"Paul Mensonides" <[EMAIL PROTECTED]> writes:
> ----- Original Message -----
> From: "David Abrahams" <[EMAIL PROTECTED]>
>
>> The latter. It was just an experiment. Fortunately, nothing I'm
>> doing depends very much on it. It was prompted by the fact that
>> Borland 5.51 can handle enable_if, but not in a templated constructor:
>>
>> template <class T>
>> struct X
>> {
>> template <class U>
>> X(X<U> const&, typename enable_if<some_property_of<U>::value,
> int*>::type = 0);
> ^
> It chokes here?
Sorry, no. I mean the 2nd '<' in that line.
> Is it legal to use the template parameter 'U' in a non-deduced
> context like that? I'm curious because I thought that it wasn't
> (though I could be wrong).
Why do you say that's non-deduced? Up to the point of enable_if, it's
standard stuff; that's how shared_ptr conversion works.
--
David Abrahams
[EMAIL PROTECTED] * http://www.boost-consulting.com
Boost support, enhancements, training, and commercial distribution
_______________________________________________
Unsubscribe & other changes: http://lists.boost.org/mailman/listinfo.cgi/boost
