HashSet源码&为什么在去重集合中加入自定义对象需要重写equals方法和hashCode方法
写在前面：20221010复习的时候查到了这个问题，在这里记录下
HashSet源码部分
我们知道，HashSet底层是HashMap，所以它的源码其实是HashMap的源码；这里只涉及相关的方法-->put() containsKey() remove()
- put()
- 简单讲解：
- 调用put方法后，其实是再次调用了putVal方法
- 在putVal方法中，首先会判断承载hash点的数组是否为空，如果为空则创建新的
- 保证数组可用之后，会先判断对应hash值得位置是否为空，如果为空则直接加进去
- 如果不为空则开始判断加入结点的哈希码是否等于对应hash位置的哈希码，如果不等则进一步判断key的地址和内容是否相同-->其中地址使用==，内容使用equals（这里其实就揭示了为什么要重写hashCode和equals方法，如果不重写则会调用Object类的方法，而所有新生成的对象Object类中的HashCode都不相同，所以会在加入过程中加入重复的元素）
- 如果都不相等，则赋值然后覆盖老的value（这里有一个onluIfAbsent参数，如果这个值为true，则只有当老的value为null时才会覆盖）
- 如果上一个步骤的判断不成立，则会开始判断是否是TreeNode，如果是TreeNode则将工作交给Tree的putTreeVal方法（说明已经超过八个元素了，变换为红黑树，而红黑树插入过程中会判重）
- 如果上一个步骤还不成立，则开始遍历该hash结点的链表，如果遍历到尾部还没有找到相同key的node，就直接插入，插入后长度如果大于设置的阈值，则开始转化为红黑树，如果找到则直接退出循环，开始赋值过程
- 最后插入完毕，开始内部的计数，如果大小大于阈值，也会触发resize
/**
* Associates the specified value with the specified key in this map.
* If the map previously contained a mapping for the key, the old
* value is replaced.
*
* @param key key with which the specified value is to be associated
* @param value value to be associated with the specified key
* @return the previous value associated with <tt>key</tt>, or
* <tt>null</tt> if there was no mapping for <tt>key</tt>.
* (A <tt>null</tt> return can also indicate that the map
* previously associated <tt>null</tt> with <tt>key</tt>.)
*/
public V put(K key, V value) {
return putVal(hash(key), key, value, false, true);
}
/**
* Implements Map.put and related methods.
*
* @param hash hash for key
* @param key the key
* @param value the value to put
* @param onlyIfAbsent if true, don't change existing value
* @param evict if false, the table is in creation mode.
* @return previous value, or null if none
*/
final V putVal(int hash, K key, V value, boolean onlyIfAbsent,
boolean evict) {
Node<K,V>[] tab; Node<K,V> p; int n, i;
if ((tab = table) == null || (n = tab.length) == 0)
n = (tab = resize()).length;
if ((p = tab[i = (n - 1) & hash]) == null)
tab[i] = newNode(hash, key, value, null);
else {
Node<K,V> e; K k;
if (p.hash == hash &&
((k = p.key) == key || (key != null && key.equals(k))))
e = p;
else if (p instanceof TreeNode)
e = ((TreeNode<K,V>)p).putTreeVal(this, tab, hash, key, value);
else {
for (int binCount = 0; ; ++binCount) {
if ((e = p.next) == null) {
p.next = newNode(hash, key, value, null);
if (binCount >= TREEIFY_THRESHOLD - 1) // -1 for 1st
treeifyBin(tab, hash);
break;
}
if (e.hash == hash &&
((k = e.key) == key || (key != null && key.equals(k))))
break;
p = e;
}
}
if (e != null) { // existing mapping for key
V oldValue = e.value;
if (!onlyIfAbsent || oldValue == null)
e.value = value;
afterNodeAccess(e);
return oldValue;
}
}
++modCount;
if (++size > threshold)
resize();
afterNodeInsertion(evict);
return null;
}
- containsKey()
- 简单讲解：
- 内部再次调用getNode方法，如果返回不为空，则返回true，否则为false
- 先判断数组是否为空，如果为空直接返回null
- 如果不为空，则开始判断hash对应结点和其下的链表结点是否有重复的值，如果有则返回它，如果没有则返回null
/**
* Returns <tt>true</tt> if this map contains a mapping for the
* specified key.
