当 std::move_iterator 取消引用右值引用时，为什么它可以将自己宣传为前向（或更强）迭代器？

Question

According to cppreference, std::move_iterator sets its ::iterator_category to the category of its underlying iterator ¹ .

But I reckon it can be an input/output iterator at best, since for forward iterators reference must be an lvalue reference, while move_iterator sets reference (and the return type of operator* ) to an rvalue reference ² .

Is this a blatant mistagging of the iterator with a wrong category?

Being able to do this for my own iterators is undoubtedly convenient. Is there any reason I shouldn't do this, if even the standard library does so?

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¹ But anything stronger than random_access_iterator_tag is truncated to random_access_iterator_tag , which is weird, since contiguous_iterator_tag is only supposed to be used for ::iterator_concept .

² Or leaves it untouched if it's not a reference, but then the underlying iterator shouldn't advertise itself as a forward iterator either.

Answer 1

Cppreference article on forward iterator requirements was wrong (already fixed by someone). reference must be any reference ( & or && ), not specifically lvalue reference ( & ). Meaning move_iterator does conform.

See [forward.iterators]/1.3 :

if X is a mutable iterator, reference is a reference to T ; if X is a constant iterator, reference is a reference to const T

See also LWG1211 (from 2009), which raised the same issue, and was resolved by N3066 (in 2010), which changed the wording from "lvalue reference" to "any reference". (Thanks @康桓瑋 for the links).