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Question

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The Result of Execute
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import java.util.HashMap;
import java.util.Map;
public class Test {
     * @param args
    public static void main(String[] args) throws Exception{
        Map map = new HashMap();
        Map in1 = new HashMap();
        in1.put("1", "test1");
        in1.put("2", "test2");
        in1.put("3", "test3");
        map.put("1", in1);
        Map in2 = new HashMap();
        in2 = ((HashMap)map.get("1"));
        map.put("2", in2);
        HashMap out1 = (HashMap)map.get("1");
        out1.put("2","testusotukijin");
        map.put("1", out1);
        System.out.println(map);
}       ��RESULT�F{2={3=test3, 2=testusotukijin, 1=test1}, 1={3=test3, 2=testusotukijin, 1=test1}}
I want result of
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RESULT�F{2={3=test3, 2=test2, 1=test1}, 1={3=test3, 2=testusotukijin, 1=test1}}�@�@�@
Help me!

Answer 1

hi
in that program finally in1,in2,out1 has same
reference.
values "test1" "test2" "test3" (strings) will be
stored in string pool.whenever we do something in any
one object ,it will reflect in string pool
so only we are getting that resultThis has nothing to do with the string pool. It has to do with the fact that both "1" and "2" reference the same HashMap.
(S)He does the following:
HashMap map = new HashMap();
HashMap in1 = new HashMap();
// assignments into in 1
map.put("1", in1);
HashMap in2 = new HashMap();
in2 = map.get("1");  // this throws away the previously (in2) created HashMap
/* in2 now references the same HashMap as in1.  Changes to the
* object (HashMap) referenced by either one of these variables
* affects the object (HashMap) referenced by both variables because
* they are THE SAME OBJECT.
*/