c++11 在列表之间移动元素以映射(或其他容器)

Question

有没有简单的方法来move不同容器之间的元素？
我找不到任何简单的方法(使用 <algorithm> )来执行以下操作:

不可复制类

class NonCopyable {
public:
    NonCopyable() {};
    ~NonCopyable() {};
    NonCopyable(const NonCopyable&) = delete;
    NonCopyable& operator=(const NonCopyable&) = delete;
    NonCopyable(NonCopyable&& that) {}
};

移动操作:

std::list<NonCopyable> eList;
std::map<int, NonCopyable> eMap;

eList.push_back(NonCopyable());

// Move from list to map
{
    auto e = std::move(eList.back());
    eList.pop_back();
    eMap.insert(std::make_pair(1, std::move(e)));
}

// Move from map to list
{
    auto it = eMap.find(1);
    if (it != eMap.end()) {
        eList.push_back(std::move(it->second));
        auto e = eMap.erase(it);
    }
}

// Move all
// Iterate over map?...

我看过 std::list::splice但这对我没有帮助，因为我有一个 list和一个 map , 而不是两个 list小...

谢谢

Answer 1

std::move_iterator怎么样？ ？这是从 vector 移动到 std::string

的示例

#include <iostream>
#include <algorithm>
#include <vector>
#include <iterator>
#include <numeric>
#include <string>

int main()
{
    std::vector<std::string> v{"this", "is", "an", "example"};

    std::cout << "Old contents of the vector: ";
    for (auto& s : v)
        std::cout << '"' << s << "\" ";

    typedef std::vector<std::string>::iterator iter_t;
    std::string concat = std::accumulate(
                             std::move_iterator<iter_t>(v.begin()),
                             std::move_iterator<iter_t>(v.end()),
                             std::string());  // Can be simplified with std::make_move_iterator

    std::cout << "\nConcatenated as string: " << concat << '\n'
              << "New contents of the vector: ";
    for (auto& s : v)
        std::cout << '"' << s << "\" ";
    std::cout << '\n';
}

输出:

Old contents of the vector: "this" "is" "an" "example"
Concatenated as string: thisisanexample
New contents of the vector: "" "" "" ""