首先,看一下源码:
public V put(K paramK, V paramV)
{
if (paramK == null) {
return putForNullKey(paramV);
}
int i = hash(paramK.hashCode());
int j = indexFor(i, this.table.length);
for (Entry localEntry = this.table[j]; localEntry != null; localEntry = localEntry.next)
{
Object localObject1;
if ((localEntry.hash == i) && (((localObject1 = localEntry.key) == paramK) || (paramK.equals(localObject1))))
{
Object localObject2 = localEntry.value;
localEntry.value = paramV;
localEntry.recordAccess(this);
return localObject2;
}
}
this.modCount += 1;
addEntry(i, paramK, paramV, j);
return null;
}
好吧,看不懂没关系,下面看例子:
import java.util.HashMap;
import java.util.Map;
public class Test {
public static void main(String[] args) {
Map<String, String> map = new HashMap<String, String>();
String p1 = map.put("11", "22");
System.out.println("p1:" + p1);
String p2 = map.put("33", "44");
System.out.println("p2:" + p2);
String value1 = map.get("11");
System.out.println("value1:" + value1);
String p3 = map.put("11", "44");
System.out.println("p3:" + p3);
String value2 = map.get("11");
System.out.println("value2:" + value2);
}
}
输出结果:
p1:null
p2:null
value1:22
p3:22
value2:44
说明:put方法返回值为null或者value;
如果key没有重复,put成功,则返回null,如p1、p2;
如果key重复了,返回的是map.get(key),也就是当前这个key对应的value,如上面的p3,key=”11″,而p1的key也是”11″,p1与p3重复,返回的是p1的value=”22″,并且将p3覆盖掉p1