C++ template specialization via a base class

Question

I want to be able to make the compiler shout when i call a constructor of foo with a class that is NOT derived from _base*. The current code allows only for foo<_base*> itself. Any easy solution ?

class _base
{
public:
    // ...
};

class _derived: public _base
{
public:
    // ...
};

template <typename T>
class foo
{
public:
    foo ()      { void TEMPLATE_ERROR; }
};

template <> foo<_base*>::foo () 
{
    // this is the only constructor
}

main-code:

foo<_base*>    a;    // should work 
foo<_derived*> b;    // should work (but doesnt)
foo<int*>      c;    // should not work (and infact doesnt)

Answer 1

Without Boost you can use something like the following to determine whether a pointer-to-type can be implicitly cast to another pointer-to-type:

template <class Derived, class Base>
struct IsConvertible
{
    template <class T>
    static char test(T*);

    template <class T>
    static double test(...);

    static const bool value = sizeof(test<Base>(static_cast<Derived*>(0))) == 1;
};

To make it trigger an error at compile-time, you can now use value in an expression that causes an error if it is false, for example typedef a negative-sized array.

template <typename T>
class foo
{
public:
    foo ()
    {
        typedef T assert_at_compile_time[IsConvertible<T, _base>::value ? 1 : -1];
    }
};