How to get the separate digits of an int number?

I noticed that there are few example of using Java 8 stream to solve your problem but I think that this is the simplest one:

int[] intTab = String.valueOf(number).chars().map(Character::getNumericValue).toArray();

To be clear: You use String.valueOf(number) to convert int to String, then chars() method to get an IntStream (each char from your string is now an Ascii number), then you need to run map() method to get a numeric values of the Ascii number. At the end you use toArray() method to change your stream into an int[] array.

I see all the answer are ugly and not very clean.

I suggest you use a little bit of recursion to solve your problem. This post is very old, but it might be helpful to future coders.

public static void recursion(int number) {
    if(number > 0) {
        recursion(number/10);
        System.out.printf("%d   ", (number%10));
    }
}

Output:

Input: 12345

Output: 1   2   3   4   5 

simple solution

public static void main(String[] args) {
    int v = 12345;
    while (v > 0){
        System.out.println(v % 10);
        v /= 10;
    }
}

Since I don't see a method on this question which uses Java 8, I'll throw this in. Assuming that you're starting with a String and want to get a List<Integer>, then you can stream the elements like so.

List<Integer> digits = digitsInString.chars()
        .map(Character::getNumericValue)
        .boxed()
        .collect(Collectors.toList());

This gets the characters in the String as a IntStream, maps those integer representations of characters to a numeric value, boxes them, and then collects them into a list.

Java 9 introduced a new Stream.iterate method which can be used to generate a stream and stop at a certain condition. This can be used to get all the digits in the number, using the modulo approach.

int[] a = IntStream.iterate(123400, i -> i > 0, i -> i / 10).map(i -> i % 10).toArray();

Note that this will get the digits in reverse order, but that can be solved either by looping through the array backwards (sadly reversing an array is not that simple), or by creating another stream:

int[] b = IntStream.iterate(a.length - 1, i -> i >= 0, i -> i - 1).map(i -> a[i]).toArray();

or

int[] b = IntStream.rangeClosed(1, a.length).map(i -> a[a.length - i]).toArray();

As an example, this code:

int[] a = IntStream.iterate(123400, i -> i > 0, i -> i / 10).map(i -> i % 10).toArray();
int[] b = IntStream.iterate(a.length - 1, i -> i >= 0, i -> i - 1).map(i -> a[i]).toArray();
System.out.println(Arrays.toString(a));
System.out.println(Arrays.toString(b));

Will print:

[0, 0, 4, 3, 2, 1]
[1, 2, 3, 4, 0, 0]

Why don't you do:

String number = String.valueOf(input);
char[] digits = number.toCharArray();

Try this one.

const check = (num) => {
  let temp = num
  let result = []
  while(temp > 0){
    let a = temp%10;
    result.push(a);
    temp = (temp-a)/10;
  }
  return result;
}

check(98) //[ 8, 9 ]

neither chars() nor codePoints() — the other lambda

String number = Integer.toString( 1100 );

IntStream.range( 0, number.length() ).map( i -> Character.digit( number.codePointAt( i ), 10 ) ).toArray();  // [1, 1, 0, 0]

As a noob, my answer would be:

String number = String.valueOf(ScannerObjectName.nextInt()); 
int[] digits = new int[number.length()]; 
for (int i = 0 ; i < number.length() ; i++)
    int[i] = Integer.parseInt(digits.substring(i,i+1))

Now all the digits are contained in the "digits" array.

if digit is meant to be a Character

String numstr = Integer.toString( 123 );
Pattern.compile( "" ).splitAsStream( numstr ).map(
  s -> s.charAt( 0 ) ).toArray( Character[]::new );  // [1, 2, 3]

and the following works correctly
numstr = "000123" gets [0, 0, 0, 1, 2, 3]
numstr = "-123"    gets [-, 1, 2, 3]

A .NET solution using LINQ.

List<int> numbers = number.ToString().Select(x => x - 48).ToList();

import java.util.Scanner;

class  Test 
{  
    public static void main(String[] args)   
    {  
        Scanner sc = new Scanner(System.in); 


    int num=sc.nextInt(); 
    System.out.println("Enter a number (-1 to end):"+num);
    int result=0;
    int i=0;
    while(true) 
    { 
      int n=num%10;
      if(n==-1){
        break;
      }
      i++;
      System.out.println("Digit"+i+" = "+n);
      result=result*10+n;
      num=num/10; 


      if(num==0) 
      { 
        break; 
      } 
    }
    }
}