java中为何重写equals时必须重写hashCode方法详解

/**
 * Returns a hash code value for the  . This method is
 * supported for the benefit of hash tables such as those provided by
 * {@  java.util.HashMap}.
 * <p>
 * The general contract of {@code hashCode} is:
 * <ul>
 * <li>Whenever it is invoked on the same   more than once during
 * an execution of a Java application, the {@code hashCode} method
 * must consistently return the same integer, provided no information
 * used in {@code equals} comparisons on the   is modified.
 * This integer need not remain consistent from one execution of an
 * application to another execution of the same application.
 * <li>If two  s are equal according to the {@code equals( )}
 * method, then calling the {@code hashCode} method on each of
 * the two  s must produce the same integer result.
 * <li>It is <em>not</em> required that if two  s are unequal
 * according to the {@  java.lang. #equals(java.lang. )}
 * method, then calling the {@code hashCode} method on each of the
 * two  s must produce distinct integer results. However, the
 * programmer should be aware that producing distinct integer results
 * for unequal  s may improve the performance of hash tables.
 * </ul>
 * <p>
 * As much as is reasonably practical, the hashCode method defined by
 * class {@code  } does return distinct integers for distinct
 *  s. (This is typically implemented by converting the internal
 * address of the   into an integer, but this implementation
 * technique is not required by the
 * Java™ programming language.)
 *
 * @return a hash code value for this  .
 * @see java.lang. #equals(java.lang. )
 * @see java.lang.System#identityHashCode
 */
 public native int hashCode();

/**
 * Indicates whether some other   is \"equal to\" this one.
 * <p>
 * The {@code equals} method implements an equivalence relation
 * on non-null   references:
 * <ul>
 * <li>It is <i>reflexive</i>: for any non-null reference value
 * {@code x}, {@code x.equals(x)} should return
 * {@code true}.
 * <li>It is <i>symmetric</i>: for any non-null reference values
 * {@code x} and {@code y}, {@code x.equals(y)}
 * should return {@code true} if and only if
 * {@code y.equals(x)} returns {@code true}.
 * <li>It is <i>transitive</i>: for any non-null reference values
 * {@code x}, {@code y}, and {@code z}, if
 * {@code x.equals(y)} returns {@code true} and
 * {@code y.equals(z)} returns {@code true}, then
 * {@code x.equals(z)} should return {@code true}.
 * <li>It is <i>consistent</i>: for any non-null reference values
 * {@code x} and {@code y}, multiple invocations of
 * {@code x.equals(y)} consistently return {@code true}
 * or consistently return {@code false}, provided no
 * information used in {@code equals} comparisons on the
 *  s is modified.
 * <li>For any non-null reference value {@code x},
 * {@code x.equals(null)} should return {@code false}.
 * </ul>
 * <p>
 * The {@code equals} method for class {@code  } implements
 * the most discriminating possible equivalence relation on  s;
 * that is, for any non-null reference values {@code x} and
 * {@code y}, this method returns {@code true} if and only
 * if {@code x} and {@code y} refer to the same  
 * ({@code x == y} has the value {@code true}).
 * <p>
 * Note that it is generally necessary to override the {@code hashCode}
 * method whenever this method is overridden, so as to maintain the
 * general contract for the {@code hashCode} method, which states
 * that equal  s must have equal hash codes.
 *
 * @param obj the reference   with which to compare.
 * @return {@code true} if this   is the same as the obj
 *  argument; {@code false} otherwise.
 * @see #hashCode()
 * @see java.util.HashMap
 */
 public boolean equals(  obj) {
 return (this == obj);
 }

/**
 * @see Person
 * @param args
 */
 public static void main(String[] args)
 {
 HashMap<Person, Integer> map = new HashMap<Person, Integer>();

 Person p = new Person(\"jack\",22,\"男\");
 Person p1 = new Person(\"jack\",22,\"男\");

 System.out.println(\"p的hashCode:\"+p.hashCode());
 System.out.println(\"p1的hashCode:\"+p1.hashCode());
 System.out.println(p.equals(p1));
 System.out.println(p == p1);

 map.put(p,888);
 map.put(p1,888);
 map.forEach((key,val)->{
  System.out.println(key);
  System.out.println(val);
 });
 }

public class Person
{
 private String name;

 private int age;

 private String sex;

 Person(String name,int age,String sex){
 this.name = name;
 this.age = age;
 this.sex = sex;
 }
}

p的hashCode:356573597
p1的hashCode:1735600054
false
false
com.blueskyli.练习.Person@677327b6
com.blueskyli.练习.Person@1540e19d

public class Person
{
 private String name;

 private int age;

 private String sex;

 Person(String name,int age,String sex){
 this.name = name;
 this.age = age;
 this.sex = sex;
 }

 @Override public boolean equals(  obj)
 {
 if(obj instanceof Person){
  Person person = (Person)obj;
  return name.equals(person.name);
 }
 return super.equals(obj);
 }
}

p的hashCode:356573597
p1的hashCode:1735600054
true
false
com.blueskyli.练习.Person@677327b6
com.blueskyli.练习.Person@1540e19d

public class Person
{
 private String name;

 private int age;

 private String sex;

 Person(String name,int age,String sex){
 this.name = name;
 this.age = age;
 this.sex = sex;
 }

 @Override public boolean equals(  obj)
 {
 if(obj instanceof Person){
  Person person = (Person)obj;
  return name.equals(person.name);
 }
 return super.equals(obj);
 }

 @Override public int hashCode()
 {
 return name.hashCode();
 }
}

p的hashCode:3254239
p1的hashCode:3254239
true
false
com.blueskyli.练习.Person@31a7df