What is the difference between Comparable and Comparator Interface?


In Java, both the Comparable and Comparator interfaces are used for comparing objects, but they have different ways of achieving this.

Comparable: The Comparable interface is used to define a natural ordering of objects. When a class implements Comparable, it defines a default ordering of objects of that class. The Comparable interface contains a single method, compareTo(), which takes an object of the same type as the implementing class and returns an integer value indicating whether the object is greater than, equal to, or less than the object being compared to. The compareTo() method is used by collections, such as TreeSet and TreeMap, to sort elements.

Comparator: The Comparator interface is used to define a custom ordering of objects that do not implement Comparable or when a different ordering is required. The Comparator interface contains two methods, compare() and equals(). The compare() method takes two objects of the type being compared and returns an integer value indicating whether the first object is greater than, equal to, or less than the second object. The equals() method is used to determine if two comparators are equal.

In summary, the Comparable interface provides a natural ordering of objects, while the Comparator interface provides a custom ordering of objects. The Comparable interface is implemented by the class being compared, while the Comparator interface is implemented by a separate class.

Comparable InterfaceComparator Interface
OrderingNatural orderingCustom ordering
Implemented byThe class being comparedA separate class
MethodcompareTo()compare()
PurposeTo define a default ordering of objectsTo define a custom ordering of objects
UsageUsed by collections like TreeSet and TreeMapUsed to sort objects in collections like ArrayList or arrays
Number of methodsOne - compareTo()Two - compare() and equals()
Return typeintint

Example Program for Comparable

import java.util.*; 
class Person implements Comparable<Person> { 
private String name;
private int age;

    

public Person(String name, int age) {
        
this.name = name;
        
this.age = age;

}

    

public String getName() {
        
return this.name;

}

public int getAge() {
        
return this.age;
}


@Override
    
public int compareTo(Person other) {
        
return Integer.compare(this.age, other.age);

}

}



public class ComparableExample {
    
public static void main(String[] args) {

List<Person> people = new ArrayList<>();

people.add(new Person("Alice", 25));
people.add(new Person("Bob", 30));

people.add(new Person("Charlie", 20));


Collections.sort(people);

        

for (Person person : people) {

System.out.println(person.getName() + " - " + person.getAge());

}

}
}    

Example Program for Comparator


import java.util.*;



class Person {
    
private String name;
    
private int age;

    

public Person(String name, int age) {
        
this.name = name;
        
this.age = age;

}

    

public String getName() {
        
return this.name;

}

    

public int getAge() {
        
return this.age;

}
}



class AgeComparator implements Comparator<Person> {


@Override
    public int compare(Person p1, Person p2) {
        
return Integer.compare(p1.getAge(), p2.getAge());

}

}



public class ComparatorExample {
    

public static void main(String[] args) {


List<Person> people = new ArrayList<>();

people.add(new Person("Alice", 25));
people.add(new Person("Bob", 30));

people.add(new Person("Charlie", 20));


AgeComparator ageComparator = new AgeComparator();


Collections.sort(people, ageComparator);

        

for (Person person : people) {

System.out.println(person.getName() + " - " + person.getAge());

}

}
}


The output for both the program is:
Charlie - 20 Alice - 25 Bob - 30

Comments

Post a Comment

Popular posts from this blog

Why Pointers are Eliminated in Java?

Image
 This is rare question but it is good to get prepared with such type of Questions as it may be asked in Beginner level. The first motivation is to make Java programming less complicated. Java is an easy-to-learn, object-oriented programming language. As is well known, the pointer idea is very confusing, challenging to understand (especially for beginners), and leads to coding errors.  The pointers concept was eliminated by the designers of Java in order to make it less complicated and more straightforward. Memory can be allocated and deallocated via pointers. Although this is a useful aspect of pointers, manually managing memory can occasionally be challenging for programmers. Pointers are not required because Java has an automatic garbage collection mechanism, which simplifies memory management. One of the major problems in the field of programming is security. Direct access to a memory address is provided through a pointer. As a result, it is simple to access any memory region and pe
Read more

What is the advancement made in Hashmap in Java 8?

Image
We'll talk about an intriguing update on HashMap performance Improvement Changes in Java 8. Due to a performance issue, the Oracle changed HashMap.  What is HashMap and How it is Implemented? It keeps the key-value pair in storage where it may be retrieved with the use of a key. You must be familiar with how HashMap is Implemented only then we can understand what was the performance issue we had earlier and how that has been overcome by Oracle in Java 8. As its name suggests, HashMap operates on the concept of hashing. Equals() and hashcode() are two of the methods included in the Java Object class. We need to appropriately override these methods in order to get our HashMap to behave as expected. Any problems with these approaches could even invalidate the entire purpose since there is a chance that the results won't be retrieved even if the key-value pair is still there. HashMap use the Basic Implementation of Map Interface. When there is a hash collision, the entry's Link
Read more

Integer.parseInt(“text”)

Image
  What will happen if we execute the program int a = Integer.parseInt(“text”) ? class HelloWorld { public static void main(String[] args) { int a = Integer.parseInt(“text”); System.out.print(a); } } When this program is executed this will throw the following Exception Exception in thread "main" java.lang.NumberFormatException: For input string: "text"at java.base/java.lang.NumberFormatException.forInputString(NumberFormatException.java:65)at java.base/java.lang.Integer.parseInt(Integer.java:652)at java.base/java.lang.Integer.parseInt(Integer.java:770) at HelloWorld.main(HelloWorld.java:7) This is because parseInt() is used to get the primitive data type of the given string. It need to accepts the Integer in the form of String. But here we have given "text" as an Input in place of Integer. So it has thrown Number for Exception. It can be corrected by giving any Integer as an Input. class HelloWorld { public static void main(String
Read more