*
* @param key The key whose presence in this map is to be tested
* @return <tt>true</tt> if this map contains a mapping for the specified
* key.
*/
public boolean containsKey(Object key) {
return getNode(hash(key), key) != null;
}
/**
* Implements Map.get and related methods.
*
* @param hash hash for key
* @param key the key
* @return the node, or null if none
*/
final Node<K,V> getNode(int hash, Object key) {
Node<K,V>[] tab; Node<K,V> first, e; int n; K k;
if ((tab = table) != null && (n = tab.length) > 0 &&
(first = tab[(n - 1) & hash]) != null) {
if (first.hash == hash && // always check first node
((k = first.key) == key || (key != null && key.equals(k))))
return first;
if ((e = first.next) != null) {
if (first instanceof TreeNode)
return ((TreeNode<K,V>)first).getTreeNode(hash, key);
do {
if (e.hash == hash &&
((k = e.key) == key || (key != null && key.equals(k))))
return e;
} while ((e = e.next) != null);
}
}
return null;
}
- remove()
- 还是再次调用了removeNode方法，内部和上述两个差不多，无非就是循环判断链表，如果节点类型为树节点，则转为循环判断红黑树
/**
* Removes the mapping for the specified key from this map if present.
*
* @param key key whose mapping is to be removed from the map
* @return the previous value associated with <tt>key</tt>, or
* <tt>null</tt> if there was no mapping for <tt>key</tt>.
* (A <tt>null</tt> return can also indicate that the map
* previously associated <tt>null</tt> with <tt>key</tt>.)
*/
public V remove(Object key) {
Node<K,V> e;
return (e = removeNode(hash(key), key, null, false, true)) == null ?
null : e.value;
}
/**
* Implements Map.remove and related methods.
*
* @param hash hash for key
* @param key the key
* @param value the value to match if matchValue, else ignored
* @param matchValue if true only remove if value is equal
* @param movable if false do not move other nodes while removing
* @return the node, or null if none
*/
final Node<K,V> removeNode(int hash, Object key, Object value,
boolean matchValue, boolean movable) {
Node<K,V>[] tab; Node<K,V> p; int n, index;
if ((tab = table) != null && (n = tab.length) > 0 &&
(p = tab[index = (n - 1) & hash]) != null) {
Node<K,V> node = null, e; K k; V v;
if (p.hash == hash &&
((k = p.key) == key || (key != null && key.equals(k))))
node = p;
else if ((e = p.next) != null) {
if (p instanceof TreeNode)
node = ((TreeNode<K,V>)p).getTreeNode(hash, key);
else {
do {
if (e.hash == hash &&
((k = e.key) == key ||
(key != null && key.equals(k)))) {
node = e;
break;
}
p = e;
} while ((e = e.next) != null);
}
}
if (node != null && (!matchValue || (v = node.value) == value ||
(value != null && value.equals(v)))) {
if (node instanceof TreeNode)
((TreeNode<K,V>)node).removeTreeNode(this, tab, movable);
else if (node == p)
tab[index] = node.next;
else
p.next = node.next;
++modCount;
--size;
afterNodeRemoval(node);
return node;
}
}
return null;
}
为什么在去重集合中加入自定义对象需要重写equals方法和hashCode方法
- 其实在上面的put方法讲解中已经提到了，这两个方法其实是支持我们自定义判重的条件的，如果不重写，则会使用Object的HashCode判断，则不一定符合我们的预期
以上
希望对后来者有所帮助