Java Map object copy small example-javaTutorial-php.cn

Java Map object copy small example

黄舟

Release： 2017-01-17 15:16:19

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Simple copying problem of Map object? ? ? ?

Solving the problem

Example 1: Copying the map object reference, just a simple reference, cannot solve the problem

[code]package com.evada.de;

import java.util.HashMap;
import java.util.Map;

/**
 * Created by Ay on 2016/5/11.
 */
public class RedisTest {

    public static void main(String[] args) {

        Map<String,String> mapAA = new HashMap<>();
        mapAA.put("A", "A");
        mapAA.put("AA","AA");
        mapAA.put("AAA","AAA");
        System.out.println(mapAA);
        //用mapBB对象去引用mapAA
        Map<String,String> mapBB = mapAA;

        mapBB.put("B","B");

        System.out.println(mapBB);
    }
}

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Result:

[code]{AA=AA, A=A, AAA=AAA}
{AA=AA, A=A, AAA=AAA, B=B}

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Example 2: Map putAll in map implements simple type of copy

[code]package com.evada.de;

import java.util.HashMap;
import java.util.Map;

/**
 * Created by Ay on 2016/5/11.
 */
public class RedisTest {

    public static void main(String[] args) {

        Map<String,String> mapAA = new HashMap<>();
        mapAA.put("A", "A");
        mapAA.put("AA","AA");
        mapAA.put("AAA","AAA");
        System.out.println(mapAA);

        Map<String,String> mapBB = new HashMap<>();
        //把mapAA的元素复制到mapBB中
        mapBB.putAll(mapAA);
        mapBB.put("B","B");
        //打印出的mapAA应该不受影响
        System.out.println(mapAA);
        //打印出的mapBB应该多了元素B
        System.out.println(mapBB);
    }
}

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Result:

[code]{AA=AA, A=A, AAA=AAA}
{AA=AA, A=A, AAA=AAA}
{AA=AA, A=A, AAA=AAA, B=B}

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Example 3: putAll in map is only shallow copy

[code]package com.evada.de;

import java.util.HashMap;
import java.util.Map;

class Person{

    private String id,name;
    Person(String id,String name){
        this.id = id;
        this.name = name;
    }

    public String getId() {
        return id;
    }

    public void setId(String id) {
        this.id = id;
    }

    public String getName() {
        return name;
    }

    public void setName(String name) {
        this.name = name;
    }
}

/**
 * Created by Ay on 2016/5/11.
 */
public class RedisTest {

    public static void main(String[] args) {

        Map<String,Person> mapAA = new HashMap<>();

        mapAA.put("A",new Person("AID","AY"));
        mapAA.put("B",new Person("BID","AL"));

        System.out.println(mapAA);

        Map<String,Person> mapBB = new HashMap<>();
        /** 把mapAA中的元素复制到mapBB中 **/
        mapBB.putAll(mapAA);
        /** 修改mapBB中A元素值，如果mapAA中的元素A受影响，说明是浅复制 **/
        Person person = mapBB.get("A");
        person.setName("Ay_New");

        System.out.println(mapBB);
        System.out.println("mapAA  的A元素value值：" + mapAA.get("A").getName());

    }
}

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Result: As can be seen from the result, print mapAA The result is the same as mapBB, indicating that the copy of putAll is a simple shallow copy.

From the last result, it can be verified again, because changing the value of the A element in mapBB directly affects the element in mapAA The value of

[code]{A=com.evada.de.Person@71bc1ae4, B=com.evada.de.Person@6ed3ef1}
{A=com.evada.de.Person@71bc1ae4, B=com.evada.de.Person@6ed3ef1}
mapAA  的A元素value值：Ay_New

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