> Good point. On a few ocasions I have use optional<> to pass optional
> parameters.
> However, I've came to the following:
>
> Take you example for instance:
>
> void fn(int iImportant, optional<int> iNotImportant = optional<int>())
> {
> if ( !!iNotImportant )
> {
> // not important argument recieved, use it.
> foo ( * iNotImportant ) ;
> }
> }
>
> Since optional<> uses pointer semantics, the above code could
> have been
> written using a true pointer with nearly the same syntatic
> usage of the
> optional parameter:
>
> void fn(int iImportant, int* iNotImportant = NULL )
> {
> if ( !!iNotImportant )
> {
> // not important argument recieved, use it.
> foo ( * iNotImportant ) ;
> }
> }
We already talked about this: pointer will add extra memory access, optional
should not (in fact it should be inlined and won't be different from by
value parameter)
> The constructor must be explicit in order to disable
> unexpected implicit
> conversions.
Some may still prefer sometime
foo(3)
to
foo(&lvalue(3));
foo(optional<T>(3));
implicit conversion to optional should not be dangerous anyway.
> Actually, optional<T> does not require T to be default constructable.
> The reason why you can't have optional<T&> is -at least- that
> you cannot
> have a reference to a reference,
> and optional<T> uses "T {const} &" (for example, in the constructor).
use add_reference instead?
Gennadiy.
_______________________________________________
Unsubscribe & other changes: http://lists.boost.org/mailman/listinfo.cgi/boost
