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

Plus Minus HackerRank Solution in Java | Programming Blog

Java Solution for HackerRank Plus Minus Problem Given an array of integers, calculate the ratios of its elements that are positive , negative , and zero . Print the decimal value of each fraction on a new line with 6 places after the decimal. Example 1 : array = [1, 1, 0, -1, -1] There are N = 5 elements, two positive, two negative and one zero. Their ratios are 2/5 = 0.400000, 2/5 = 0.400000 and 1/5 = 0.200000. Results are printed as:  0.400000 0.400000 0.200000 proportion of positive values proportion of negative values proportion of zeros Example 2 : array = [-4, 3, -9, 0, 4, 1]  There are 3 positive numbers, 2 negative numbers, and 1 zero in array. Following is answer : 3/6 = 0.500000 2/6 = 0.333333 1/6 = 0.166667 Lets see solution Solution 1 import java.io.*; import java.math.*; import java.security.*; import java.text.*; import java.util.*; import java.util.concurrent.*; import java.util.function.*; import java.util.regex.*; import java.util.stream.*; import static java.util.st

Flipping the Matrix HackerRank Solution in Java with Explanation

Java Solution for Flipping the Matrix | Find Highest Sum of Upper-Left Quadrant of Matrix Problem Description : Sean invented a game involving a 2n * 2n matrix where each cell of the matrix contains an integer. He can reverse any of its rows or columns any number of times. The goal of the game is to maximize the sum of the elements in the n *n submatrix located in the upper-left quadrant of the matrix. Given the initial configurations for q matrices, help Sean reverse the rows and columns of each matrix in the best possible way so that the sum of the elements in the matrix's upper-left quadrant is maximal.  Input : matrix = [[1, 2], [3, 4]] Output : 4 Input : matrix = [[112, 42, 83, 119], [56, 125, 56, 49], [15, 78, 101, 43], [62, 98, 114, 108]] Output : 119 + 114 + 56 + 125 = 414 Full Problem Description : Flipping the Matrix Problem Description   Here we can find solution using following pattern, So simply we have to find Max of same number of box like (1,1,1,1). And last