Implicit constructor available for all types derived from Base excepted the current type?
- by Vincent
The following code sum up my problem :
template<class Parameter>
class Base {};
template<class Parameter1, class Parameter2, class Parameter>
class Derived1 : public Base<Parameter>
{ };
template<class Parameter1, class Parameter2, class Parameter>
class Derived2 : public Base<Parameter>
{
public :
// Copy constructor
Derived2(const Derived2& x);
// An EXPLICIT constructor that does a special conversion for a Derived2
// with other template parameters
template<class OtherParameter1, class OtherParameter2, class OtherParameter>
explicit Derived2(
const Derived2<OtherParameter1, OtherParameter2, OtherParameter>& x
);
// Now the problem : I want an IMPLICIT constructor that will work for every
// type derived from Base EXCEPT
// Derived2<OtherParameter1, OtherParameter2, OtherParameter>
template<class Type, class = typename std::enable_if</* SOMETHING */>::type>
Derived2(const Type& x);
};
How to restrict an implicit constructor to all classes derived from the parent class excepted the current class whatever its template parameters, considering that I already have an explicit constructor as in the example code ?
EDIT :
For the implicit constructor from Base, I can obviously write :
template<class OtherParameter> Derived2(const Base<OtherParameter>& x);
But in that case, do I have the guaranty that the compiler will not use this constructor as an implicit constructor for Derived2<OtherParameter1, OtherParameter2, OtherParameter> ?
EDIT2:
Here I have a test : (LWS here : http://liveworkspace.org/code/cd423fb44fb4c97bc3b843732d837abc)
#include <iostream>
template<typename Type> class Base {};
template<typename Type> class Other : public Base<Type> {};
template<typename Type> class Derived : public Base<Type>
{
public:
Derived() {std::cout<<"empty"<<std::endl;}
Derived(const Derived<Type>& x) {std::cout<<"copy"<<std::endl;}
template<typename OtherType> explicit Derived(const Derived<OtherType>& x) {std::cout<<"explicit"<<std::endl;}
template<typename OtherType> Derived(const Base<OtherType>& x) {std::cout<<"implicit"<<std::endl;}
};
int main()
{
Other<int> other0;
Other<double> other1;
std::cout<<"1 = ";
Derived<int> dint1; // <- empty
std::cout<<"2 = ";
Derived<int> dint2; // <- empty
std::cout<<"3 = ";
Derived<double> ddouble; // <- empty
std::cout<<"4 = ";
Derived<double> ddouble1(ddouble); // <- copy
std::cout<<"5 = ";
Derived<double> ddouble2(dint1); // <- explicit
std::cout<<"6 = ";
ddouble = other0; // <- implicit
std::cout<<"7 = ";
ddouble = other1; // <- implicit
std::cout<<"8 = ";
ddouble = ddouble2; // <- nothing (normal : default assignment)
std::cout<<"\n9 = ";
ddouble = Derived<double>(dint1); // <- explicit
std::cout<<"10 = ";
ddouble = dint2; // <- implicit : WHY ?!?!
return 0;
}
The last line worry me. Is it ok with the C++ standard ? Is it a bug of g++ ?
