将映射值复制到 STL 中的向量

2022-01-07 00:00:00 containers c++ stl

目前正在通过 Effective STL 工作.第 5 条建议通常最好使用范围成员函数而不是它们的单个元素对应物.我目前希望将地图中的所有值(即 - 我不需要键)复制到向量中.

Working my way through Effective STL at the moment. Item 5 suggests that it's usually preferable to use range member functions to their single element counterparts. I currently wish to copy all the values in a map (i.e. - I don't need the keys) to a vector.

最干净的方法是什么?

推荐答案

在这里你不能轻易使用范围，因为你从地图中得到的迭代器是指一个 std::pair，你将用来插入迭代器的地方into a vector 指的是存储在 vector 中的类型的对象，它(如果您丢弃密钥)不是一对.

You can't easily use a range here because the iterator you get from a map refers to a std::pair, where the iterators you would use to insert into a vector refers to an object of the type stored in the vector, which is (if you are discarding the key) not a pair.

我真的不认为它比显而易见的更干净:

I really don't think it gets much cleaner than the obvious:

#include <map>
#include <vector>
#include <string>
using namespace std;

int main() {
    typedef map <string, int> MapType;
    MapType m;  
    vector <int> v;

    // populate map somehow

    for( MapType::iterator it = m.begin(); it != m.end(); ++it ) {
        v.push_back( it->second );
    }
}

如果我要多次使用它，我可能会将其重写为模板函数.类似的东西:

which I would probably re-write as a template function if I was going to use it more than once. Something like:

template <typename M, typename V> 
void MapToVec( const  M & m, V & v ) {
    for( typename M::const_iterator it = m.begin(); it != m.end(); ++it ) {
        v.push_back( it->second );
    }
}

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