c++ - 如何在自定义容器中支持范围适配器？

Question

我创建了一个名为 goldbox 的自定义容器只包含算术类型，我也实现了begin和 end成员函数来迭代元素。
我的完整源代码:

#include <algorithm>
#include <vector>
#include <initializer_list>
#include <iostream>
#include <type_traits>
#include <ranges>

template <typename T>
concept Arithmetic = std::is_arithmetic_v<T>;

template <Arithmetic T = int>
class goldbox {
private:
    template <Arithmetic Base_t>
    struct Node {
        Base_t data;
        Node<Base_t>* prev;
        Node<Base_t>* next;
    };
    Node<T>* head;
    Node<T>* current_node;

    Node<T>*& __at_node(size_t index) {
        auto temp = head;
        size_t count {0};

        while (count < index) {
            temp = temp->next;
            count++;
        }
        current_node = temp;

        return current_node;
    }

    Node<T>*& __get_tail() {
        return __at_node(length() - 1);
    }

public:
    using value_type = T;

    goldbox() : head{nullptr}, current_node{nullptr} {}
    goldbox(std::initializer_list<T>&& list_arg) : goldbox() {
        decltype(auto) list_1 = std::forward<decltype(list_arg)>(list_arg);
        T temp[list_1.size()];
        std::copy(list_1.begin(), list_1.end(), temp);
        std::reverse(temp, temp + list_1.size());
        for (const auto& elem : temp)
            push_front(elem);
    }

    class iterator {
        private:
        Node<T>* node;
    public:
        iterator(Node<T>* arg) noexcept : node{arg} {}

        iterator& operator=(Node<T>* arg) {
            node = arg;
            return *this;
        }

        iterator operator++() {
            if (node)
                node = node->next;
            return *this;
        }

        iterator operator++(int) {
            iterator iter = *this;
            ++(*this);
            return iter;
        }

        iterator operator--() {
            if (node)
                node = node->prev;
            return *this;
        }

        iterator operator--(int) {
            iterator iter = *this;
            --(*this);
            return iter;
        }

        bool operator==(const iterator& other) {
            return (node == other.node);
        }

        bool operator!=(const iterator& other) {
            return (node != other.node);
        }

        T& operator*() {
            return node->data;
        }
    };

    iterator begin() {
        return iterator{head};
    }

    iterator end() {
        return iterator{nullptr};
    }

    size_t length() const {
        auto temp = head;
        size_t count {0};
        while (temp != nullptr) {
            ++count;
            temp = temp->next;
        }
        return count;
    }

    goldbox& push_front(T arg) {
        auto new_node = new Node<T>;

        new_node->data = arg;
        new_node->prev = nullptr;
        new_node->next = head;

        if (head != nullptr)
            head->prev = new_node;

        head = new_node;
        return *this;
    }

    goldbox& push_back(T arg) {
        auto new_node = new Node<T>;
        auto last = head;

        new_node->data = arg;
        new_node->next = nullptr;

        if (head == nullptr){
            new_node->prev = nullptr;
            head = new_node;
            return *this;
        }

        while (last->next != nullptr)
            last = last->next;
        last->next = new_node;
        new_node->prev = last;

        return *this;
    }

    goldbox& clear() {
        auto temp = head;
        Node<T>* next_temp;

        while (temp != nullptr) {
            next_temp = temp->next;
            delete temp;
            temp = next_temp;
        }

        head = nullptr;

        return *this;
    }

    goldbox& pop_back() {
        if (head != nullptr) {
            if (length() != 1) {
                delete std::move(__get_tail());
                __at_node(length() - 2)->next = nullptr;
            } else {
                this->clear();
            }
        }
        return *this;
    }

    goldbox& pop_front() {
        if (head != nullptr) {
            auto temp = head;
            head = head->next;
            delete temp;
        }
        return *this;
    }
};

int main() {
    goldbox goldbox_1 {2, 3, 5, 6, 7, 9};
    goldbox goldbox_2;

    for (const auto& elem : goldbox_1) {
        std::cout << elem << ' ';
    } std::cout << '\n';

    std::transform(goldbox_1.begin(), goldbox_1.end(), 
                   std::back_inserter(goldbox_2),
                   [](auto x){return 2 * x - 1; }
    );

    for (const auto& elem : goldbox_2) {
        std::cout << elem << ' ';
    } std::cout << '\n';

    return 0;
}

输出:

2 3 5 6 7 9
3 5 9 11 13 17

但我想使用范围来使用它，这样我就不必创建新实例。
一旦我申请了goldbox在基于范围的 for 循环内:

for (const auto& elem: goldbox_1 | std::ranges::views::transform([](auto x){return 2 * x - 1;})) {
     std::cout << elem << ' ';
} std::cout << '\n';

它会引发错误，因为我没有提供 operator| .
如果我使用非管道语法:

for (const auto& elem: std::ranges::views::transform(goldbox_1, [](auto x){return x + 1;})) {
     std::cout << elem << ' ';
} std::cout << '\n';

它仍然会抛出一个错误，这两个 begin和 end未在范围内声明。

Answer 1

TLDR
你的类(class)不满足 std::ranges::input_range因为你的迭代器不满足 std::ranges::input_iterator .在您的迭代器类中，您需要:

添加默认构造函数

添加公众difference_type别名

使预递增和递减运算符返回引用

加差运算符

制作 operator==常量

添加公众value_type别名

添加 T& operator*() const

it will still throw an error that both begin and end were not declared in the scope.

