Java - Overriding hashCode and toString

Question

When two objects have same value of ele in class A then those two objects are equal. So I have overridden toString and hashCode to return the object's ele (not considering the value of s anywhere).

public class A {
    private int ele;
    private String s;

    public int getEle() {
        return ele;
    }

    public void setEle(int ele) {
        this.ele = ele;
    }

    public String getS() {
        return s;
    }

    public void setS(String s) {
        this.s = s;
    }

    @Override
    public int hashCode(){
        return ele;
    }

    @Override
    public String toString() {
        return String.valueOf(ele);
    }
}

public static void main(String[] args) {

    Map<A, String> map = new HashMap<>();
    A a1 = new A();
    a1.setEle(10);
    a1.setS("abc");

    A a2 = new A();
    a2.setEle(10);
    a2.setS("efg");

    map.put(a1, "val1");
    map.put(a2, "val2");

    System.out.println(map.get(a1));
    System.out.println(map.get(a2));

}

Output:

val1
val2

But if I put value of a1 and a2 in a map, I was expecting either val1 or val2 to be returned for both map.get(a1) and map.get(a2).

score 7 · Accepted Answer · answered Mar 29 '16 at 17:06

7

Sure, a1 and a2 have the same hash code, but they weren't considered equal because you didn't override equals to consider two A objects with the same ele to be equal. A map will use equals to the final ruler on equality after it uses the hash code. The map will place both objects in the same bucket, but because they aren't equal, it will keep both.

Override equals so that it returns true if the other object is an A and they both have the same ele. Then you will see that val2 will be returned for both get calls.

answered Mar 29 '16 at 17:06

rgettman

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Oh ya.. My bad. I thought it uses toString to get the string representation and then compare – Exuberant Mar 29 '16 at 17:12
But I guess it uses a list when chaining. But I enter val1 first. But I get val2 for both. Why? – Exuberant Mar 29 '16 at 17:12
When you `put` `a2` into the map, the map considers it equal to `a1`, because `equals` returns `true`. The map then _replaces_ the key-value pair (an `Entry`) that was there, `a1 -> "val1"`, with `a2` -> "val2". Both your `get` calls will result in `equals` returning `true`, so the current value, `"val2"` is retrieved each time. – rgettman Mar 29 '16 at 17:15
1

Historically, `HashMap` has always used a linked list of nodes that hash to the same bucket. Starting in Java 8, `HashMap` switches over to a tree map-like structure when the number of entries in the bucket reaches a threshold. My Java 8 source code indicates that this threshold is currently 8. – rgettman Mar 29 '16 at 17:26

score 1 · Answer 2 · answered Mar 29 '16 at 17:09

You need to implement equals() to take ele value into consideration when adding to a map, i.e.:

public class A {

    private int ele;
    private String s;

    public int getEle() {
        return ele;
    }

    public void setEle(int ele) {
        this.ele = ele;
    }

    public String getS() {
        return s;
    }

    public void setS(String s) {
        this.s = s;
    }

    @Override
    public boolean equals(Object o) {
        if (this == o) return true;
        if (o == null || getClass() != o.getClass()) return false;

        A a = (A) o;

        return ele == a.ele;

    }

    @Override
    public int hashCode() {
        return ele;
    }
}

This will make you return only one value as you want.

score 0 · Answer 3 · answered Mar 29 '16 at 17:09

0

every time you use new keyword it makes a new object in heap Memory. So, a1 and a2 both are different Object in actual.

Please Refer this for more info about new keyword What New keyword do Internally in Java

answered Mar 29 '16 at 17:09

Vikrant Kashyap

6,398
3
32
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Java - Overriding hashCode and toString

3 Answers3