嵌套 class "does not name a type"

Question

我收到以下错误：

main.cpp:18:5: error: 'Iterator' does not name a type
   18 |     Iterator begin() {
      |     ^~~~~~~~

使用此代码：

#include <iostream>
#include <iostream>
#include <memory>
#include <fstream>
#include <filesystem>

using namespace std;

class Numbers {
    private:
    int current;
    int end;

    public:

    Numbers(int end) : current(0), end(end) {}

    Iterator begin() {
        return Iterator(this);
    }

    bool operator==(const Numbers& other) const {
        return current == other.current && end == other.end;
    }

    bool operator!=(const Numbers& other) const {
        return !(other == *this);
    }

    class Iterator {
        private:
        Numbers* range;

        public:
        using value_type = int;
        using difference_type = ptrdiff_t;
        using pointer = int*;
        using reference = int&;
        using iterator_category = input_iterator_tag;

        Iterator(Numbers* range) : range(range) {}

        int operator*() const {
            return range->current;
        }

        int* operator->() const {
            return &range->current;
        }

        bool operator==(const Iterator& other) const {
            return other.range == range;
        }

        bool operator!=(const Iterator& other) const {
            return !(*this == other);
        }

        Iterator& operator++() {
            range->current++;
            return *this;
        }


    };
};

事实证明，将 begin 函数 移动到 嵌套的 Iterator class 下可以编译。

但这很奇怪 - 嵌套的 class 是否遵循与任何其他成员相同的访问规则，这意味着不需要前向引用？

我在网站上搜索了关于这个确切问题的其他问题，似乎没有找到答案。

Answer 1

从评论到问题

Also - I have no problem calling member function f in member function q where f is defined after q. Can you explain why the latter example is different than the situation described in this question?

根据 C++ 20 标准（11.4 Class 个成员）

6 A complete-class context of a class is a

> (6.1) — function body (9.5.1),

(6.2) — default argument (9.3.3.6),

(6.3) — noexcept-specifier (14.5), or

(6.4) — default member initializer

within the member-specification of the class

因此在完整的 class 上下文中，函数 f 的名称在函数 q.

的主体中可见

然而，内部 class 是在完整的 class 上下文之外声明的。所以根据 C++ 20 Standard (6.5.1 Unqualified name lookup)

7 A name used in the definition of a class X23 outside of a complete-class context (11.4) of X shall be declared in one of the following ways:

> (7.1) — before its use in class X or be a member of a base class of X (11.8), or ...

所以内部 class Iterator 的名称必须在用作 return 类型的成员函数之前声明。

您可以使用占位符 auto 作为 return 类型来代替名称 Iterator。

auto begin() {
    return Iterator(this);
}

嵌套 class "does not name a type"

Nested class "does not name a type"

c++

scope

nested

class

declaration