Re: shared_ptr; derived classes; ambiguitity in overloaded functions
Fokko Beekhof wrote:
Hello all,
please consider the following code:
--------------------------------------------------
#include <tr1/memory>
struct BaseA
{
int x;
};
struct BaseB
{
double x;
};
struct DerivA : public BaseA
{
int y;
};
struct DerivB : public BaseB
{
double y;
};
struct S
{
S(std::tr1::shared_ptr<BaseA> pa_) : pa(pa_) {}
S(std::tr1::shared_ptr<BaseB> pb_) : pb(pb_) {}
std::tr1::shared_ptr<BaseA> pa;
std::tr1::shared_ptr<BaseB> pb;
};
int main()
{
// S s(std::tr1::shared_ptr<BaseA>(new DerivA()) ); // works
// S s( new DerivA() ); // Doesn't work, SP constructor is explicit
S s(std::tr1::shared_ptr<DerivA>(new DerivA()) ); // breaks
return 0;
}
--------------------------------------------------
As you see, there are 3 possibilities in main(). #1 uses
std::tr1::shared_ptr<BaseA>, which works, like I would expect it to.
Then, #2 tries to pass a pointer where a shared_ptr is required, so that
doesn't work - I can understand that too, pointers cannot be implicitly
converted to shared_ptr.
My question:
Option #3, using a std::tr1::shared_ptr<DerivA>, does NOT work:
$ g++ -Wall ambiguousSP.cpp
ambiguousSP.cpp: In function ?int main()?:
ambiguousSP.cpp:37: error: call of overloaded
?S(std::tr1::shared_ptr<DerivA>)? is ambiguous
ambiguousSP.cpp:28: note: candidates are:
S::S(std::tr1::shared_ptr<BaseB>) ambiguousSP.cpp:27: note:
S:S(std::tr1::shared_ptr<BaseA>)
If I use "regular" pointers, it would works fine:
---------------------------------------------
struct BaseA
{
int x;
};
struct BaseB
{
double x;
};
struct DerivA : public BaseA
{
int y;
};
struct DerivB : public BaseB
{
double y;
};
struct S
{
S(BaseA * pa_) : pa(pa_), pb(0) {}
S(BaseB * pb_) : pa(0), pb(pb_) {}
BaseA * pa;
BaseB * pb;
};
int main()
{
S s(new DerivA()); // works fine
delete s.pa;
return 0;
}
---------------------------------------------
Could anyone comment on this ? Should option #3 with shared_ptr<DerivA>
work or not ?
According to the technical report TR1, it should not work.
Conceptually, I believe so. I can understand that the platform could
have some trouble with it, if it doesn't recognize that the template
parameter DerivA in shared_ptr<DerivA> is derived of BaseA.
You have a point, but it would require changing the specs of shared_ptr.
To simplify the exposition, let us consider the following class:
template < typename T >
struct pointer_to {
T * the_ptr;
pointer_to ( T * ptr )
: the_ptr ( ptr )
{}
T & operator* ( void ) const {
return ( *the_ptr );
}
T * operator-> ( void ) const {
return ( the_ptr );
}
template < typename D >
pointer_to ( pointer_to<D> const & d_pointer )
: the_ptr ( d_pointer.the_ptr )
{}
};
There is a conversion operator that allows to copy construct a pointer_to<T>
from a pointer to any derived class. Attempts to copy construct from
non-derived classes will fail when the compiler encounters the body of the
conversion operator. With this setup, the following will be ambiguous:
struct X {};
struct XD : public X {};
struct Y {};
struct YD : public Y {};
void f ( pointer_to<X> xp ) {}
void f ( pointer_to<Y> yp ) {}
int main ( void ) {
pointer_to<YD> ydp ( new YD );
f( ydp );
}
The reason is that the compiler sees two possible conversions and it is not
supposed to check whether only one of them can be compiled cleanly.
Now, there is a way to guide the compiler in these issues. But it requires
some serious scaffolding:
struct yes_type { char dummy; };
struct no_type { yes_type a; yes_type b; };
template < typename From, typename To >
class is_convertible {
static
From* dummy ( void );
static
yes_type check ( To );
static
no_type check ( ... );
public:
static bool const value =
sizeof( check( *dummy() ) ) == sizeof( yes_type );
}; // is_convertible
template < bool b, typename T >
struct enable_if;
template < typename T >
struct enable_if<true,T> { typedef T type; };
template < typename T >
struct pointer_to {
T * the_ptr;
pointer_to ( T * ptr )
: the_ptr ( ptr )
{}
T & operator* ( void ) const {
return ( *the_ptr );
}
T * operator-> ( void ) const {
return ( the_ptr );
}
template < typename D >
pointer_to ( pointer_to<D> const & d_pointer,
typename enable_if< is_convertible<D*,T*>::value, void* >::type
p = 0 )
: the_ptr ( d_pointer.the_ptr )
{}
};
With this setup, the above snippet will compile cleanly since the signature
of the conversion operator is enough to tell the compiler that there is
only one possible conversion.
Best
Kai-Uwe Bux