Data<T>{{T{}, T{}}};
}
template <>
template <typename T>
std::vector<T> Get<std::vector<T>>() {
return std::vector<T>(3);
}
int main() {
std::cout << Get<Data<int>>().data.size() << std::endl; // expected output is 2
std::cout << Get<std::vector<int>>().size() << std::endl; // expected output is 3
return 0;
}
Overloading won't help in this case, since call to Get<...>() will be ambiguious (see):
template <typename T>
Data<T> Get() {
return Data<T>{{T{}, T{}}};
}
template <typename T>
std::vector<T> Get() {
return std::vector<T>(3);
}
Any direction on how to overcome this are welcome.

Answer 1

There is workaround, that gives you something like this: do not specialize - overload:
#include <iostream>
#include <vector>
#include <string>
using namespace std;
template <typename T>
size_t Get(const T& data)
{
return 444;
}
template <typename T>
struct Data
{
std::vector<T> data;
};
template <typename T>
size_t Get(const Data<T>& data) {
return data.data.size();
}
int main() {
std::cout << Get<>(0) << std::endl; // expected output is 444
std::cout << Get<>(Data<int>{}) << std::endl; // expected output is 0
return 0;
}
Output:
444
0
Note, that size_t Get(const Data<T>& data) is not a specialization - it is completely "different" Get(), that is called for argument of type Data<T> for any T.
Here you can see working sample.
EDIT
I see you changed your question completely. However, I will still try to answer it. There is a standard workaround for lack of partial function specialization - using delegation to structs/classes.
Here is what you need:
#include <iostream>
#include <vector>
using namespace std;
template <typename T>
struct GetImpl;
template <typename T>
struct Data {
std::vector<T> data;
};
template <typename T>
struct GetImpl< Data<T> >
{
static Data<T> Get() {
return Data<T>{ {T{}, T{}} };
};
};
template <typename T>
struct GetImpl< std::vector<T> >
{
static std::vector<T> Get() {
return std::vector<T>(3);
};
};
int main() {
std::cout << GetImpl< Data<int> >::Get().data.size() << std::endl; // expected output is 2
std::cout << GetImpl< std::vector<int> >::Get().size() << std::endl; // expected output is 3
return 0;
}
Output:
2
3
Working sample can be found here.
If you don't like the syntax, you can make it a little bit shorter, by changing static function Get() to function call operator:
template <typename T>
struct Get< Data<T> >
{
Data<T> operator()() {
return Data<T>{ {T{}, T{}} };
};
};
template <typename T>
struct Get< std::vector<T> >
{
std::vector<T> operator()() {
return std::vector<T>(3);
};
};
And then:
Get< Data<int> >()().data.size();
Get< std::vector<int> >()().size();
You have only two extra characters - (). This is the shortest solution I can think of.

Answer 2

As Columbo mentioned in his comment, you should apply the standard workaround for lack of partial specialization support for functions: delegation to a partially specialized class:
template <typename T>
struct GetImpl;
template <typename T>
T Get() { return GetImpl<T>::Do(); }
and now use partial specialization on struct GetImpl<T> { static T Do(); } instead of Get<T>()

Answer 3

But it would be impossible for compiler to distinguish Get<Data<int>> from Get<Data<Data<int>>>.
It's not impossible. If that's something you need to do, we can add separate overloads:
template <typename T>
size_t Get(const Data<T>& data);
template <typename T>
size_t Get(const Data<Data<T>>& data); // preferred for Data<Data<int>>
Or if what you want is to only overload for the non-nested case, we can add a type trait and use SFINAE:
template <typename T> struct is_data : std::false_type { };
template <typename T> struct is_data<Data<T>> : std::true_type { };
template <typename T>
enable_if_t<!is_data<T>::value, size_t>
Get(const Data<T>& data);
That way, the call with Data<Data<int>> would call the generic Get(const T&). Or, if you want that case to not compile at all:
template <typename T>
size_t Get(const Data<T>& data) {
static_assert(!is_data<T>::value, "disallowed");
...
}
So overloading gives you lots of options. Specialization gives you none, since it's disallowed anyway.

Answer 4

Following delegation to the struct's way you can implement more general approach: you can use structs to check the container type and inner type like this:
#include <iostream>
#include <vector>
template <typename T>
struct Data {
std::vector<T> data;
};
template <template <typename...> class Container, typename>
struct get_inner;
template <template <typename...> class Container, typename T>
struct get_inner<Container, Container<T>>
{
typedef T type;
};
template <typename T, typename U = typename get_inner<Data, T>::type>
Data<U> Get() {
return Data<U>{ {U{}, U{}} };
}
template <typename T, typename U = typename get_inner<std::vector, T>::type>
std::vector<U> Get() {
return std::vector<U>(3);
}
int main() {
std::cout << Get<Data<int>>().data.size() << std::endl; // expected output is 2
std::cout << Get<std::vector<int>>().size() << std::endl; // expected output is 3
return 0;
}
http://coliru.stacked-crooked.com/a/90b55767911eff0e