Java中hashCode和equals相关问题阐述

本文主要针对如下三个问题进行解释：

• 默认情况下hashCode相同是不是意味着equals方法相等？
• 默认情况下equals方法相等是不是意味着hashCode相同？
• 重写equals方法是不是需要重写hashCode方法？为什么？

默认情况下hashCode相同是不是意味着equals方法相等和问题?equals方法相等是不是意味着hashCode相同?

之所以将这两个问题放在一起，是因为两个问题可以联系在一起回答，在Object类中的hashCode和equals方法中已经有了该问题的答案

image.png

图中红色区域的意思为：如果两个对象根据equals方法判定相等，那么这两个对象的hashCode方法必定是相同的integer的整形值。其中暗含了两层意思：

1. equals相等的两个对象，其hashCode必定相等
2. 通过equals判定前，必定有hashCode值比较判断的步骤

看到这里我们自然疑惑hashCode方法是如何得到某个对象的hash值的，我们再看如下这句话

image.png

其中说到hashCode方法一种典型的实现是将对象在堆内的地址通过某种手段转成一个integer整形值，但是该方法是native修饰的，需要通过查阅openjdk的源码得到，查阅相关资料得到真正对应的hashCode生成方法如下

intptr_t ObjectSynchronizer::FastHashCode (Thread * Self, oop obj) {
  if (UseBiasedLocking) {
    // NOTE: many places throughout the JVM do not expect a safepoint
    // to be taken here, in particular most operations on perm gen
    // objects. However, we only ever bias Java instances and all of
    // the call sites of identity_hash that might revoke biases have
    // been checked to make sure they can handle a safepoint. The
    // added check of the bias pattern is to avoid useless calls to
    // thread-local storage.
    if (obj->mark()->has_bias_pattern()) {
      // Box and unbox the raw reference just in case we cause a STW safepoint.
      Handle hobj (Self, obj) ;
      // Relaxing assertion for bug 6320749.
      assert (Universe::verify_in_progress() ||
              !SafepointSynchronize::is_at_safepoint(),
             biases should not be seen by VM thread here);
      BiasedLocking::revoke_and_rebias(hobj, false, JavaThread::current());
      obj = hobj() ;
      assert(!obj->mark()->has_bias_pattern(), biases should be revoked by now);
    }
  }
 
  // hashCode() is a heap mutator ...
  // Relaxing assertion for bug 6320749.
  assert (Universe::verify_in_progress() ||
          !SafepointSynchronize::is_at_safepoint(), invariant) ;
  assert (Universe::verify_in_progress() ||
          Self->is_Java_thread() , invariant) ;
  assert (Universe::verify_in_progress() ||
         ((JavaThread *)Self)->thread_state() != _thread_blocked, invariant) ;
 
  ObjectMonitor* monitor = NULL;
  markOop temp, test;
  intptr_t hash;
  markOop mark = ReadStableMark (obj);
 
  // object should remain ineligible for biased locking
  assert (!mark->has_bias_pattern(), invariant) ;
 
  if (mark->is_neutral()) {
    hash = mark->hash();              // this is a normal header
    if (hash) {                       // if it has hash, just return it
      return hash;
    }
    hash = get_next_hash(Self, obj);  // allocate a new hash code
    temp = mark->copy_set_hash(hash); // merge the hash code into header
    // use (machine word version) atomic operation to install the hash
    test = (markOop) Atomic::cmpxchg_ptr(temp, obj->mark_addr(), mark);
    if (test == mark) {
      return hash;
    }
    // If atomic operation failed, we must inflate the header
    // into heavy weight monitor. We could add more code here
    // for fast path, but it does not worth the complexity.
  } else if (mark->has_monitor()) {
    monitor = mark->monitor();
    temp = monitor->header();
    assert (temp->is_neutral(), invariant) ;
    hash = temp->hash();
    if (hash) {
      return hash;
    }
    // Skip to the following code to reduce code size
  } else if (Self->is_lock_owned((address)mark->locker())) {
    temp = mark->displaced_mark_helper(); // this is a lightweight monitor owned
    assert (temp->is_neutral(), invariant) ;
    hash = temp->hash();              // by current thread, check if the displaced
    if (hash) {                       // header contains hash code
      return hash;
    }
    // WARNING:
    //   The displaced header is strictly immutable.
    // It can NOT be changed in ANY cases. So we have
    // to inflate the header into heavyweight monitor
    // even the current thread owns the lock. The reason
    // is the BasicLock (stack slot) will be asynchronously
    // read by other threads during the inflate() function.
    // Any change to stack may not propagate to other threads
    // correctly.
  }
 
