c++ - 迭代器需求满足

Question

考虑以下类:

template <class T>
struct X {
    T& operator*() & { return t; }
    T& operator*() && = delete; 
    X& operator++() { return *this; }
    T t;
};

这个类是否满足 C++ 标准对迭代器概念的要求？此类的对象是可递增和可取消引用的。但右值对象的取消引用是被禁止的。

int main() {
    X<int> x;
    *x; // ok
    *X<int>(); // fail. But is it really necessary for Iterator by Standard?
}

Answer 1

严格阅读标准； [iterator.iterators]/2 (§24.2.2/2)确实暗示 X 类型符合迭代器的条件；

...a and b denote values of type X or const X

...r denotes a value of X&...

A type X satisfies the Iterator requirements if:

• X satisfies the CopyConstructible, CopyAssignable, and Destructible requirements ([utility.arg.requirements]) and lvalues of type X are swappable ([swappable.requirements]), and

• the expressions in Table (below) are valid and have the indicated semantics.

  • *r (r is dereferenceable)
  • ++r (returns X&)

给定代码；

template <class T>
struct X {
    T& operator*() & { return t; }
    T& operator*() && = delete; 
    X& operator++() { return *this; }
    T t;
};
int main()
{
    X<int> x;
    ++x;
    int i = *x;
    X<int> y(x);
    using std::swap;
    std::swap(x, y);
}

当然看起来确实满足这些要求。

然而，故事还在继续，迭代器 概念，如上定义，并未列为标准中的迭代器类别之一 §24.2.1/2 ;

This International Standard defines five categories of iterators, according to the operations defined on them: input iterators, output iterators, forward iterators, bidirectional iterators and random access iterators...

他们都定义了一个操作*a and *r++ X 类型编译失败；

int j = *x++; // fails to compile even with the inclusion of the post increment operator
const X<int> xc {};
int ic = *x1; // no const overloads

要在定义的类别之一中使用迭代器，它需要包含更多成员以取消引用 const 值、左值和右值、后增量等。在 spirit of the standard 中;

Iterators are a generalization of pointers that allow a C++ program to work with different data structures (containers) in a uniform manner.

此处针对重载成员和运算符的指导是，可以/应该添加它们以强制遵守并优化(如果可能)通用指针语义的实现——而不是禁止语义；限制语义可能会产生意想不到的后果。