如何使用 auto 获取 const_iterator? [英] How do I get a const_iterator using auto?

问题描述

第一个问题:是否可以使用 auto 来强制"const_iterator?例如:

First question: is it possible to "force" a const_iterator using auto? For example:

map<int> usa;
//...init usa
auto city_it = usa.find("New York");

我只是想查询，而不是改变city_it指向的任何东西，所以我想让city_it成为map::const_iterator.但是通过使用auto，city_it与map::find()的返回类型相同，即map::iterator>.有什么建议吗?

I just want to query, instead of changing anything pointed by city_it, so I'd like to have city_it to be map<int>::const_iterator. But by using auto, city_it is the same to the return type of map::find(), which is map<int>::iterator. Any suggestion?

推荐答案

抱歉，我只是认为最好的建议是不要使用 auto，因为您想要执行(隐式有效)类型转换.auto 用于推导精确类型，这不是您在这里想要的.

Sorry, but I just think the best suggestion is not using auto at all, since you want to perform a (implicitly valid) type conversion. auto is meant for deducing the exact type, which is not what you want here.

就这样写:

std::map<std::string, int>::const_iterator city_it = usa.find("New York");

正如 MooingDuck 正确指出的那样，使用类型别名可以提高代码的可读性和可维护性:

As correctly pointed out by MooingDuck, using type aliases can improve the readability and maintainability of your code:

typedef std::map<std::string, int> my_map;
my_map::const_iterator city_it = usa.find("New York");

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