Comments on: std::iterator is deprecated: Why, What It Was, and What to Use Instead

By: Mirality

Mirality — Wed, 26 Sep 2018 07:21:00 +0000

How can the base class be well-formed if one of the typedefs is missing, since the base class would look exactly like libstdc++’s iterator_traits, which is not well-formed for exactly the same reason?

At least according to the comments in libc++’s implementation, it’s not actually intended to support the case where “some” typedefs are missing. It supports exactly two cases: either your iterator provides iterator_category, in which case it is expected to provide ALL of the other typedefs as well, which iterator_traits then forwards — or it does not, in which case iterator_traits provides NONE of the typedefs. If something other than this actually works then it presumably is not intended.

By: Ariath

Ariath — Thu, 10 May 2018 22:39:00 +0000

In reply to Sergio Adán. Thanks, but the question was why libc++ is well-formed, not why libstdc++ is ill-formed because it's pretty obvious.

By: Sergio Adán

Sergio Adán — Wed, 09 May 2018 09:47:00 +0000

See C++17 draft § 14.6.2.1 Dependent names
7 If a qualified-id in which the nested-name-specifier refers to the current instantiation is not a member of
the current instantiation or a member of an unknown specialization, the program is ill-formed even if the
template containing the qualified-id is not instantiated;

libstdc++ is ill-formed when at least one of the typedefs is missing. Templates are compiled on demand but the template must be well-formed and it includes the types defined in the template.

libc++ is well-formed because types are not explicitly set in the template (and they are not used in the template) but maybe in the inherited class. So until someone needs a type that is not set the program will be ok

By: Sonu Lohani

Sonu Lohani — Wed, 09 May 2018 04:14:00 +0000

Nice post. Thanks for the info.