1

I was make some code and found that objects ar eno equals - it is trivial question but not understand how default equals works.

class A {
    String id;
    public A(String id) {
        this.id = id;
    }

    public static void main(String args[])
    {
        A a = new A("1");
        A b = new A("1");
        System.out.println(a.id);
        System.out.println(b.id);
        System.out.println(a.equals(b));
    }
}

Result is:

1
1
false

But I want to have a.equals(b) == true why it is false?

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1
  • 8
    You need to override .equals() Commented Feb 12, 2013 at 1:41

4 Answers 4

Reset to default
4

Your class currently extends only Object class and in Object class equals method looks like this

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

What you need is to override this method, for example like this 

@Override
public boolean equals(Object obj) {
    if (this == obj)
        return true;
    if (obj == null)
        return false;
    if (getClass() != obj.getClass())
        return false;
    A other = (A) obj;
    if (id == other.id)
        return true;
    if (id == null)
        return false;
    if (other.id == null)
        return false;
    if (!this.id.equals(other.id))
        return false;
    return true;
}

Also when you override equals you probably should override hashCode method, but this is not subject of your question. You can read more about it here.

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1 Comment

Fixed other.id is failing before (null.id -> null exception).
4

If you don't override equals() on the object, you are comparing two different memory references. So override equals() to compare the id fields.

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1 Comment

Good hint but @Pschemo has full solution.
1

It overrides Object's equals method by default, it checks the "same object" rather than "same content". If you want to have a.equals(b) == true, you should override it:

@Override
public boolean equals (Object obj) {
    if (obj instanceof A) {
        A a = (A) obj;
        if (id == null) {
            return a.id == null;
        } else {
            return id.equals(a.id);
        }
    }
    return false;
}

----- EDITED -----

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7 Comments

Good but code is not clear (not for me) - improve code style you could be good but not readable.
Some bugs in code since null of obj + null of this.is or obj.id.
Good but still code style is ugly (could be faster too) - it is good advice for you the most important in writing large program is readability not code compression - it is major factor of quality. You focus on compression of code (side effect is complication and hard to understand code or review it) change it since it blocks development.
Some example !obj.getClass().equals(A.class) better is simple obj.getClass() != A.getClass()) (who know what do equals, why to check it, why negate result).
Some redundancy if (id == null) return id == ((A) obj).id; == if (id == null) return id == obj.id; better is if (id == null && obj.id == null) return true; else return false; more clear not need understand transition (if id == null so id == obj.id mean obj.id == null faster reading/review - avoid logic puzzles as much as possible. Maybe I help you little.
|
0

you should rewrite an equals() method for your code, as you would a toString() method.

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