CODEWITHSUNDEEP
java
Leem.fin
Leem.fin

Reputation: 42672

Why this two simple objects are not equal?

I have a School class:

public class School {

    private String name;
    private int id;
    private boolean isOpen;

    public String getName() {
        return name;
    }
    public void setName(String name) {
        this.name = name;
    }
    public int getId() {
        return id;
    }
    public void setId(int id) {
        this.id = id;
    }
    public boolean isOpen() {
        return isOpen;
    }
    public void setOpen(boolean isOpen) {
        this.isOpen = isOpen;
    }
}

Then I created two instances of School, and compare the equality of the two instances:

public static void main(String[] args) {
        //school1
        School school1 = new School();
        school1.setId(1);
        school1.setName("schoolOne");

        //school2
        School school2 = new School();
        school2.setId(1);
        school2.setName("schoolOne");

            //result is false , why?
        System.out.println("school1 == school2 ? " + school1.equals(school2));

    }

Even though I set the same id and name to school1 & school2 instances, but school1.equals(school2) returns false, why?

Upvotes: 5

Views: 1425

Answers (9)

minani huruka
minani huruka

Reputation: 21

If you do not override the public boolean equals(Object) method, the version in Object.class will be called:

public boolean equals(Object obj) {
    return (this == obj);
}

Just compare the references(If they are exactly the same object)!

So, you have to implement your own equals(Object) in the School.class. Compare these fields:

private String name;  // use String.equals(String)
private int id;  // use ==
private boolean isOpen; // use ==

Upvotes: 1

Mr. Polywhirl
Mr. Polywhirl

Reputation: 48743

You have to override the equals(Object) method:

Place this in your School class:

@Override
public boolean equals(Object other) {
    if (other == this) return true;
    if (other == null || !(other instanceof School)) return false;
    School school = (School) other;
    if (school.id != this.id) return false;
    if (!(school.name.equals(this.name))) return false;
    if (school.isOpen != this.isOpen) return false;
    if (!(school.hashCode().equals(this.hashCode()))) return false;
    return true;
}

If you are going to this, it is also wise to override the hashCode() method as well.

@Override
public int hashCode() {
    final int prime = 31;
    int result = 1;
    result = prime * result + (int) id;
    result = prime * result + (name != null ? name.hashCode() : 0);
    result = prime * result + (isOpen ? 0 : 1);
    return result;
}

Additional Information

I believe this is the best explanation of overriding hashCode().

This answer was posted by dmeister for the following post: SO: Hash Code implementation.

I reference this all the time, and it looks like this functionallity is used in Eclipse when generating the hashCode() method for a given class.

A for nearly all cases reasonable good implementation was proposed in Josh Bloch's "Effective Java" in item 8. The best thing is to look it up there because the author explains there why the approach is good.

A short version:

  1. Create a int result and assign a non-zero value.

  2. For every field tested in the equals-Method, calculate a hash code c by:

    • If the field f is a boolean: calculate (f ? 0 : 1);
    • If the field f is a byte, char, short or int: calculate (int)f;
    • If the field f is a long: calculate (int)(f ^ (f >>> 32));
    • If the field f is a float: calculate Float.floatToIntBits(f);
    • If the field f is a double: calculate Double.doubleToLongBits(f) and handle the return value like every long value;
    • If the field f is an object: Use the result of the hashCode() method or 0 if f == null;
    • If the field f is an array: See every field as separate element and calculate the hash value in a recursive fashion and combine the values as described next.

  3. Combine the hash value c with result with:

    result = 37 * result + c

  4. Return result

    This should result in a proper distribution of hash values for most use situations.

Upvotes: 5

yellowB
yellowB

Reputation: 3020

If you don't override the equals(), then the default equals() in Java.lang.Object will be called:

public boolean equals(Object obj) {
        return (this == obj);
    }

As you see, it compares the references of two object, so in your case it returns false.

If you want to compare the contents of two object, you can:

@Override
public boolean equals(Object obj) {
    // A simple impl. Pls add some checking for null/class type/.. yourself
    return this.name.equals(obj.getName()) && this.id == obj.getId() && this.isOpen == isOpen();
}

Upvotes: 0

Faiz Mohamed Haneef
Faiz Mohamed Haneef

Reputation: 3596

You have created 2 new objects. Now you are comparing 2 object references ... You are not comparing field member values.. So comparison is false.

For primitive datatypes, you wouldn't have this problem.

Upvotes: 1

Suresh Atta
Suresh Atta

Reputation: 122026

Even though I set the same id and name to school1 & school2 instances, but school1.equals(school2) returns false, why?

You need to ovveride the equals() method in your School class. Otherwise the default method implementation from Object class.

see default implementation 

public boolean More ...equals(Object obj) {
        return (this == obj);
    }

In your case it is false, since you are creating two objects. making sense ??

For solution Prefer to read.

Upvotes: 2

Anupam Saini
Anupam Saini

Reputation: 2421

The simple answer is, equals of the implicit super class Object is being used for comparison.

From the documentation:

The equals method for class Object implements the most discriminating possible equivalence relation on objects; that is, for any non-null reference values x and y, this method returns true if and only if x and y refer to the same object (x == y has the value true). Equals method

You would have to override the equals and hashcode method in your Student class.

Upvotes: 0

libik
libik

Reputation: 23049

Imagine twins (in real life), even if they have same look and same age and same name, are they equal? No they are not, they are two different "instances".

It is same in Java. Two different instances cannot be (implicitly) equal, because they each exist independently in their part of memory.

However if you want to compare them like that, you can ovveride equals() method or you can create your own new method for comparing.

Upvotes: 2

TheLostMind
TheLostMind

Reputation: 36304

By default, .equals() does "==" ie., comapring references. You have to override equals().

Upvotes: 0

Abimaran Kugathasan
Abimaran Kugathasan

Reputation: 32488

You have to override the equals() method meaningfully. The default equals() method inherited from Object class check if two reference are referring same object in the memory.

The equals method for class Object implements the most discriminating possible equivalence relation on objects; that is, for any non-null reference values x and y, this method returns true if and only if x and y refer to the same object (x == y has the value true).

Upvotes: 1

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