From: Daniel Wallin (dalwan01_at_[hidden])
Date: 2004-04-17 05:28:55
Andrea Torsello wrote:
> template<class U>
> enable_if_different<Tp, U, X<Tp>&> operator = (X<U> const &);
>
> where enable_if_different<Tp, U, X<Tp>&> converts to X<Tp>&, but then it
> would interfere with template type deduction when passing the result to a
> templated function.
No it wouldn't. It would be written:
template<class U>
typename enable_if_different<Tp, U, X<Tp>&>::type
operator = (X<U> const &);
-- Daniel Wallin