Re: Comparison operator overload via base classes.
* usenet only_tech_talk:
Hello,
Please consider :
struct A{};
template<typename T> struct B{};
struct C : B< C >{};
struct D : B< D >{};
template<typename T,typename R>
bool operator==(const B<T>&,const R&){}
template<typename T,typename L>
bool operator==(const L&,const B<T>&){}
One would hope this is all that is needed:
int main (int argc, char * const argv[]) {
A a; C c1,c2; D d1,d2;
c1 == a;
c1 == c2;
c1 == d2;
d1 == d2;
d2 == d1;
return 0;
}
but of course not (ambiguous overload). So in a second round of naive
hope, one tries:
template<typename T,typename T1>
bool operator==(const B<T>&,const B<T1>&){}
before accepting the ugly truth:
bool operator==(const C&,const C&){}
bool operator==(const D&,const D&){}
bool operator==(const C&,const D&){}
bool operator==(const D&,const C&){}
Is this the end of the road, or is there something more scalable?
<code>
#include <stdio.h>
void say( char const s[] ) { printf( "%s\n", s ); }
struct A{};
template< typename T > struct B{};
struct C: B< C >{};
struct D: B< D >{};
template< typename T, typename T1 >
void operator==( B< T > const&, B< T1 > const& ) { say( "BB" ); }
template< typename T >
void operator==( A const&, B< T > const& ) { say( "AB" ); }
template< typename T >
void operator==( B< T > const& b, A const& a )
{ return (a == b); }
int main()
{
A a; C c1,c2; D d1,d2;
c1 == a; // AB
c1 == c2; // BB
c1 == d2; // BB
d1 == d2; // BB
d2 == d1; // BB
}
</code>
Cheers & hth.,
- Alf
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