请问各位大牛list和set调用contains方法时是比较两个元素的地址，还是调用他们的equals方法呢？
public class Foo {
int value;
public Foo(int value) {
this.value = value;
}
public boolean equals(Object obj) {
if (obj instanceof Foo) {
Foo foo = (Foo) obj;
return value == foo.value;
} else {
return false;
}
}
public static void main(String[] args) {
//运行下面程序段：
ArrayList list = new ArrayList();
HashSet set = new HashSet();
list.add(new Foo(1));
set.add(new Foo(1));
System.out.println(list.contains(new Foo(1)) + ","+ set.contains(new Foo(1)));
//或者
System.out.println(new Foo(1).equals (new Foo(1)) + ","+ set.contains(new Foo(1)));
}
}
我想问的就是为什么这个输出的结果是true,false
public boolean contains(Object o) {
Iterator<E> it = iterator();
if (o==null) {
while (it.hasNext())
if (it.next()==null)
return true;
} else {
while (it.hasNext())
if (o.equals(it.next()))
return true;
}
return false;
}
这个是AbstractCollection的实现, AbstractList, AbstractSet都是以此为父类实现的.
equals的方法，但是值得注意的是二楼说的是String下的equals方法，因为String的equals方法是重写过的，如果题主希望通过contains方法对于一般Object的比较的话，还是得和String一样重写equals方法，确定何种规则
我查了下源码，我觉得HashSet的Hash这里有文章。
1.首先看添加元素的过程
//HashSet代码
/**
* Adds the specified element to this set if it is not already present.
* More formally, adds the specified element <tt>e</tt> to this set if
* this set contains no element <tt>e2</tt> such that
* <tt>(e==null ? e2==null : e.equals(e2))</tt>.
* If this set already contains the element, the call leaves the set
* unchanged and returns <tt>false</tt>.
*
* @param e element to be added to this set
* @return <tt>true</tt> if this set did not already contain the specified
* element
*/
public boolean add(E e) {
return map.put(e, PRESENT)==null;
}向HashSet内部维护的map添加新元素
//HashMap代码
/**
* 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) {
if (table == EMPTY_TABLE) {
inflateTable(threshold);
}
if (key == null)
return putForNullKey(value);
int hash = hash(key);
int i = indexFor(hash, table.length);
for (Entry<K,V> e = table[i]; e != null; e = e.next) {
Object k;
if (e.hash == hash && ((k = e.key) == key || key.equals(k))) {
V oldValue = e.value;
e.value = value;
e.recordAccess(this);
return oldValue;
}
}
modCount++;
addEntry(hash, key, value, i);
return null;
}可以看出会对Element做哈希，所以如果添加Foo(1)的话，保存到HashSet过程中也少不了这个环节的。
再来看比较的情况
//HashSet代码
/**
* Returns <tt>true</tt> if this set contains the specified element.
* More formally, returns <tt>true</tt> if and only if this set
* contains an element <tt>e</tt> such that
* <tt>(o==null ? e==null : o.equals(e))</tt>.
*
* @param o element whose presence in this set is to be tested
* @return <tt>true</tt> if this set contains the specified element
*/
public boolean contains(Object o) {
return map.containsKey(o);
}//HashMap代码
/**
* 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 getEntry(key) != null;
} /**
* Returns the entry associated with the specified key in the
* HashMap. Returns null if the HashMap contains no mapping
* for the key.
*/
final Entry<K,V> getEntry(Object key) {
if (size == 0) {
return null;
}
int hash = (key == null) ? 0 : hash(key);
for (Entry<K,V> e = table[indexFor(hash, table.length)];
e != null;
e = e.next) {
Object k;
if (e.hash == hash &&
((k = e.key) == key || (key != null && key.equals(k))))
return e;
}
return null;
}结论
插入HashSet容器时对元素做哈希，判断容器是否含有某个元素时也要对元素做哈希，他们比较的是元素的哈希值。