我如何得到一个const_iterator使用汽车？ [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强制一个 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");

我只想查询，而不是改变 city_it ，所以我想让 city_it 为 map< int> :: const_iterator 。但是通过使用auto， city_it 与 map :: find（）的返回类型相同， code> map< int> :: iterator 。任何建议？

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

很抱歉，我认为最好的建议是不 $ c> auto ，因为您想要执行（隐式有效的）类型转换。 auto 是用于推导确切类型，这不是您想要的。

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.

只需这样写：

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

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

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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