Compile time type determination in C++

Question

A coworker recently showed me some code that he found online. It appears to allow compile time determination of whether a type has an "is a" relationship with another type. I think this is totally awesome, but I have to admit that I'm clueless as to how this actually works. Can anyone explain this to me?

template<typename BaseT, typename DerivedT>
inline bool isRelated(const DerivedT&)
{
    DerivedT derived();
    char test(const BaseT&); // sizeof(test()) == sizeof(char)
    char (&test(...))[2];    // sizeof(test()) == sizeof(char[2])
    struct conversion 
    { 
        enum { exists = (sizeof(test(derived())) == sizeof(char)) }; 
    };
    return conversion::exists;
} 

Once this function is defined, you can use it like this:

#include <iostream>

class base {};
class derived : public base {};
class unrelated {};

int main()
{
    base b;
    derived d;
    unrelated u;

    if( isRelated<base>( b ) )
        std::cout << "b is related to base" << std::endl;

    if( isRelated<base>( d ) )
        std::cout << "d is related to base" << std::endl;

    if( !isRelated<base>( u ) )
        std::cout << "u is not related to base" << std::endl;
} 

Answer 1

It declares two overloaded functions named test, one taking a Base and one taking anything (...), and returning different types.

It then calls the function with a Derived and checks the size of its return type to see which overload is called. (It actually calls the function with the return value of a function that returns Derived, to avoid using memory)

Because enums are compile-time constants, all of this is done within the type system at compile-time. Since the functions don't end up getting called at runtime, it doesn't matter that they have no bodies.