是否可以使用STL复制功能与地图

Question

我想知道是否有任何技巧可以使用地图复制将地图内容复制到数组中.因为STL映射是通过键值和映射值的组合,所以映射的元素形成键值对.这阻止我们使用标准算法,如std :: copy.例如,以下代码给出错误:

#include <iostream>
#include <algorithm>
#include <vector>
#include <map>

int
main()
{
  std::map <int, double> test(4);
  test[0] = 11;
  test[2] = 1.23;
  test[3] = 23.29;
  test[1] = 12.12;
  double *test_arr = (double *) malloc(4 * sizeof(double));
  std::copy(test.begin(), test.end(), test_arr);
  std::cout << test_arr[3] << std::endl;
  return 0;
}

错误:

stl_copy_tests.cpp: In function ‘int main()’:
stl_copy_tests.cpp:9:32: error: no matching function for call to ‘std::map<int, double>::map(int)’
/usr/include/c++/4.5/bits/stl_map.h:170:7: note: candidates are: std::map<_Key, _Tp, _Compare, _Alloc>::map(const std::map<_Key, _Tp, _Compare, _Alloc>&) [with _Key = int, _Tp = double, _Compare = std::less<int>, _Alloc = std::allocator<std::pair<const int, double> >, std::map<_Key, _Tp, _Compare, _Alloc> = std::map<int, double>]
/usr/include/c++/4.5/bits/stl_map.h:159:7: note:                 std::map<_Key, _Tp, _Compare, _Alloc>::map(const _Compare&, const allocator_type&) [with _Key = int, _Tp = double, _Compare = std::less<int>, _Alloc = std::allocator<std::pair<const int, double> >, allocator_type = std::allocator<std::pair<const int, double> >]
/usr/include/c++/4.5/bits/stl_map.h:150:7: note:                 std::map<_Key, _Tp, _Compare, _Alloc>::map() [with _Key = int, _Tp = double, _Compare = std::less<int>, _Alloc = std::allocator<std::pair<const int, double> >]
In file included from /usr/include/c++/4.5/bits/char_traits.h:41:0,
                 from /usr/include/c++/4.5/ios:41,
                 from /usr/include/c++/4.5/ostream:40,
                 from /usr/include/c++/4.5/iostream:40,
                 from stl_copy_tests.cpp:1:
/usr/include/c++/4.5/bits/stl_algobase.h: In static member function ‘static _OI std::__copy_move<<anonymous>, <anonymous>, <template-parameter-1-3> >::__copy_m(_II, _II, _OI) [with _II = std::_Rb_tree_iterator<std::pair<const int, double> >, _OI = double*, bool <anonymous> = false, bool <anonymous> = false, <template-parameter-1-3> = std::bidirectional_iterator_tag]’:
/usr/include/c++/4.5/bits/stl_algobase.h:404:70:   instantiated from ‘_OI std::__copy_move_a(_II, _II, _OI) [with bool _IsMove = false, _II = std::_Rb_tree_iterator<std::pair<const int, double> >, _OI = double*]’
/usr/include/c++/4.5/bits/stl_algobase.h:442:39:   instantiated from ‘_OI std::__copy_move_a2(_II, _II, _OI) [with bool _IsMove = false, _II = std::_Rb_tree_iterator<std::pair<const int, double> >, _OI = double*]’
/usr/include/c++/4.5/bits/stl_algobase.h:474:18:   instantiated from ‘_OI std::copy(_II, _II, _OI) [with _II = std::_Rb_tree_iterator<std::pair<const int, double> >, _OI = double*]’
stl_copy_tests.cpp:15:47:   instantiated from here
/usr/include/c++/4.5/bits/stl_algobase.h:319:6: error: cannot convert ‘std::pair<const int, double>’ to ‘double’ in assignment

是否有任何简单的技巧/黑客来克服这个问题.

免责声明:对于在for循环中迭代map并将元素添加到数组的解决方案不感兴趣.

Answer 1

你可以std::transform改用:

template <typename T, typename U>
const U &extract_second(const std::pair<T,U> &p)
{
    return p.second;
}

std::transform(test.begin(), test.end(), test_arr, extract_second<int,double>);

正如@Andre在下面的评论中指出的那样,如果你想要稍微冗长的开销,你可以避免通过仿函数显式地陈述模板参数:

struct extract_second
{
    template <typename T, typename U>
    const U operator() (const std::pair<T,U> &p) const
    {
        return p.second;
    }
};

std::transform(test.begin(), test.end(), test_arr, extract_second());

我确信使用Boost活页夹有一个不那么冗长的解决方案,但是我不记得我头脑中的语法.