Considering the following code:
class MyClass
{
...
};
template <typename Object>
class List
{
public:
void insert(const Object & x)
{
// call when Object is MyClass
}
void insert(const Object & x)
{
// call when Object is MyClass*
}
}
int main()
{
MyClass a;
List<MyClass> lst;
List<MyClass*> plst;
lst.insert(a);
plst.insert(new Myclass);
return 0;
}
How to tell the compiler call different methods based on if the template is a class or a pointer?
How to fix the code above?
You can use a combination of std::is_pointer
and std::enable_if
:
#include <type_traits>
#include <iostream>
class MyClass
{
};
template <typename Object>
class List
{
public:
template<class T=Object>
void insert(T t, typename std::enable_if<std::is_pointer<T>::value >::type* = 0)
{
std::cout << "insert pointer" << std::endl;
}
template<class T=Object>
void insert(T t, typename std::enable_if<!std::is_pointer<T>::value >::type* = 0)
{
std::cout << "insert non-pointer" << std::endl;
}
};
int main()
{
MyClass a;
List<MyClass> lst;
List<MyClass*> plst;
lst.insert(a);
plst.insert(new MyClass());
return 0;
}
Live example: https://ideone.com/CK8Zdo
This will allow you to insert both pointers and non-pointers into a pointer or non-pointer list. If you want to restrict that, you can use this:
#include <type_traits>
#include <iostream>
class MyClass
{
};
template <typename Object>
class List
{
public:
template<class T=Object>
void insert(T t, typename std::enable_if<std::is_same<T,Object>::value&&std::is_pointer<T>::value >::type* = 0)
{
std::cout << "insert pointer" << std::endl;
}
template<class T=Object>
void insert(const T& t, typename std::enable_if<std::is_same<T,Object>::value&&!std::is_pointer<T>::value >::type* = 0)
{
std::cout << "insert non-pointer" << std::endl;
}
};
int main()
{
MyClass a;
List<MyClass> lst;
List<MyClass*> plst;
lst.insert(a);
// plst.insert(a); // compiler error
// lst.insert(new MyClass()); // compiler error
plst.insert(new MyClass());
return 0;
}
Live example: https://ideone.com/3DtBfr
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