I wanted to have type traits which will help me to get the type of the class from a member function pointer. I looked into this answer and found my half way to the aim.
It looks like this:
#include <iostream>
// example class
struct MyClass {
void funct() { std::cout << "funct has been called....\n"; }
};
// traits
template<typename Class> struct get_class{};
template<typename ReType, typename Class, typename... Args>
struct get_class<ReType(Class::*)(Args...)>
{
using type = Class;
};
template<typename Type> using get_class_t = typename get_class<Type>::type;
int main()
{
get_class_t<decltype(&MyClass::funct)> myObj;
//^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ---> this is a lot of typing
myObj.funct();
return 0;
}
But, as shown in the code I need every time to write get_class_t<decltype(&MyClass::funct)>
or in the case of
auto ptr = &MyClass::funct;
get_class_t<decltype(ptr)> myObj;
// ^^^^^^^^^^^^^^
which is a lot of decltype()
ing. I would like to write instead
class_t<ptr> obj;
or
class_t<&MyClass::funct> myObj;
which is more convenient.
I did the following function, which will return a resulting object of the class and maybe I could do, want I wanted to.
template<typename Type>
auto helper_function(Type ptr)->get_class_t<Type>
{
return get_class_t<Type>{};
}
template<typename Type>
using class_t = /* decltype(helper_function(Type ptr));*/
// ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ // what could be here?
I do not know how to complete this. My goal is to extend the traits in such a way that I could create an object like
auto ptr = &MyClass::funct;
class_t<ptr> myObj;
// or
class_t<&MyClass::funct> myObj;
Is there any other way to do this? or should I have to stick with decltype()
ing?
As I have tagged, I would like to see whether it possible with C++11?
auto is a keyword in C++11 and later that is used for automatic type deduction. The decltype type specifier yields the type of a specified expression. Unlike auto that deduces types based on values being assigned to the variable, decltype deduces the type from an expression passed to it.
The decltype type specifier yields the type of a specified expression. The decltype type specifier, together with the auto keyword, is useful primarily to developers who write template libraries. Use auto and decltype to declare a template function whose return type depends on the types of its template arguments.
Decltype keyword in C++ Decltype stands for declared type of an entity or the type of an expression. It lets you extract the type from the variable so decltype is sort of an operator that evaluates the type of passed expression. SYNTAX : decltype( expression )
decltype is a compile time evaluation (like sizeof ), and so can only use the static type.
(Answer archived for future visitors; this solution requires C++17!)
You're really close!
The trick is auto
template arguments, and the fact that pointers-to-members can be used as template arguments, like so:
template <auto thing>
using class_t = get_class_t<decltype(thing)>;
int main()
{
class_t<&MyClass::funct> myObj;
myObj.funct();
}
Of course if you can write this then you already know the type so you'd just write MyClass
, so that's not very useful.
Sadly you won't be able to make it accept ptr
as a template argument, though; you're stuck with get_class_t
for that:
int main()
{
auto ptr = &MyClass::funct;
get_class_t<decltype(ptr)> myObj;
myObj.funct();
}
In the latter case, a nice type alias can help you a bit:
auto ptr = &MyClass::funct;
using ClassType = get_class_t<decltype(ptr)>;
ClassType myObj;
myObj.funct();
Personally I think this level of verbosity is pretty reasonable.
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