  // Inflate the monitor to set hash code
  monitor = ObjectSynchronizer::inflate(Self, obj);
  // Load displaced header and check it has hash code
  mark = monitor->header();
  assert (mark->is_neutral(), invariant) ;
  hash = mark->hash();
  if (hash == 0) {
    hash = get_next_hash(Self, obj);
    temp = mark->copy_set_hash(hash); // merge hash code into header
    assert (temp->is_neutral(), invariant) ;
    test = (markOop) Atomic::cmpxchg_ptr(temp, monitor, mark);
    if (test != mark) {
      // The only update to the header in the monitor (outside GC)
      // is install the hash code. If someone add new usage of
      // displaced header, please update this code
      hash = test->hash();
      assert (test->is_neutral(), invariant) ;
      assert (hash != 0, Trivial unexpected object/monitor header usage.);
    }
  }
  // We finally get the hash  
  return hash;

重写equals方法是不是需要重写hashCode方法？为什么？

首先该问题的答案仍然在Object中的equals方法注释中写的很清楚，如下图所示

image.png

按红框处的官方解释来说，只要equals方法被重写了就必须重写hashCode方法，此处也解释了“必须”的原因，要维持hashCode方法的contract约定，hashCode方法中申明了相同的对象必须有相同的hash code。
为了进一步加深对该“必须”的理解，这里又从两个方面举例说明：
<li> 自己创建一个自定义对象，只重写equals方法而不重写hashCode方法，看看有什么问题
<li> 从HashMap源码的角度分析一下，如果只重写equals而不重写hashCode有什么问题

<p> 自己创建一个Person对象，有pname和age字段，如下所示

public class Person implements Serializable {
    private static final long serialVersionUID = 7592930394427200495L;

    private String pname;
    private int age;

    public Person() {

    }

    public Person(String pname, int age) {
        this.pname = pname;
        this.age = age;
    }

    public String getPname() {
        return pname;
    }

    public void setPname(String pname) {
        this.pname = pname;
    }

    public int getAge() {
        return age;
    }

    public void setAge(int age) {
        this.age = age;
    }

    @Override
    public boolean equals(Object o) {
        if (this == o) return true;
        if (!(o instanceof Person)) return false;

        Person person = (Person) o;

        if (age != person.age) return false;
        return !(pname != null ? !pname.equals(person.pname) : person.pname != null);

    }

//    @Override
//    public int hashCode() {
//        int result = pname != null ? pname.hashCode() : 0;
//        result = 31 * result + age;
//        return result;
//    }
}

<p> 进行测试

@Test
    public void fun() {
        Person p1 = new Person("lisi", 15);
        Person p2 = new Person("lisi", 15);
        Assert.assertEquals(false, p1.equals(p2));
    }

<p> 结果如下

image.png

p1和p2在java堆中肯定分属不同的Person实例对象，其地址必定不相同，但因为我们仅仅重写了equals方法，只对pname和age的值进行了比对从而导致了结果的错误，如果重写了hashCode方法，根据两个实例地址的相关算法进行判断就会避免这个问题

同样的我们再来分析HashMap中的一段源码再次说明hashCode和equals方法同时重写的重要性，其中的关键点在于put操作时的逻辑

image.png

当新元素放入HashMap时，会首先计算出该元素对应放在哪一个Entry链表上（HashMap原理不了解的请查阅相关文档），然后通过和链表上的每一个元素比较，来判断新加入元素是否是重复元素，而判断重复元素的思路就体现了两个方法协同的重要性，首先会判断两个元素的hash值是否相等，再判断两个元素equals是否相等，设想一下，如果没有重写元素的hashCode方法，那么就有可能存在这种可能，两个元素不等，但hash code相等，重写的equals也相等（如Person例中），从而导致错误的覆盖