C++ named requirements: LegacyForwardIterator

A LegacyForwardIterator is a LegacyIterator that can read data from the pointed-to element.

Unlike LegacyInputIterator and LegacyOutputIterator, it can be used in multipass algorithms.

If a LegacyForwardIterator it originates from a Container, then it's value_type is the same as the container's, so dereferencing (*it) obtains the container's value_type.

Requirements

The type It satisfies LegacyForwardIterator if

• The type It satisfies LegacyInputIterator
• The type It satisfies DefaultConstructible
• Objects of the type It provide multipass guarantee described below
• Let T be the value type of It. The type std::iterator_traits<It>::reference must be either

• T& or T&&(since C++11) if It satisfies LegacyOutputIterator (It is mutable), or
• const T& or const T&&(since C++11) otherwise (It is constant),

(where T is the type denoted by std::iterator_traits<It>::value_type)

• Equality and inequality comparison is defined over all iterators for the same underlying sequence and the value initialized-iterators(since C++14).

And, given

• i, dereferenceable lvalue of type It
• reference, the type denoted by std::iterator_traits<It>::reference

The following expressions must be valid and have their specified effects

A mutable LegacyForwardIterator is a LegacyForwardIterator that additionally satisfies the LegacyOutputIterator requirements.

Multipass guarantee

Given a and b, dereferenceable iterators of type It:

• If a and b compare equal (a == b is contextually convertible to true) then either they are both non-dereferenceable or *a and *b are references bound to the same object.
• If *a and *b refer to the same object, then a == b.
• Assignment through a mutable ForwardIterator iterator cannot invalidate the iterator (implicit due to reference defined as a true reference).
• Incrementing a copy of a does not change the value read from a (formally, either It is a raw pointer type or the expression (void)++It(a), *a is equivalent to the expression *a).
• a == b implies ++a == ++b.

template<class It>
concept __LegacyForwardIterator =
    __LegacyInputIterator<It> && std::constructible_from<It> &&
    std::is_reference_v<std::iter_reference_t<It>> &&
    std::same_as<
        std::remove_cvref_t<std::iter_reference_t<It>>,
        typename std::indirectly_readable_traits<It>::value_type> &&
    requires(It it) {
        {  it++ } -> std::convertible_to<const It&>;
        { *it++ } -> std::same_as<std::iter_reference_t<It>>;
    };

Notes

Unlike the std::forward_iterator concept, the LegacyForwardIterator requirements requires dereference to return a reference.

Defect reports

The following behavior-changing defect reports were applied retroactively to previously published C++ standards.

1. In the scenario where a and b use the same underlying iterator, evaluating the expression ++a == ++b actually increments the underlying container twice, but the result is still true.
2. Formally also requires implying ++b != a.

See also