Skip to main content

BiFunction in Java with examples

BiFunction in Java

What is BiFunction in Java 8?

 BiFunction is a functional interface.
It is used for method reference or as a assignment target for Lambda Expression.

If you want to learn about Functional Interface check it out this post.

BiFunction is a special function who takes two argument and produce result.

Syntax of BiFunction
@FunctionalInterface
public interface BiFunction<T, U, R>

Let's see simple example of BiFunction.

Example 1
import java.util.function.BiFunction;

public class _BiFunction {

    public static void main(String[] args) {
        // BiFunction with two int parameter and produce int result
        BiFunction<Integer, Integer, Integer> biFunction = (arg1, arg2) -> {
            return arg1 + arg2;
        };
       
        // call biFunction and print output
        System.out.println( biFunction.apply(5, 5));
       
    }
}

Output :-
10

In above example we take BiFunction that takes two integer values and produce output based on that.

We can also pass different argument and produce result. like the following example.
Example 2
import java.util.function.BiFunction;

public class BiFunction {

    public static void main(String[] args) {
       
        // BiFunction with one int and another string parameter and produce string result.
        BiFunction<Integer, String, String> biFunction = (arg1, arg2) -> {
            return arg1 + arg2;
        };

         // call biFunction and print output
        System.out.println( biFunction.apply(5, "5"));  
    }
}

Output :-
55

Now we pass two Integer parameter and produce Float result. so lets see how its looks like.
Example 3
import java.util.function.BiFunction;

public class BiFunction {

    public static void main(String[] args) {
       
        // BiFunction with two int value and produce floating value.   
        BiFunction<Integer, Integer, Float> biFunction = (arg1, arg2) -> {
            return (float) (arg1 + arg2);
        };
       
         // call biFunction and print output   
        System.out.println( biFunction.apply(5, 5));
   
    }
}

Output :-
10.0

We pass two int values and produce floating point result. But for that we have to casting values into float like in above example.

Lets see another example which split string using BiFunction.

Example 4
public class BiFunction {

    public static void main(String[] args) {
   
        String name = "Welcome to Programming Blog";
       
        // BiFunction that split string from space and give array of string.
        BiFunction<String, Character, String[]> splitString =
            (nameAsString, splitFrom) -> {
                String[] arrayOfString = nameAsString.split(splitFrom.toString());
                return arrayOfString;
        };
       
        // call BiFunction with string and how to we want to split(using space).
        String[] finalArray = splitString.apply(name, (' '));
       
        // print final array of string.
        for (String string : finalArray) {
            System.out.println(string);
        }
       
    }
}

Output :-
Welcome
to
Programming
Blog

In above example we split the string with space. we can also split with other number or character or anything that is in string.

You know how to use BiFunction that gives result as object? so if you does not know then lets see example of that.

Example 5 :-
public class Student {

    private String name;
    private Integer age;
   
    // Constructor
    public Student(String name, Integer age) {
        this.name = name;
        this.age = age;
    }

    // getter and setter method for name and age.
    public String getName() {
        return name;
    }

    public void setName(String name) {
        this.name = name;
    }

    public Integer getAge() {
        return age;
    }

    public void setAge(Integer age) {
        this.age = age;
    }

    public static void main(String[] args) {
             
        BiFunction<String, Integer, Student> biFunction =
            (name, age) -> {
                return new Student(name, age);
        };
       
        Student studentObj = biFunction.apply("Programmer", 20);
       
        System.out.println("Student Name : "+ studentObj.getName());
        System.out.println("Student Age : "+ studentObj.getAge());
       
    }
}

Output :-
Student Name : Programmer
Student Age : 20

So in above example we created student object with two properties
1. name
2. age
After we created constructor with our two properties.
Created getter and setter for both.
In BiFunction we pass three parameter. first two as input and third with Student object that return student object.

So we had seen so many example now see what is advatages of BiFunction. here is link
Learn more about BiFunction here is link for Java document.


Comments

Popular posts from this blog

Sales by Match HackerRank Solution | Java Solution

HackerRank Sales by Match problem solution in Java   Problem Description : Alex works at a clothing store. There is a large pile of socks that must be paired by color for sale. Given an array of integers representing the color of each sock, determine how many pairs of socks with matching colors there are. For example, there are n=7 socks with colors socks = [1,2,1,2,1,3,2]. There is one pair of color 1 and one of color 2 . There are three odd socks left, one of each color. The number of pairs is 2 .   Example 1 : Input : n = 6 arr = [1, 2, 3, 4, 5, 6] Output : 0 Explanation : We have 6 socks with all different colors, So print 0. Example 2 : Input : n = 10 arr = [1, 2, 3, 4, 1, 4, 2, 7, 9, 9] Output : 4 Explanation : We have 10 socks. There is pair of color 1, 2, 4 and 9, So print 4. This problem easily solved by HashMap . Store all pair of socks one by one in Map and check if any pair is present in Map or not. If pair is present then increment ans variable by 1 ...

Queen's Attack II HackerRank Solution in Java with Explanation

Queen's Attack II Problem's Solution in Java (Chessboard Problem)   Problem Description : You will be given a square chess board with one queen and a number of obstacles placed on it. Determine how many squares the queen can attack.  A queen is standing on an n * n chessboard. The chess board's rows are numbered from 1 to n, going from bottom to top. Its columns are numbered from 1 to n, going from left to right. Each square is referenced by a tuple, (r, c), describing the row r and column c, where the square is located. The queen is standing at position (r_q, c_q). In a single move, queen can attack any square in any of the eight directions The queen can move: Horizontally (left, right) Vertically (up, down) Diagonally (four directions: up-left, up-right, down-left, down-right) The queen can move any number of squares in any of these directions, but it cannot move through obstacles. Input Format : n : The size of the chessboard ( n x n ). k : The number of obstacles...