使用迭代器调用擦除与const_iterator [英] Calling erase with iterator vs const_iterator

问题描述

为什么调用 const_iterator 的容器的擦除成员函数失败？

Why does calling the erase member function of a container with a const_iterator fail?

它适用于一个非const iterator 。

It works with a non const iterator.

推荐答案

这不编译，因为 container :: iterator 和 container :: const_iterator 是两种不同的类型并且唯一（一个参数）的擦除版本是： iterator erase（iterator）;

This doesn't compile because container::iterator and container::const_iterator are two distinct types and the only (one-argument) version of erase is: iterator erase(iterator);

不接受 const_iterator 可以被视为语言标准的缺陷： http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2007/n2350.pdf

Not accepting a const_iterator can be viewed as a defect in the language standard: http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2007/n2350.pdf

没有这个限制的特殊原因。迭代器仅用于在（可修改）容器中表示位置，并且在插入或 是迭代器修改的pintee（在 erase 的情况下，它只是在概念上不再存在，这是一个正常的事情const对象）

There is no particular reason for this restriction. The iterator is only used to indicate a position in the (modifiable) container, and neither in case of insert or erase is the "pointee" of the iterator modified (in case of erase it just conceptually goes out of existence, which is a normal thing to do for const objects).

当前标准表示迭代器常量和容器常量之间的混淆（和其他答案一样），似乎 const_iterator 可能会被C ++ 0x中的 erase 所接受。

Current standard indicates a confusion between "iterator constness and container constness" (as do other answers here), and it seems const_iterator might become acceptable for erase in C++0x.

作为解决方法，您可以从 const_iterator 中有效地获取 iterator ，因为容器有首先是可变的。

As a workaround, you can validly obtain an iterator from a const_iterator because the container has to be mutable in the first place.

下面的函数只能用于随机访问迭代器，因为使用其他类型的迭代器可能会太慢。

The function below is only compilable for random access iterators, as it might be a bit too slow to do this with other types of iterators.

#include <vector>

template <class Container>
typename Container::iterator to_mutable_iterator(Container& c, typename Container::const_iterator it)
{
    return c.begin() + (it - c.begin());
}

int main()
{
    int arr[] = {1, 5, 2, 5, 3, 4, 5, 1};
    std::vector<int> vec(arr, arr + sizeof(arr) / sizeof(*arr));
    for (std::vector<int>::const_iterator it = vec.begin(); it != vec.end(); ) {
        //if (*it = 5) {  //const_iterator prevents this error
        if (*it == 5) {
            it = vec.erase(to_mutable_iterator(vec, it));
        }
        else {
            ++it;
        }
    }
}

但是，可能会更好重组代码，以便您首先不需要 const_iterator 。在这种情况下，最好使用 std :: remove 算法。如果您需要在删除之前进行更多的非突变工作，则可以将其提取到单独的方法等中。

However, it might be better to restructure code so that you don't need a const_iterator in the first place. In this case, it would be better to use the std::remove algorithm. If you need to do more non-mutating work before erasing, you can extract that into a separate method etc.

这篇关于使用迭代器调用擦除与const_iterator的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！