std::map、std::unordered_map - 在初始化列表中缩小转换

Question

Is it bug or standart permits it?

#include <iostream>
#include <map>
#include <unordered_map>

int main() {
    std::unordered_map<int,int> mm {{44,44}, {33.3, 54}, {222.2,222.2}};
    for(auto& [f,s] :mm) {
        std::cout<<f<<" - "<<s<<std::endl;
    }

    std::map<int,int> m {{44,44}, {33.3, 54}, {222.2,222.2}};
    for(auto& [f,s] :m) {
       std::cout<<f<<" - "<<s<<std::endl;
    }
}

Test it on wandbox.org with clang10 and gcc10. There is not such problem with std::set and std::unordered_set .

Answer 1

The element type of std::map and std::unordered_map is std::pair . The problem is that std::pair has a templated constructor,

 template< class U1, class U2 > constexpr pair( U1&& x, U2&& y );

Initializes first with std::forward<U1>(x) and second with std::forward<U2>(y) .

Eg given {33.3, 54} , U1 is deduced as double and U2 is deduced as int , note that this is an exact match and no conversions are required for this constructor to be used to construct the std::pair , then no narrowing conversion happens either.

On the other hand, for std::set , given std::set<int> s {33.3}; , the constructor of std::set taking std::initializer_list<int> will be used, and the std::initializer_list<int> is initialized from {33.3} , narrow conversion happens.