hashcode and equality

[Fedry Kemilau]

public class MyList {    
    public static void main(String[] args) {
    Map<Object, Object> m = new HashMap<Object, Object>();
    Dog d = new Dog("clover");
    
    m.put(d, "This is clover dog");
    System.out.println("\n===========");
    System.out.println(m.get(d));// line 1
    
    d.name = "magnolia";
    System.out.println("\n===========");
    System.out.println(m.get(d));// line 2
    
    d.name = "argume";
    System.out.println("\n===========");
    System.out.println(m.get(new Dog("argume")));  // line 3 
    }
}
 
class Dog {
String name;

public Dog(String name){
this.name = name;
}

public boolean equals(Object o){
System.out.println("equality OK");
if (o instanceof Dog && (((Dog) o).name == this.name))
return true;
else
return false;
}

public int hashCode(){
System.out.println("hashcode OK");
return name.length();
}
}

and the result is:

hashcode OK

===========
hashcode OK
This is clover dog

===========
hashcode OK
null

===========
hashcode OK
equality OK
This is clover dog

Question:
Why in the line 1, the code does not check for equality?

Thank you
[ August 11, 2007: Message edited by: fedry kemilau ]

[Srinivasan thoyyeti]

Hi fedry kemilau,

I got you. but after carefully observing your code 15 mins.
Its not that easy for me.

Now into details,

1. Dog d = new Dog("clover");
This line looks simple; but key to whole explanation;

here we are creating Dog in heap with state name = clover;



2. m.put(d, "This is clover dog");
Here d.hashCode() calulated and it will be placed in corresponding bucket.

hashCode() = 6 (length of string).
So its placed in 6th bucket.

Here what need to understood is
Object will only be there in Heap.
just reference will be put in map.


System.out.println("\n===========");

3. System.out.println(m.get(d));// line 1

Here to find which bucket d belongs, it will calculate hashCode on it.
Here d still points to "clover"; so hashCode = 6;

Now we are in 6th bucket;
it will iterate on elements in bucket 6 and takes fisrt object reference.

*** here both the references we passed are same.
so no need to call equals.

Thats why we got the value "This is clover Dog"

4. d.name = "magnolia";

// here the object state on heap changed to "magnolia".
System.out.println("\n===========");


5.System.out.println(m.get(d));// line 2
// hashCode(d) --> 8(length of "magnolia".)
but it founds nothing in the 8th bucket return null.
***Remember, we have only one element in 6th element.


6. d.name = "argume";
Again changing the state of object;
**** Here you are placing d reference in pool. useful in step 7

System.out.println("\n===========");

7. System.out.println(m.get(new Dog("argume"))); // line 3

// This is very complex to understand; prepare for this;

you are now creating another Dog in heap with state "argume";
hashCode returns 6;

so in 6th bucket it will iterate and fetches the first element
here map element's d reference and new Dog("argume")'s references are differnt.

So executes equals() method.
// your equals definition is very poor. not efficient.
// you are comparing both dog object's references. not contents.
Here you need String's pool concept to understand
why reference( Map element's d.name ) == reference(new Dog("argume").name)

reference( Map element's d.name ) >>>
1. Map has only reference to Dog Object; its state changed in 6th step as d.name = "argume";

"argume" will be created in heap and reference placed in pool.
reference(new Dog("argume").name)
2. second time when you refer "argume" it will check in pool and gives that reference[ no new string "argume" created here]

Hence reference( Map element's d.name ) == reference(new Dog("argume").name)

Thanks for your post. which made me think.
[ August 11, 2007: Message edited by: Srinivasan thoyyeti ]

[Fedry Kemilau]

So, the point is if we are using the same reference variable as the get() method's argument, it will not check for the equality, isn't it?
Hmm... I think I have to review again the String pool concept.

Thanks a lot for the explanation.

[Srinivasan thoyyeti]

You are welcome.

But whats that
Solve this code:

100101011000011110110110000110000011100101101111110001111010111110011100001

How did you get this ?
we can only use shift operators on integers.
highest available length is Long 8 bytes.

But that is more than 8 bytes.
How did you get that?
I guess you simply wrote that
Any hidden trick ?
[ August 11, 2007: Message edited by: Srinivasan thoyyeti ]

[Fedry Kemilau]

lol, I did that stuff when i was bored.

Those actually come from String --> char[] --> int[] --> then these ints are converted into binary. Try it if you got nothing to do.

[Manfred Klug]

Originally posted by fedry kemilau:
Those actually come from String --> char[] --> int[] --> then these ints are converted into binary.

Tip: There is an easier way to work with big integers

[Fedry Kemilau]

Hmm.. what is big integer? I think it is not related at all.
What i am doing is just convert; let's say 'a' ==> 1100001, ' ' ==> 0100000.

But looks like Big Integer is a useful thing, gotta read the API.

[Srinivasan thoyyeti]

Hi fedry,

Your bit representation:

f ----> 01100110
e ----> 01100101
d ----> 01100100
r ----> 01110010
y ----> 01111001
  ----> 00100000
k ----> 01101011
e ----> 01100101
m ----> 01101101
i ----> 01101001
l ----> 01101100
a ----> 01100001
u ----> 01110101
"fedry kemilau" in bits  :)  >>>011001100110010101100100011100100
101011011001010110110101101001011011000110000101110101
 
code:
class  BitManipulator{
 
        public  static String nameToBits(String name){
               String bits="";
 
              byte[]  parts = name.getBytes();
               
              for (int partCount =0; partCount < parts.length; partCount++){
            String charInBits =  showBits(parts[partCount]);
                         bits += charInBits;
                         System.out.println(name.charAt(partCount) + " ----> "+ charInBits );
              } 
 
              return( bits );
        }
                 
 
        private static String showBits(byte bits){
               String resultBits="";        
               for (int i = 7;i >=0 ;i--){
 
                        int result = bits & (1 << i);
                        resultBits  += (result >0) ? "1": "0"; 
               } 
               return resultBits;
 
        }       
 
}
 
public class TestBitManipulator{
 
         public static void main(String[] args){
 
                    System.out.println("\"fedry kemilau\" in bits  :)  >>>"+BitManipulator.nameToBits("fedry kemilau"));    
 
         }
 
}

Rejoice.