如何删除 const_iterator 的常量性? [英] How to remove constness of const_iterator?

问题描述

作为这个问题的扩展const_iterators 是否更快?，我还有一个关于 const_iterators 的问题.如何删除 const_iterator 的常量性?虽然迭代器是指针的一般形式，但 const_iterator 和 iterator 仍然是两个不同的东西.因此，我相信，我也不能使用 const_cast<> 从 const_iterator 转换到 iterators.

As an extension to this question Are const_iterators faster?, I have another question on const_iterators. How to remove constness of a const_iterator? Though iterators are generalised form of pointers but still const_iterator and iterators are two different things. Hence, I believe, I also cannot use const_cast<> to covert from const_iterator to iterators.

一种方法可能是您定义一个迭代器，它移动 '直到 const_iterator 指向的元素.但这看起来是一个线性时间算法.

One approach could be that you define an iterator which moves 'til the element to which const_iterator points. But this looks to be a linear time algorithm.

知道实现这一目标的最佳方法是什么吗?

Any idea on what is the best way to achieve this?

推荐答案

在 C++11 中有一个具有恒定时间复杂度的解决方案:对于任何序列、关联或无序关联容器(包括所有标准库容器)，可以调用range-erase成员函数，范围为空:

There is a solution with constant time complexity in C++11: for any sequence, associative, or unordered associative container (including all of the Standard Library containers), you can call the range-erase member function with an empty range:

template <typename Container, typename ConstIterator>
typename Container::iterator remove_constness(Container& c, ConstIterator it)
{
    return c.erase(it, it);
}

范围擦除成员函数有一对const_iterator 参数，但它们返回一个iterator.因为提供了一个空范围，所以对erase 的调用不会改变容器的内容.

The range-erase member functions have a pair of const_iterator parameters, but they return an iterator. Because an empty range is provided, the call to erase does not change the contents of the container.

感谢 Howard Hinnant 和 Jon Kalb 的技巧.

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