String重写了hashcode()方法，String类型的hash值是根据字符串的内容来决定的，并不是内存地址，只要两个String类型的字符串内容一致，那么两者的hashcode就相同。String为了节约内存比较特殊，因为很常用，所以定义为值一样的hashcode就一样，不同对象的hashCode 值一般来说不会相同，同一个对象的hashCode值肯定相同
package com.web;
public class Test55 {
public static void main(String[] args) {
String s1 = "abc";
String s2 = "abc";
String s3 = new String("abc");//new String()虽然灰色不可省略
String s4 = new String("abc");//new String()虽然灰色不可省略
boolean b2= s1==s2;//写法注意
System.out.println("b2 = " + b2);//true
System.out.println("s1==s2 = "+(s1==s2));//true 没有开辟新的内存空间，abc被s1，s2共享
System.out.println("s1==s3 = "+(s1==s3));//false 开辟了新的内存空间
System.out.println("s3==s4 = "+(s3==s4));//false
System.out.println(s1.equals(s2));//true 只比较值大小
System.out.println(s1.equals(s3));//true 只比较值大小
System.out.println("s1.hashCode() = " + s1.hashCode());//96354
System.out.println("s2.hashCode() = " + s2.hashCode());//96354
System.out.println("s3.hashCode() = " + s3.hashCode());//96354
System.out.println("s4.hashCode() = " + s4.hashCode());//96354
System.out.println("-------------------------------------");
Student stu1 = new Student("wang1", 11);
Student stu2 = new Student("wang1", 11);
Student stu0 = stu2;
System.out.println("stu1.hashCode() = " + stu1.hashCode());//1580066828
System.out.println("stu2.hashCode() = " + stu2.hashCode());//491044090
System.out.println("stu0.hashCode() = " + stu0.hashCode());//491044090
Student stu3 = new Student();
Student stu4 = new Student();
Student stu5 = stu4;
System.out.println("stu3.hashCode() = " + stu3.hashCode());//644117698
System.out.println("stu4.hashCode() = " + stu4.hashCode());//1872034366
System.out.println("stu5.hashCode() = " + stu5.hashCode());//1872034366
System.out.println("stu3 = " + stu3);
System.out.println("stu4 = " + stu4);
System.out.println("stu5 = " + stu5);
Integer i1 = new Integer(100);
System.out.println("i1.hashCode() = " + i1.hashCode());//100
Integer i2 = 100;
System.out.println("i2.hashCode() = " + i2.hashCode());//100
boolean b = i1==i2;//包装类不可以这么比较
System.out.println("b = " + b);//false
System.out.println(i1.equals(i2));//true
}
}
