public class Test {
public static void main(String[] args) {
Map
map = new HashMap<>();
String test = map.get("hello");
System.out.println(test);
运行结果为：
从结果可以看出,HashMap集合中,获取不存在的key时并不会报异常.
在Map的实现类HashMap中有这样一段代码
* Returns the value to which the specified key is mapped,
* or {@code null} if this map contains no mapping for the key.
* <p>More formally, if this map contains a mapping from a key
* {@code k} to a value {@code v} such that {@code (key==null ? k==null :
* key.equals(k))}, then this method returns {@code v}; otherwise
* it returns {@code null}. (There can be at most one such mapping.)
* <p>A return value of {@code null} does not <i>necessarily</i>
* indicate that the map contains no mapping for the key; it's also
* possible that the map explicitly maps the key to {@code null}.
* The {@link #containsKey containsKey} operation may be used to
* distinguish these two cases.
* @see #put(Object, Object)
public V get(Object key) {
Node<K,V> e;
return (e = getNode(hash(key), key)) == null ? null : e.value;
* 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);
if (e.hash == hash &&
((k = e.key) == key || (key != null && key.equals(k))))
return e;
} while ((e = e.next) != null);
return null;
在get方法中并没有向上抛出异常,注释也说明了
返回节点或者null
