在地图上的remove_if试图将const作为非const传递 - 为什么？

Question

这里有一些代码可以过滤掉满足谓词的地图元素到新地图(MCVE-fied):

#include <algorithm>
#include <unordered_map>
#include <iostream>

using namespace std;

int main() {
    unordered_map<string, int> m = { { "hello", 1 }, { "world", 2 } };
    auto p = [](const decltype(m)::value_type& e) { return e.second == 2; };
    const auto& m2(m);
    auto m3(m2);
    auto it = remove_if(m3.begin(), m3.end(), p);
    m3.erase(it, m3.end());
    cout << "m3.size() = " << m3.size() << endl;
    return 0;
}

remove_if()行编译失败,我得到:

In file included from /usr/include/c++/4.9/utility:70:0,
                 from /usr/include/c++/4.9/algorithm:60,
                 from /tmp/b.cpp:1:
/usr/include/c++/4.9/bits/stl_pair.h: In instantiation of ‘std::pair<_T1, _T2>& std::pair<_T1, _T2>::operator=(std::pair<_T1, _T2>&&) [with _T1 = const std::basic_string<char>; _T2 = int]’:
/usr/include/c++/4.9/bits/stl_algo.h:868:23:   required from ‘_ForwardIterator std::__remove_if(_ForwardIterator, _ForwardIterator, _Predicate) [with _ForwardIterator = std::__detail::_Node_iterator<std::pair<const std::basic_string<char>, int>, false, true>; _Predicate = __gnu_cxx::__ops::_Iter_pred<main()::<lambda(const value_type&)> >]’
/usr/include/c++/4.9/bits/stl_algo.h:937:47:   required from ‘_FIter std::remove_if(_FIter, _FIter, _Predicate) [with _FIter = std::__detail::_Node_iterator<std::pair<const std::basic_string<char>, int>, false, true>; _Predicate = main()::<lambda(const value_type&)>]’
/tmp/b.cpp:12:48:   required from here
/usr/include/c++/4.9/bits/stl_pair.h:170:8: error: passing ‘const std::basic_string<char>’ as ‘this’ argument of ‘std::basic_string<_CharT, _Traits, _Alloc>& std::basic_string<_CharT, _Traits, _Alloc>::operator=(const std::basic_string<_CharT, _Traits, _Alloc>&) [with _CharT = char; _Traits = std::char_traits<char>; _Alloc = std::allocator<char>]’ discards qualifiers [-fpermissive]
  first = std::forward<first_type>(__p.first);
        ^

为什么会这样？remove_if不应该需要非const映射键(在这种情况下是字符串) - 如果我没有弄错的话.也许autos假设我想要非const迭代器？如果是这样,我该怎么做才能拼出这个类型(我想避免这种情况,因为这个代码需要模板化).

Answer 1

即使没有所有中间变量,您的示例也会失败.

unordered_map<string, int> m = { { "hello", 1 }, { "world", 2 } };
auto p = [](const decltype(m)::value_type& e) { return e.second == 2; };  
auto it = remove_if(m.begin(), m.end(), p);

上面的代码将失败并出现相同的错误.您不能使用remove_if关联容器,因为该算法通过将满足谓词的元素移动到容器的末尾来工作.但你怎么重新排序unordered_map？

写一个擦除元素的循环

for(auto it = m.begin(); it != m.end();)
{
  if(p(*it)) it = m.erase(it);
  else ++it;
}

或者您可以将其打包成算法

template<typename Map, typename Predicate>
void map_erase_if(Map& m, Predicate const& p)
{
    for(auto it = m.begin(); it != m.end();)
    {
      if(p(*it)) it = m.erase(it);
      else ++it;
    }
}

如果您有一个实现统一容器擦除库基础扩展的标准库实现,那么您有一个类似于std::experimental命名空间中的算法.