Subject: Re: [boost] [optional] operator<(optional<T>, T) -- is it wrong?
From: Gottlob Frege (gottlobfrege_at_[hidden])
Date: 2014-11-29 16:13:06
On Sat, Nov 29, 2014 at 5:58 AM, Vicente J. Botet Escriba
> We have worked a lot with implicit conversion as C++98 did have explicit
> ones. I would say that the conversion should be explicit by default.
> Implicit conversion should be allowed only when there is a sub-type
> relationships <: between the types. This sub-type relationship should
> If we have types R, S such that R <: S with implicit conversions from R to S
> , the conversion from S to R can not be implicit (no implicit cycles),
> however there should be an explicit conversion from S to R (coercion).
Does the coercion need to be via explicit constructor, or can it be a
function? ie shared_ptr::get() ?
As "test cases", I think shared_ptr and unique_ptr need explicit
from-ptr constructors (for safety),
but (IMO) dumb_ptr does not.
Do your rules agree?
> If we have types R, S, T such that R <: S <: T with implicit conversions
> from R to S and from S to T, there should be also an implicit conversion
> from R to T. That is R <: T. Note that C++ conversions are not transitive.
> Identity: R Â° S = identity
> If we have types R, S such that R <: S
> For any r:R R(S(r)) == r.
> That is, there is no loss of information.
> The cost of the implicit conversion from the subtype to the super type is
> almost null. Conversions that implies a high conversion cost must be
> The conversions from/to the super type must be defined on the sub-type (this
> is in line with OO sub-typing, we can refine, but not generalize).
> Defining implicit conversion on the super-type side would invalidate valid
> programs. If we had
> class S
> void f(S);
> and now we define a subtype R of S, R <: S, with an implicit conversion from
> R to S.
> R r;
> f(r); // well formed
> Now if we define a super type of T fo S, S <: T, that defines implicit
> conversions from S to T and from R to S, and add a generalization for f
> void f(T);
> The previous program would not be any more well formed
> R r;
> f(r); // ambiguous conversion f(S) and f(T)
> That is, super typing is not allowed. This point should be the most
> controversial, but I think it is the most important :(
>>> Currently we have std::less, not std::order and the STL ordered
>>> are using as default comparator std::less. So let define
>>> std::less<optional<T>> using std::less<T>.
>> Yes. We currently have that much - std::less<optional<T>> is built
>> with std::less<T>, not op<(T,T).
>> Even if only for the sake of pointers. On almost-non-existent
>> hardware, (and maybe future hardware).
> I'm missing the wording for the definition of std::less<optional<T>> in
> function of std::less<T>. Could you point me where this is described?
Grrrrrrrrrrrrr. I give up. It was there when I last argued for it.
Not sure when it got removed. :-(
Tony Van Eerd
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