Re: automatic conversion

Alberto Ganesh Barbati <>
Tue, 29 Jan 2008 19:20:17 CST
Mathieu Lacage ha scritto:


The code below fails to compile with my g++ 4.1.1:

class Base
   virtual ~Base () = 0;
Base::~Base () {}
class Derived : public Base
   virtual ~Derived () {}

void Do (const Base &base)

class A
   operator Derived () const {return Derived ();}

int main (int argc, char *argv[])
   Derived derived = A ();
   const Base &base = A ();
   Do (A ());

   return 0;

with the following error: In function 'int main(int, char**)': error: cannot allocate an object of abstract type 'Base' note: because the following virtual functions are pure within
'Base': note: virtual Base::~Base()

Which points to the line which calls function Do (A ()). I have to confess
that I am a bit fumbled by this: I would have expected the line above
const Base &base = A (); to also fail or to have both lines not fail.

Looks like a bug in g++ to me. According to my interpretation of 8.5.3/5
  both lines are ill-formed, because this rule applies:

"a temporary of type ?cv1 T1? is created and initialized from the
initializer expression using the rules for a non-reference copy
initialization (8.5)"

(where T1 is Base). Of course you cannot create a temporary of type Base
because it's an abstract class. The funny thing is that g++ 3.4.5 agrees
with my interpretation and chokes on both lines.

Anyhow, the question then is to know whether there is a way to make this
code to compile and run to allow the compiler a way to guess the right
type conversion when calling Do (const Base &). I could, of course, make A
derive from Base but this (A) class (in my code) is really not expected to
be polymorphic so, it kind of defeats the whole purpose of the exercise.

Any hints ?

The main problem here is that the A is only convertible to an rvalue of
type Derived. If it could be made convertible to an *lvalue* of type
Derived, then a different rule would apply, i.e.: the reference
would bound directly to the lvalue result of the conversion.

For example with this definition of A:

   class A
      operator const Derived& () const
        static Derived d;
        return d;

both lines compile correctly.



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