How to count the number of digits in an int value?

Question

Given the following code.

 int[] eg = {21,20,1,12,2}
 public numbers(int[] eg)
 {
     for(int i=0; i < eg.length; i++)

I would expect eg.length to be 5, where eg[0] would = 21 eg[1] would = 20, eg[3] would = 1

I would like to check each element of eg[i] to see if it is a 1 digit or 2 digit number I tried eg[0].length to no avail, but am I correct in assuming that is only for Array Lists?

MChaker · Answer 1 · 2015-05-23T09:36:23.123

21

Convert the number to a String and get the length.

int length = String.valueOf(eg[i]).length();

Or

int length = Integer.valueOf(eg[i]).toString().length();

Use this mathematical equation (assuming that all the data in the array are strictly positive numbers).

int length = (int)(Math.log10(eg[i])+1);

edited May 23 '15 at 09:36

answered May 23 '15 at 08:52

MChaker

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Thank you. cant up vote as im new but this helped. – 30iT May 23 '15 at 08:59
Log function will probably fall short if the integer is something like two zeros (00). In that case, the string method is better. – Advait Saravade Jun 28 '17 at 06:06

score 8 · Answer 2 · answered May 23 '15 at 08:51

8

eg[i] is of type int, which doesn't have the length property.

There are many ways of telling how many digits the number has:

write your own method that accepts an int and keeps dividing by 10 until the number reaches 0.
using math properties:

int length = (int) (Math.log10(number) + 1);
converting the number to String and using its methods

answered May 23 '15 at 08:51

Maroun

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Ok thanks a lot understand it now, would upvote but cant atm!!! – 30iT May 23 '15 at 08:59
2

@30iT I'm glad it helped. Never mind the upvote, when you'll get the privilege you'll be able to. It's more important that you understood the solution. – Maroun May 23 '15 at 09:11

score 2 · Answer 3 · answered May 23 '15 at 09:40

If you want to see what is the number length, MChaker's solution is the correct one. However, if you want to see if the number contains one digit or more than one digit, then you might consider the following solution:

public class NumberTest {
public static void main(String[] args) {
    int[] eg = {21,20,1,12,2, -3, -8, -20};
    System.out.println("Array Length: " + eg.length);

    for(int i=0; i < eg.length; i++) {
        System.out.println("#" + i + ": Value: " + eg[i] + " one digit number: " + isOneDigitNumber(eg[i]));
    }
}
static private boolean isOneDigitNumber(int number) {
    return (-10 < number && number < 10);
}

}

and here is the result of above code:

Array Length: 8
#0: Value: 21 one digit number: false
#1: Value: 20 one digit number: false
#2: Value: 1 one digit number: true
#3: Value: 12 one digit number: false
#4: Value: 2 one digit number: true
#5: Value: -3 one digit number: true
#6: Value: -8 one digit number: true
#7: Value: -20 one digit number: false
#7: Value: -20 one digit number: false

score 1 · Answer 4 · answered May 23 '15 at 08:54

1

There's an algorithm used in JDK which is probably the fastest one:

final static int [] sizeTable = { 9, 99, 999, 9999, 99999, 999999, 9999999,
                                  99999999, 999999999, Integer.MAX_VALUE };

// Requires positive x
static int stringSize(int x) {
    for (int i=0; ; i++)
        if (x <= sizeTable[i])
            return i+1;
}

answered May 23 '15 at 08:54

Tagir Valeev

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looks like a few ways to skin a cat...but i am not sure i fully understand this implementation – 30iT May 23 '15 at 09:01

How to count the number of digits in an int value?

4 Answers4