equal() and hashcode()..

[Preethi Dev]

Hi,
I got it from Master Exam.
11.x=0;
12.if(x1.hashcode()!=x2.hashcode())x=x+1;
13.if(x3.equals(x4)==false)x=x+10;
14.if(x5.equals(x6)==true)x=x+100;
15.if(x7.hashcode()==x8.hashcod())x=x+1000;
16.System.out.println("x=" + x);

if the output is "x=1111", which of the following will always be true?

Answer given: x2.equals(x1)==true
I think this answer violates the contract.the correct answer would be x5.hashcode()==x6.hascode().
other than "x2.equals(x1)" the rest are coming under "can be true" so not suitable for the question asked.

please do correct me if i am wrong.

Thanks
Preetha

[M Srilatha]

Hi,

I guess it would be better if you can give all the options.
Then it would be easier!

[Ankit Garg]

Under the given circumstances, Arun is right. Since the hashCodes don't match, so the objects can never be equal...

[subhasish nag]

Hi Pritha,
you are right.

[geeta vemula]

Hi preetha,

yes the answer is x5.hashcode==x6.hashcode and its given correct in master exam.

[patrick avery]

Like M Srilatha, I would also like to see all of the original choices for the answers to the question.

There is a good Mock exam (7 questions) on the hashCodes and equals contract for Inquisition (http://enigma.vm.bytemark.co.uk/inquisition/index.php/Main_Page).

I find that there are really only 2 rules to remmber, i.e.
Rule 1 if x1.equals(x2)==true then x1.hashCode()==x2.hashCode()
Rule 2 if x1.hashCode()!=x2.hashCode() then x1.equals(x2)==false

because if x1.equals(x2)==false
then nothing can be deduced about hashCode()
and if x1.hashcode()==x2.hascode()
then nothing can be deduced about equals()

Please someone correct me if I have incorrectly oversimplified it.

Using these rules on the problem as given,

line 12,
12. if(x1.hashcode()!=x2.hashcode())x=x+1;
easily disproves the answer:
x2.equals(x1)==true

and line 14,
14. if(x5.equals(x6)==true)x=x+100;
easily proves the answer:
x5.hashcode()==x6.hascode()

[Fabio Nascimento]

The rules to remember are:

If x.equals(y)==true then x.hashCode()==y.hashCode() MUST be true;
If x.equals(y)==false then x.hashCode()==u.hashCode() MIGHT be true;

And these rules are applied on Collections that uses Hash Codes for fast searching, because they first search the bucket (hashCode) before search the object (equals), otherwise equals() have nothing to do with hashCode().

For instance:

    
class T1 {

int id;

T1 (int id) {
this.id = id;
}

public boolean equals(Object c) {
return ((T1) c).id == this.id;
}
 
public int hashCode() {

Random r = new Random();

return Math.abs(r.nextInt());

}

}
 
public class Test {
 
public static void main(String[] args) {
 
T1 t1 = new T1(1);

ArrayList<T1> arrayList = new ArrayList<T1>();
arrayList.add(t1);

T1 t2 = new T1(1);

for (T1 t : arrayList) {
if (t.equals(t2)) System.out.println("Achei");
}

System.out.println(t1.hashCode());
System.out.println(t2.hashCode());
 
}
 
}

They will be equals even with differente hashCodes, but they will not be found in a collection that uses hash codes for fast searching using its own implemented methods.
[ December 05, 2008: Message edited by: Fabio Nascimento ]

[Asanka Vithanage]

hay ranchers,

this is not a big problem if you know little pure mathematics

If A implies B

we can say (Not)B implies (Not)A
thats all to remember

we can think A as equals() method return true and
B as two hashcode values are same


hopes you can understand