我不知道您使用的编译器会给出这种误导性错误消息。如果你用 gcc 编译，你会得到一个正确的错误诊断:

note: the required expression 'std::ranges::__cust::begin(__t)' is invalid

  581 |         ranges::begin(__t);
      |         ~~~~~~~~~~~~~^~~~~

cc1plus: note: set '-fconcepts-diagnostics-depth=' to at least 2 for more detail

设置更高的-fconcepts-diagnostics-depth=我们可以看到根本原因:

note: no operand of the disjunction is satisfied

  114 |         requires is_array_v<remove_reference_t<_Tp>> || __member_begin<_Tp>
      |                  ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
  115 |           || __adl_begin<_Tp>
      |           ^~~~~~~~~~~~~~~~~~~

你的容器不是一个数组，你没有使用 adl begin 而是 member begin 所以我们需要看看为什么你的类不满足 __member_begin :

note: 'std::__detail::__decay_copy(__t.begin())' does not satisfy return-type-requirement, because

  939 |           { __detail::__decay_copy(__t.begin()) } -> input_or_output_iterator;
      |             ~~~~~~~~~~~~~~~~~~~~~~^~~~~~~~~~~~~

问题是您的迭代器类实际上不是一个合适的迭代器。让我们看看为什么:

note: the expression 'is_constructible_v<_Tp, _Args ...> [with _Tp = goldbox::iterator; _Args = {}]' evaluated to 'false'

  139 |       = destructible<_Tp> && is_constructible_v<_Tp, _Args...>;
      |                              ^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

第一个修复是使您的迭代器默认可构造。这样做然后重新编译我们看到您的类进一步不满足迭代器的概念:

note: the required type 'std::iter_difference_t<_Iter>' is invalid, because

  601 |         typename iter_difference_t<_Iter>;
      |         ~~~~~~~~~^~~~~~~~~~~~~~~~~~~~~~~~~

和

note: '++ __i' does not satisfy return-type-requirement, because

  603 |         { ++__i } -> same_as<_Iter&>;
      |           ^~~~~

您需要添加difference_type公共(public)别名并使预递增和预递减运算符返回对迭代器的引用。
修复这个然后重新编译我们看到你的类进一步不满足迭代器的概念:

note: 'std::__detail::__decay_copy(__t.end())' does not satisfy return-type-requirement, because

  136 |           { __decay_copy(__t.end()) }
      |             ~~~~~~~~~~~~^~~~~~~~~~~

error: deduced expression type does not satisfy placeholder constraints

  136 |           { __decay_copy(__t.end()) }
      |           ~~^~~~~~~~~~~~~~~~~~~~~~~~~
  137 |             -> sentinel_for<decltype(_Begin{}(std::forward<_Tp>(__t)))>;
      |             ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

note: the required expression '(__t == __u)' is invalid, because

  282 |           { __t == __u } -> __boolean_testable;
      |             ~~~~^~~~~~

这意味着 begin() 返回的迭代器与 end 返回的迭代器不具有可比性.向下查看诊断深度，您可以看到您的 operator==不考虑，因为它不是 const。
修复此问题然后重新编译，我们看到您的类进一步不满足 input_iterator 的概念:

note: the required type 'std::iter_value_t<_In>' is invalid, because

  514 |         typename iter_value_t<_In>;
      |         ~~~~~~~~~^~~~~~~~~~~~~~~~~~

通过添加公共(public) value_type 来解决此问题别名。
下一个:

note: nested requirement 'same_as<std::iter_reference_t, std::iter_reference_t<_Tp> >' is not satisfied, because

  517 |         requires same_as<iter_reference_t<const _In>,
      |         ~~~~~~~~~^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
  518 |                          iter_reference_t<_In>>;
      |                          ~~~~~~~~~~~~~~~~~~~~~~

这意味着 operator*应该为 iterator 返回相同的引用类型和 const iterator .这可以通过添加 const operator* 来解决。 :

T& operator*();
T& operator*() const;

现在所有编译错误都已修复，并且两个版本(管道和非管道)都可以编译。请注意我已经修复了编译错误，没有检查你的语义。