Digit sum loops+random positive number generator

Question

I have to prepare program that summing the digits of two given by user numbers and then give back the random number which is

• natural number
• the digit sum of this number is bigger then the digit sum of given by user numbers

It's look like everything is ok but i dont know if it is just luck in random number generator or it's working well. Thank you for review.

import java.util.Random;
import java.util.Scanner;
public class Ex1 {
public static void main(String[] args) {
 Random rnd = new Random();
 Scanner scn = new Scanner(System.in);
 int sum1 = 0;
 int sum2 = 0;
 int sum3 = 0;
 int sum4 = 0;
 System.out.println("1TH NUMBER : ");
 int a1 = scn.nextInt();
 System.out.println("2ND NUMBER : ");
 int a2 = scn.nextInt();
 System.out.println((0 > a1 || 0 > a2 ? "ERROR-NEGATIVE NUMBER" : "OK"));
 while (a1 > 0) {
 sum1 += a1 % 10;
 a1 /= 10;
 }
 //System.out.println(sum1);
 while (a2 > 0) {
 sum2 += a2 % 10;
 a2 /= 10;
 }
 //System.out.println(sum2);
 int temp = sum1 + sum2; //temporary-for storage /=
 while (temp > 0) {
 sum3 += (temp) % 10;
 (temp) /= 10;
 }
 // System.out.println(sum3);
 while (true) {
 int a3 = rnd.nextInt(Integer.MAX_VALUE);
 sum4 += (a3) % 10;
 (a3) /= 10;
 // System.out.println(sum4);
 if (sum4 > sum3) {
 System.out.println(a3 + " this is my number");
 break;
 }
 }
}
}

• akuzminykh: Could you provide some pairs of examples? (input -> expected output)
• AdamK: Ex1 : Input: a1 = 222 - so digit sum = 6; a2 = 333 - so digit sum = 9; sum of digits from a1 and a2=わ 15 =わ 1 +たす 5 =わ 6 so i expect the number with higher digit sum like Output 4545 because 4+5+4+5 = 18 Ex2 Input - a1 - 23234 (digit sum= 14) a2 - 454545(digit sum=27), sum of digits a1+a2 =わ 14+たす27 =わ 41 =わ 4 +たす 1 =わ 5 Output: a3=61= 6+1 = 7 (higher digit sum)
• Doi9t: @AdamK I suggest that you edit your answer and add the inputs and outputs, that you gave, in a table / formatted.

Answer 1

I have some suggestion for you.

Code duplication

In your code, you have some code duplication that can be extracted in a method. By extracting the code, the code will become shorter, be less error-prone and easier to read.

1. I suggest that you make a method to print a question and read the user input.

public static void main(String[] args) {
 //[...]
 int a1 = askQuestionAndReceiveAnswer(scn, "1TH NUMBER : ");
 int a2 = askQuestionAndReceiveAnswer(scn, "2ND NUMBER : ");
 //[...]
}
private static int askQuestionAndReceiveAnswer(Scanner scn, String s) {
 System.out.println(s);
 return scn.nextInt();
}

2. Since the logic is the same to handle the sum, you can extract both of the while into a method. This extraction will remove lots of code!

public static void main(String[] args) {
 //[...]
 int sum1 = getSum(a1);
 int sum2 = getSum(a2);
 int temp = sum1 + sum2; //temporary-for storage /=
 int sum3 = getSum(temp);
 //[...]
}
private static int getSum(int userInput) {
 int currentSum = 0;
 while (userInput > 0) {
 currentSum += userInput % 10;
 userInput /= 10;
 }
 return currentSum;
}

Other observations

1. In my opinion, I would extract the last calculation in a method and return the result.

public static void main(String[] args) {
 //[...]
 int number = findNumber(rnd, sum3);
 System.out.println(number + " this is my number");
 //[...]
}
private static int findNumber(Random rnd, int sum3) {
 int sum4 = 0;
 while (true) {
 int a3 = rnd.nextInt(Integer.MAX_VALUE);
 sum4 += (a3) % 10;
 (a3) /= 10;
 if (sum4 > sum3) {
 return a3;
 }
 }
}

Refactored code

public static void main(String[] args) {
 Random rnd = new Random();
 Scanner scn = new Scanner(System.in);
 int a1 = askQuestionAndReceiveAnswer(scn, "1TH NUMBER : ");
 int a2 = askQuestionAndReceiveAnswer(scn, "2ND NUMBER : ");
 System.out.println((0 > a1 || 0 > a2 ? "ERROR-NEGATIVE NUMBER" : "OK"));
 int sum1 = getSum(a1);
 int sum2 = getSum(a2);
 int temp = sum1 + sum2; //temporary-for storage /=
 int sum3 = getSum(temp);
 int number = findNumber(rnd, sum3);
 System.out.println(number + " this is my number");
}
private static int findNumber(Random rnd, int sum3) {
 int sum4 = 0;
 while (true) {
 int a3 = rnd.nextInt(Integer.MAX_VALUE);
 sum4 += (a3) % 10;
 (a3) /= 10;
 if (sum4 > sum3) {
 return a3;
 }
 }
}
private static int getSum(int userInput) {
 int currentSum = 0;
 while (userInput > 0) {
 currentSum += userInput % 10;
 userInput /= 10;
 }
 return currentSum;
}
private static int askQuestionAndReceiveAnswer(Scanner scn, String s) {
 System.out.println(s);
 return scn.nextInt();
}

Answer 2

Welcome to Code Review, you are not closing the Scanner resource and this is a resource leak: to avoid this you can use from java 8 the construct try-with-resource like below:

try (Scanner scn = new Scanner(System.in)) {
 //here your code
}

You can write your line:

System.out.println((0 > a1 || 0 > a2 ? "ERROR-NEGATIVE NUMBER" : "OK"));

using Math.min in the following way:

System.out.println(Math.min(a1, a2) < 0 ? "ERROR-NEGATIVE NUMBER" : "OK");