Eric wrote in
Say you have the following:
template <typename _T>
Note that identifiers that begin with an underscore followed by one
of A through Z are reserved for the implementation, for eg your
std library implementation my define _T as a macro, don't use such
identifiers,
struct A {
typedef _T T;
A (T t);
operator T () const;
};
How do you define the T conversion operator?
template <typename Type >
struct A
{
typedef Type T;
A (T t) {}
operator T () const;
};
template < typename Type >
A< Type >:
perator typename A< Type >::T () const
{
return T();
}
You could also in this case of writen this,
template < typename Type >
A< Type >:
perator Type () const
{
return T(); // or Type();
}
Or (if inline is Ok) inside the class/struct,
template <typename Type >
struct A
{
typedef Type T;
A (T t) {}
operator T () const
{
return T();
}
};
HTH.
Rob.
