CODEWITHSUNDEEP
c++templatespointers
Loay
Loay

Reputation: 601

passing the type of a pointer as a template parameter

I'll explain in an example:

template<typename T>
class Base{}

class A: public Base<A>{}

class foo(){
public:
   template<typename T>
   bool is_derived(){
      /* static check if T is derived from Base<T> */
   }
}

I found this trait to determine whether a class is a base of another.

My question is, how can I send template arguments to is_base_of from T if T is a pointer without specializing the boo template function?

I'd like to do something like this: if T is a pointer then if (is_base_of<Base<*T>,*T>) return true; if T is not a pointer then if (is_base_of<Base<T>,T>) return true;

Upvotes: 0

Views: 1169

Answers (2)

Jan Korous
Jan Korous

Reputation: 666

Basically you have already answered yourself.

C++14

bool is_derived(){
    static_assert( std::is_base_of<Base<T>, std::remove_pointer_t<T>>::value );
}

C++11

bool is_derived(){
    static_assert( std::is_base_of<Base<T>, typename std::remove_pointer<T>::type>::value );
}

C++03 - wrapper over is_base_of you mentioned (using Boost.StaticAssert)

template<class Base, class Derived> struct my_is_base_of : is_base_of<Base, Derived> { };
template<class Base, class Derived> struct my_is_base_of<Base*, Derived*> : is_base_of<Base, Derived> { };

// ...

bool is_derived(){
    BOOST_STATIC_ASSERT( my_is_base_of<Base<T>, T>::value );
}

Upvotes: 1

Jarod42
Jarod42

Reputation: 217085

You may use std::remove_pointer traits:

class foo(){
public:
   template<typename T>
   bool is_derived() const {
      using type = std::remove_pointer_t<T>;
      static_assert(std::is_base_of<Base<type>, type>::value,
                    "type should inherit from Base<type>");
   }
};

Upvotes: 3

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