CodeGym/Help with Java Tasks/How to ensure that the hashset of Solutions behaves corre...

Level 41

10.06.2020
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How to ensure that the hashset of Solutions behaves correctly?

Question about the task

Under discussion

I'm failing the 2nd last requirement, but I am not sure why. The task description says that the main method isn't tested, so I don't know how to go about passing this test. I'm getting the correct answer as far as I can tell. I'm just spinning in circles trying to solve this.

Solution objects are not being compared correctly. Find the bug and fix it.
The main method is not tested.

Requirements:

The hashcodes of identical objects must be equal.
The equals method should check whether the passed object is equal to the current object (comparison using ==).
The equals method should check whether the passed object is a Solution object.
The equals method must return true if the first and last fields are equal to the passed object and the current one (don't forget that they might be null).
You must ensure that the HashSet of Solution elements behaves correctly.
The Solution class must implement the hashCode method.

package com.codegym.task.task21.task2105; import java.util.HashSet; import java.util.Set; /* Fix the bug. Comparing objects */ public class Solution { private final String first, last; public Solution(String first, String last) { this.first = first; this.last = last; } @Override public boolean equals(Object o) { if (o == this) { return true; } if (o == null) { return false; } if (!(o instanceof Solution)) { return false; } Solution n = (Solution) o; return (n.first == this.first) && (n.last == this.last); } @Override public int hashCode() { return this.first.length() + this.last.length(); } public static void main(String[] args) { Set<Solution> s = new HashSet<>(); s.add(new Solution("Mickey", "Mouse")); System.out.println(s.contains(new Solution("Mickey", "Mouse"))); } }

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Jiyongie

Level 28 , Zielona Gora, Poland

29 March, 10:08

I think public int hashcode() might be the problem I wrote: @Override public int hashCode() { return Objects.hash(first, last); } and it worked

matemate123

Level 50 , Kraków, Poland

28 February 2023, 18:27

I know it's 2,5 years later, but I think this line may be problem:

if (o == null) {
            return false;
        }

I think conditions assume that two object can be "null", and we compare them with "==" because we can't use "equals" to compare "nulls".

return (n.first == this.first) && (n.last == this.last);

Thomas

Level 41 , Bayreuth, Germany

28 February 2023, 18:49

nope, you're confusing something. When you're checking n == null, then you're inside the equals method and have a valid this reference (you can call equals only on a valid Solution object, that object is referenced inside the Solution class as this). If the object you compare against is null, then both can't be equal (this != null). But the solution object has properties (first and last) and those can be null. In both objects (if you reach this point), this and n. You can not compare them using == because it just compares references. That would mean this and n are just equal if they share properties at the same memory address. And that you do not want. You want them to be equal if the strings are equal. Means you need to check if this.first == null and n.first == null, if the are, the properties are equal. If only one is null, then the are not equal. If both are not null, then you need to compare them using equals. That looks in code (easy way Objects.equals)

return Objects.equals(first, n.first) && Objects.equals(last, n.last);

or the hard way

if (first != null ? !n.first.equals(first) : n.first != null) return false;
return last != null ? n.last.equals(last) : n.last == null;

matemate123

Level 50 , Kraków, Poland

28 February 2023, 20:23

Yes I was thought about this two properties but was not precise. But it's good that I wrote because you illuminate subject a little more. Now I see that this Objects.equals is nice method but previously I was confuse because I validate tasks without use nor Objects. nor equals. Why is that?

public boolean equals(Object o) {
        if (this == o)
            return true;
        if (!(o instanceof Solution))
            return false;

        Solution solution = (Solution) o;

        return this.first == solution.first && this.last == solution.last;
    }

    public int hashCode() {
        if (first == null)
            return 0;
        if (last == null)
            return 0;
        return 31 * first.hashCode() + last.hashCode();
    }

Fix the bug. Comparing objects

public boolean equals(Object o) {
        if (o == this) {
            return true;
        }
        if (!(o instanceof Solution))
            return false;
        Solution n = (Solution) o;
        return n.first == this.first && n.last == this.last;
    }

    @Override
    public int hashCode() {
        return Objects.hash(first, last);
    }

Bug in equals/hashCode

@Override
    public boolean equals(Object o) {
        if (this == o) return true;
        if (!(o instanceof Solution)) return false;

        Solution solution1 = (Solution) o;

        if (Double.compare(solution1.aDouble, aDouble) != 0) return false;
        if (anInt != solution1.anInt) return false;
        if (date != null)
            if (date != solution1.date)
                return false;
        if (solution != null)
            if (solution != solution1.solution)
                return false;
        if (string != null)
            if (string != solution1.string)
                return false;
        return true;
    }

Thomas

Level 41 , Bayreuth, Germany

28 February 2023, 21:02

My guess: The person who wrote the test case didn't pay attention to it. You have here string properties. That means to get references pointing to different addresses you have to work with the new operator and not just assign a literal. The latter way would always point to the same address because of string interning.

String s1 = "test";
String s2 = "test";
String s3 = new String("test");

System.out.println(s1 == s2); // true
System.out.println(s1.equals(s2)); // true

System.out.println(s1 == s3); // false
System.out.println(s1.equals(s3)); // true

If they had used objects that have different addresses for each instance by default you wouldn't have been that lucky ;)

matemate123

Level 50 , Kraków, Poland

1 March 2023, 04:57

Yes I get it, I was little confused because I thought "Now when I'm in equals() method I need use == because I can't use equals()" (but fast I remembered that I can use equals() on Strings because this method use override for Strings). I have read now more and I think I understand not bad idea about overriding this two guys. Thanks!

Justin Johnson

Level 31 , Clearwater, United States

11 June 2020, 21:47

return Objects.equals() for the first name && Objects.equals() for the last name

Thomas

Level 41 , Bayreuth, Germany

10 June 2020, 07:42

first and last are strings, so you just compare references with the equals operator