Conditional typing based on pointer or value type

Question

I wonder if I can do some conditional typing based on if a template parameter is a pointer or not?

So for instance I want the get method below to return T itself if it is a pointer(i.e. T*). Or T* if is not a pointer (i.e. T).

template<typename T>
class MyContainer {
    T get(); // If T is a pointer
    T* get(); // If T is not a pointer
}

@MichaelChourdakis it is almost the same, but it doesn't help me that much. I think it is basing the overload resolution based on the parameter. My `get()` method doesn't have any parameters. — einstein, Jul 14 '19 at 08:21
`I think it is basing the overload resolution based on the parameter.` - no, it is not. — KamilCuk, Jul 14 '19 at 08:24
@MichaelChourdakis you are right it was probably a duplicate. — einstein, Jul 14 '19 at 09:53

score 2 · Answer 1 · answered Jul 14 '19 at 08:23

2

Just condition if it's a pointer and if not:

#include <type_traits>

template<typename T>
class MyContainer {
    std::conditional_t<std::is_pointer<T>::value, T, T*> get();
};

answered Jul 14 '19 at 08:23

KamilCuk

1 Answers1