Way to get number of digits in an int?

Question

Is there a neater way for getting the number of digits in an int than this method?

int numDigits = String.valueOf(1000).length();

Answer 1

    int num = 02300;
    int count = 0;
    while(num>0){
         if(num == 0) break;
         num=num/10;
         count++;
    }
    System.out.println(count);

Answer 2

Curious, I tried to benchmark it ...

import org.junit.Test;
import static org.junit.Assert.*;


public class TestStack1306727 {

    @Test
    public void bench(){
        int number=1000;
        int a= String.valueOf(number).length();
        int b= 1 + (int)Math.floor(Math.log10(number));

        assertEquals(a,b);
        int i=0;
        int s=0;
        long startTime = System.currentTimeMillis();
        for(i=0, s=0; i< 100000000; i++){
            a= String.valueOf(number).length();
            s+=a;
        }
        long stopTime = System.currentTimeMillis();
        long runTime = stopTime - startTime;
        System.out.println("Run time 1: " + runTime);
        System.out.println("s: "+s);
        startTime = System.currentTimeMillis();
        for(i=0,s=0; i< 100000000; i++){
            b= number==0?1:(1 + (int)Math.floor(Math.log10(Math.abs(number))));
            s+=b;
        }
        stopTime = System.currentTimeMillis();
        runTime = stopTime - startTime;
        System.out.println("Run time 2: " + runTime);
        System.out.println("s: "+s);
        assertEquals(a,b);


    }
}

the results are :

Run time 1: 6765
s: 400000000
Run time 2: 6000
s: 400000000

Now I am left to wonder if my benchmark actually means something but I do get consistent results (variations within a ms) over multiple runs of the benchmark itself ... :) It looks like it's useless to try and optimize this...

---

edit: following ptomli's comment, I replaced 'number' by 'i' in the code above and got the following results over 5 runs of the bench :

Run time 1: 11500
s: 788888890
Run time 2: 8547
s: 788888890

Run time 1: 11485
s: 788888890
Run time 2: 8547
s: 788888890

Run time 1: 11469
s: 788888890
Run time 2: 8547
s: 788888890

Run time 1: 11500
s: 788888890
Run time 2: 8547
s: 788888890

Run time 1: 11484
s: 788888890
Run time 2: 8547
s: 788888890

Answer 3

Or instead the length you can check if the number is larger or smaller then the desired number.

    public void createCard(int cardNumber, int cardStatus, int customerId) throws SQLException {
    if(cardDao.checkIfCardExists(cardNumber) == false) {
        if(cardDao.createCard(cardNumber, cardStatus, customerId) == true) {
            System.out.println("Card created successfully");
        } else {

        }
    } else {
        System.out.println("Card already exists, try with another Card Number");
        do {
            System.out.println("Enter your new Card Number: ");
            scan = new Scanner(System.in);
            int inputCardNumber = scan.nextInt();
            cardNumber = inputCardNumber;
        } while(cardNumber < 95000000);
        cardDao.createCard(cardNumber, cardStatus, customerId);
    }
}

}

Answer 4

I haven't seen a multiplication-based solution yet. Logarithm, divison, and string-based solutions will become rather unwieldy against millions of test cases, so here's one for ints:

/**
 * Returns the number of digits needed to represents an {@code int} value in 
 * the given radix, disregarding any sign.
 */
public static int len(int n, int radix) {
    radixCheck(radix); 
    // if you want to establish some limitation other than radix > 2
    n = Math.abs(n);

    int len = 1;
    long min = radix - 1;

    while (n > min) {
        n -= min;
        min *= radix;
        len++;
    }

    return len;
}

In base 10, this works because n is essentially being compared to 9, 99, 999... as min is 9, 90, 900... and n is being subtracted by 9, 90, 900...

Unfortunately, this is not portable to long just by replacing every instance of int due to overflow. On the other hand, it just so happens it will work for bases 2 and 10 (but badly fails for most of the other bases). You'll need a lookup table for the overflow points (or a division test... ew)

/**
 * For radices 2 &le r &le Character.MAX_VALUE (36)
 */
private static long[] overflowpt = {-1, -1, 4611686018427387904L,
    8105110306037952534L, 3458764513820540928L, 5960464477539062500L,
    3948651115268014080L, 3351275184499704042L, 8070450532247928832L,
    1200757082375992968L, 9000000000000000000L, 5054470284992937710L,
    2033726847845400576L, 7984999310198158092L, 2022385242251558912L,
    6130514465332031250L, 1080863910568919040L, 2694045224950414864L,
    6371827248895377408L, 756953702320627062L, 1556480000000000000L,
    3089447554782389220L, 5939011215544737792L, 482121737504447062L,
    839967991029301248L, 1430511474609375000L, 2385723916542054400L,
    3902460517721977146L, 6269893157408735232L, 341614273439763212L,
    513726300000000000L, 762254306892144930L, 1116892707587883008L,
    1617347408439258144L, 2316231840055068672L, 3282671350683593750L,
    4606759634479349760L};

public static int len(long n, int radix) {
    radixCheck(radix);
    n = abs(n);

    int len = 1;
    long min = radix - 1;
    while (n > min) {
        len++;
        if (min == overflowpt[radix]) break;
        n -= min;
        min *= radix;

    }

    return len;
}

Answer 5

Here's a really simple method I made that works for any number:

public static int numberLength(int userNumber) {

    int numberCounter = 10;
    boolean condition = true;
    int digitLength = 1;

    while (condition) {
        int numberRatio = userNumber / numberCounter;
        if (numberRatio < 1) {
            condition = false;
        } else {
            digitLength++;
            numberCounter *= 10;
        }
    }

    return digitLength; 
}

The way it works is with the number counter variable is that 10 = 1 digit space. For example .1 = 1 tenth => 1 digit space. Therefore if you have int number = 103342; you'll get 6, because that's the equivalent of .000001 spaces back. Also, does anyone have a better variable name for numberCounter? I can't think of anything better.

Edit: Just thought of a better explanation. Essentially what this while loop is doing is making it so you divide your number by 10, until it's less than one. Essentially, when you divide something by 10 you're moving it back one number space, so you simply divide it by 10 until you reach <1 for the amount of digits in your number.

Here's another version that can count the amount of numbers in a decimal:

public static int repeatingLength(double decimalNumber) {

    int numberCounter = 1;
    boolean condition = true;
    int digitLength = 1;

    while (condition) {
        double numberRatio = decimalNumber * numberCounter;

        if ((numberRatio - Math.round(numberRatio)) < 0.0000001) {
            condition = false;
        } else {
            digitLength++;
            numberCounter *= 10;
        }
    }
    return digitLength - 1;
}

Answer 6

The logarithm is your friend:

int n = 1000;
int length = (int)(Math.log10(n)+1);

NB: only valid for n > 0.