考慮如下代碼：

#include <iostream>
#include <map>
#include <string>
#include <boost/chrono.hpp>
#include <boost/timer/timer.hpp>

int main()
{
    std::map<int, std::string> a{ { 1, "one" }, { 2, "two" }, { 3, "buckle my shoe" } };
    std::map<int, std::string> b{ { 3
 
, "three" } };

    //計算函數調用占用的時間
    boost::timer::cpu_timer timer;

    for (int i=0;i<10000;++i){

        {
        auto ite = a.find(2);
        auto [k,v] = *ite;
        b.emplace(std::make_pair(k, v));
        a.erase(ite);    
        }
        
        {
        auto ite = b.find(2);
        auto [k,v] 
 
= *ite;
        a.emplace(std::make_pair(k, v));
        b.erase(ite);
        }
    }

    boost::timer::cpu_times time = timer.elapsed();
    int wall_time_ = time.wall / 1000000; //納秒轉毫秒
    
    std::cout << "elasped time=" << wall_time_;

}

把元素2（key==2）在a,b兩個容器之間移動。涉及到heap的內存分配和釋放。當insert時，發生malloc，當erase時，發生free。
C++17開始，支持無heap動作的元素搬移：

#include <iostream>
#include <map>
#include <string>
#include <boost/chrono.hpp>
#include <boost/timer/timer.hpp>

int main()
{
    std::map<int, std::string> a{ { 1, "one" }, { 2, "two" }, { 3, "buckle my shoe" } };
    std::map<int, std::string> b{ { 3, "three" } };

    //計算函數調用占用的時間
    boost::timer::cpu_timer timer;

    for (int i=0;i<10000;++i){
        b.insert(a.extract(2)); 
        auto ite = b.find(2);
        a.insert(b.extract(ite)); 
    }

    boost::timer::cpu_times time = timer.elapsed();
    int wall_time_ = time.wall / 1000000; //納秒轉毫秒
    
    std::cout << "elasped time=" << wall_time_;

}

關鍵在於extract函數，它返回一個node handle

C++ map node handle