如何获得非const迭代器 [英] How get non-const iterator

问题描述

std::map<char,int> dict;
...
auto pmax = dict.begin(); // here i get const iterator

我可以明确指出常数类型？

Can I "explicitly indicate" that the value obtained is a non-constant type?

推荐答案

查看您的代码，你基本上实现了 std :: max_element 。因此，您可以将最后一个输出行重写为：

Looking at your code, you are basically implementing std::max_element. So you could rewrite your last output line to:

    std::cout << std::max_element(begin(dict), end(dict),
        [](decltype(*begin(dict)) a, decltype(*begin(dict)) b) {
            return a.second < b.second;
        })->first << std::endl;

诚然， decltype（* begin（dict））是丑陋的，这希望可以通过C ++ 1y中的通用lambdas补救。

Admittedly, the decltype(*begin(dict)) is ugly, which hopefully will be remedied by generic lambdas in C++1y.

关键是，无论你是否有一个 map :: iterator 或 map :: const_iterator 当你解除引用时，结果将是 std :: pair 与 const key_type 作为第一个参数。所以，即使你有两个迭代器 s it1，it2 到可变数据（例如，通过 map :: begin（）），您不能重新分配这些迭代器引用的完整对。因此， * it1 = * it2 将无法正常工作，因为您试图覆盖 mapped_type const key_type 。

The point is, that regardless of whether you have a map::iterator or map::const_iterator when you dereference it, the result will be a std::pair with a const key_type as first argument. So, even if you have two iterators it1, it2 to mutable data (e.g., acquired via map::begin()), you can not re-assign the complete pair that is referenced by those iterators. Thus, *it1 = *it2 won't work, because you are trying to overwrite the mapped_type and the const key_type.

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