Template specialization of inherited member function - c++

Question

We would like to specialize member functions of a base class. However, it does not compile. Does anybody know of any alternative that does compile?
Here is an example
struct Base
{
template<typename T>
void Foo()
{
throw "Foo() is not defined for this type";
}
};
struct Derived : public Base
{
template<>
void Foo<int>() { cout << "Foo<int>()" << endl; } // compile error (cannot specialize members from a base class)
template<>
void Foo<double>() { cout << "Foo<double>()" << endl; } // compile error (cannot specialize members from a base class)
};

Answer 1

Eventually, we solved it using overloading.
Here is how the base class looks like
struct Base
{
template<typename T>
class OfType {}
template<typename T>
void Foo(OfType<T>) { static_assert(false, "Foo is not implemented for this type. Please look in the compiler error for more details."); }
};
struct Derived : public Base
{
using Base::Foo;
void Foo(OfType<int>) { // here comes logic for ints }
void Foo(OfType<double>) { // here comes logic for doubles }
};
Here is an example of client code that uses Foo()
template<typename S>
class ClassThatUsesFoo
{
private: S s;
template<typename T>
void Bar(T item)
{
s.Foo(Base::OfType<T>()); // this is the code that uses Foo
DoSomeStuffWithItem(item);
}
};
void main()
{
ClassThatUsesFoo<Derived> baz;
baz.Bar(12); // this will internally use Foo for ints
baz.Bar(12.0); // this will use Foo for doubles
baz.Bar("hello world"); // this will give a verbose compile error
}

Answer 2

This will compile, except for the call to Foo<char>():
#include <iostream>
#include <string>
using namespace std;
struct Base
{
template<typename T>
void Foo()
{
throw "Foo() is not defined for this type";
}
};
struct Derived : public Base
{
template<typename T> void Foo();
};
template<>
void Derived::Foo<int>() { cout << "Foo<int>()" << endl; }
template<>
void Derived::Foo<double>() { cout << "Foo<double>()" << endl; }
int main()
{
Derived derived;
// this is client code
derived.Foo<int>();
derived.Foo<double>();
derived.Foo<char>(); // this throws
}
If you want the call to Foo<char>() -- or any type not specifically specialized by you -- then this works. If you want a non-specialized implementation that works for all types, then you need to add a non-specialized implementation of Foo() as well:
template<typename T>
void Derived::Foo() { cout << "generic" << endl; }

Answer 3

In response to the discussion with Alex (see comments of the answer of John Dibling), this is what I meant (SSCCE):
#include <iostream>
using namespace std;
struct Base
{
template<typename T>
void Foo()
{
//static_assert(false, "Foo() is not defined for this type");
throw "Foo() is not defined for this type";
}
};
// you can add as many specializations in Base as you like
template <>
void Base::Foo<char>() { cout << "Base::Foo<char>()" << endl; }
struct Derived : public Base
{
// just provide a default implementation of Derived::Foo
// that redirects the call to the hidden Base::Foo
template < typename T >
void Foo()
{ Base::Foo<T>(); }
};
// the specializations for Derived
template<>
void Derived::Foo<int>() { cout << "Foo<int>()" << endl; }
template<>
void Derived::Foo<double>() { cout << "Foo<double>()" << endl; }
struct Derived_wo_specialization : public Base
{
/* nothing */
};
int main()
{
Derived d;
d.Foo<char>();
d.Foo<double>();
Derived_wo_specialization dws;
dws.Foo<char>();
dws.Foo<double>();
}