Hello coders, today we are going to solve Java Exception Handling (Try-catch) HackerRank Solution.
Problem
Exception handling is the process of responding to the occurrence, during computation, of exceptions – anomalous or exceptional conditions requiring special processing – often changing the normal flow of program execution. (Wikipedia)
Java has built-in mechanism to handle exceptions. Using the try statement we can test a block of code for errors. The catch block contains the code that says what to do if exception occurs.
This problem will test your knowledge on try-catch block.
You will be given two integers x and y as input, you have to compute x/y. If x and y are not 32 bit signed integers or if y is zero, exception will occur and you have to report it. Read sample Input/Output to know what to report in case of exceptions.
Sample Input 0
10
3
Sample Output 0
3
Sample Input 1
10
Hello
Sample Output 1
java.util.InputMismatchException
Sample Input 2
10
0
Sample Output 2
java.lang.ArithmeticException: / by zero
Sample Input 3
23.323
0
Sample Output 3
java.util.InputMismatchException
Solution – Java Exception Handling (Try-catch)
import java.io.*; import java.util.*; public class Solution { public static void main(String[] args) { /* Enter your code here. Read input from STDIN. Print output to STDOUT. Your class should be named Solution. */ try{ Scanner sc = new Scanner(System.in); int a = sc.nextInt(); int b = sc.nextInt(); int c = a/b; System.out.print(c); } catch(InputMismatchException ob){ System.out.print("java.util.InputMismatchException"); } catch(Exception e) { System.out.print(e); } } }
Disclaimer: The above Problem ( Java Exception Handling (Try-catch) ) is generated by Hacker Rank but the Solution is Provided by CodingBroz. This tutorial is only for Educational and Learning Purpose.