Re: Throwing constructor for wrong type of objects

From:
"Balog Pal" <pasa@lib.hu>
Newsgroups:
comp.lang.c++
Date:
Tue, 29 Sep 2009 18:06:54 +0200
Message-ID:
<h9tb55$2b3$1@news.ett.com.ua>
"Vladimir Jovic" <vladaspams@gmail.com>

Template example update:
/////////////////////////////
template< typename T >
class B
{
    public:
        virtual ~B()
        {
        }
};
template< typename T >
class C : public B< T >
{
    public:
        virtual ~C()
        {
        }
};
class A
{
    public:
        A(const B< int > &)
        {
        }
        template< typename T >
        A(T)
        {
            throw 111;
        }
        ~A()
        {
        }
};
int main()
{
    A a( C<int>() ); // doesn't throw


this line is a declaration, not creating an onject 'a'. Try

     A a(( C<int>() )); // throws

    B<int> *b = new C<int>;
    A e(*b); // doesn't throw

    C< int > c;
    A d(c); // throws
}
////////////////

This example really demonstrates a problem I am facing.

So, is there a way to create such a constructor that is going a take a
reference to a specific base type, and to throw for all other types?


With an overload set template function is chosed if it creates a better
match -- in this case tepmlate is instantiated for T = C<int> that matches
perfectly, while the fixed function would need a derived-to-base conversion.

Boost:: has support for type traits and conditionals in templates, I guess
you could create a template that checks whether its T is derived from B<T>
using the tools there and make it work, while the rest throw. (Btw what is
the point of throw? Why not place a compile-time error/ STATIC_ASSERT
there?)

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