Although I don't see this as problematic:
optional<int> x;
if (x) foo(x);
Or perhaps:
optional<int> x;
if (!!x) foo(x);
We already have an implicit conversion to safe_bool and an operator ! anyway. Keep it. There's nothing wrong with it:
operator unspecified-bool-type() const; bool operator!() const;
IMHO, there is something terribly wrong here because now optional<T> has two interfaces. The interface of optional itself and the interface of T. If you think that optional<T> can be used like T (having the value-interface), you are immediately fooled by if(x) as it doesn't check T's value. A pointer-interface is much cleaner as it gives the user a hint that he is using a wrapper and in practice, I always prefer to be a little more explicit on these things (even at the cost of an occasional * here and there) than to have silent bugs.
Regards, Daniel
-- Daniel Frey
aixigo AG - financial training, research and technology Schlo�-Rahe-Stra�e 15, 52072 Aachen, Germany fon: +49 (0)241 936737-42, fax: +49 (0)241 936737-99 eMail: [EMAIL PROTECTED], web: http://www.aixigo.de
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