From: "Fernando Cacciola" <[EMAIL PROTECTED]>
> From: "Peter Dimov" <[EMAIL PROTECTED]>
[...]
> > if(both uninitialized)
> > {
> > // do nothing, nothrow
> > }
> > else if(one initialized, one not)
> > {
> > lhs.reset(*rhs); // strong
> > rhs.reset(); // nothrow
> > }
> > else // both initialized
> > {
> > using std::swap;
> > swap(*lhs, *rhs);
> > }
> >
> > It doesn't even need friendship.
> >
> I see.
> Q: This code is supposed to be in boost::swap( optional<T>&,
optional<T>&)?
Yep.
> > void f(optional<T> /*const &*/ opt);
> >
> > is different than
> >
> > void f(T const * pt);
> >
> > as the latter might potentially store 'pt' while the former cannot.
> >
> ? You mean that the code inside f() could hold onto 'pt'?
> Well, yes it can... but that would be nasty.
> It is supposed to know that ownership is not being handed in.
> The use of a pointer is reserved to convey optionality.
No, functions that store an address use by-pointer, too, even when it must
not be NULL. Consider:
void f(int const & r); // stores &r
f(5L);
> If I have:
>
> void foo ( optional<T> x, optional<T> y )
> {
> ( x == y ) ;
> }
>
> Why do you suggest to be the semantic of the comnparison?
>
> I think that you suggest that it is:
>
> ( (!x) != (!y) ? false : ( !x ? true : ( *x == *y ) )
>
> but in this case, replacing optional<T> with T* changes this semantic,
which
> is what I object.
>
> OTOH, my definition as: get_pointer(x) == get_pointer(y)
>
> will not change the semantic of the comparison if optional<T> is replaced
by
> T*
Yes it will. :-)
optional<T> m;
foo(m, m); // comparison inside yields false
T * p;
foo(p, p); // comparison inside yields true
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