Difference between stl: map and boost: unorder_map, boostunorder_map

Difference between stl: map and boost: unorder_map, boostunorder_map

The map in stl is implemented based on the red and black trees. When the insert element is used, operator <is used to compare the element and locate the position where the element can be inserted. Therefore, the final traversal result is ordered.

In boost, unorder_map compares elements based on hash values. Some elements may have the same hash value but different elements. Therefore, you must first define the hash_value function and operator =. Therefore, the result of traversing unorder_map is unordered.

#include<string>#include<iostream>#include<map>#include<stdio.h>#include<boost/unordered_map.hpp>using namespace std;struct person{    string name;    int age;    person(string name, int age)    {        this->name =  name;        this->age = age;    }    bool operator < (const person& p) const    {        return this->age < p.age;    }    bool operator== (const person& p) const    {        return name==p.name && age==p.age;    }};size_t hash_value(const person& p){    size_t seed = 0;    boost::hash_combine(seed, boost::hash_value(p.name));    boost::hash_combine(seed, boost::hash_value(p.age));    return seed;}map<person,int> m;boost::unordered_map<person,int> um;int main(){    person p1("p1",20);    person p2("p2",22);    person p3("p3",22);    person p4("p4",23);    person p5("p5",24);    m.insert(make_pair(p3, 100));    m.insert(make_pair(p4, 100));    m.insert(make_pair(p5, 100));    m.insert(make_pair(p1, 100));    m.insert(make_pair(p2, 100));    um.insert(make_pair(p3, 100));    um.insert(make_pair(p4, 100));    um.insert(make_pair(p5, 100));    um.insert(make_pair(p1, 100));    um.insert(make_pair(p2, 100));    for(map<person, int>::iterator iter = m.begin(); iter != m.end(); iter++)    {        cout<<iter->first.name<<"\t"<<iter->first.age<<endl;    }    cout<<endl;    for(boost::unordered_map<person, int>::iterator iter = um.begin(); iter != um.end(); iter++)    {        cout<<iter->first.name<<"\t"<<iter->first.age<<endl;    }    cout<<(m.find(p3)!=m.end())<<endl;    return 0;}

Output:

P1 20
P3 22
P4 23
P5 24

P3 22
P1 20
P5 24
P4 23
P2 22
0

Stl: The age of p2 in map is the same as that of p3. As a result, p2 cannot find the insert position. In the end, there is no p2 in map.