Type conversion operators

[Eric]

Say you have the following:

template <typename _T>
struct A {
typedef _T T;
A (T t);
operator T () const;
};

How do you define the T conversion operator?

TIA.

[Rob Williscroft]

Eric wrote in news:2e**************************@posting.google.c om:

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 >::operator typename A< Type >::T () const
{
return T();
}

You could also in this case of writen this,

template < typename Type >
A< Type >::operator 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.
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