Finding the string length of a integer in .NET

Question

In .NET what is the best way to find the length of an integer in characters if it was represented as a string?

e.g.

1 = 1 character
10 = 2 characters
99 = 2 characters
100 = 3 characters
1000 = 4 characters

The obvious answer is to convert the int to a string and get its length but I want the best performance possible without the overhead of creating a new string.

Always balance the free microseconds you are attempting to gain against the expensive reading time of the poor sucker (yourself, I guess) who has to read and debug that code later on. — reinierpost
– reinierpost, Commented Mar 24, 2010 at 9:38
@reinierpost it's not a micro-optimization if the said method is run, like, a billion times in a loop called by thousands of different threads. — Paulo Santos
– Paulo Santos, Commented Mar 24, 2010 at 9:47
Duplicate of this one? stackoverflow.com/questions/679602/… — Stefan Steinegger
– Stefan Steinegger, Commented Mar 24, 2010 at 10:04
If the performance of conversion to base 10 is an issue, then you should consider direct decimal representations, such as BCD, for your integers. This conversion penalty is one reason why calculators and early CPUs had BCD arithemetic. — President James K. Polk
– President James K. Polk, Commented Mar 24, 2010 at 13:47
@GregS - I suspect that it's not the conversion itself that's the problem, but rather filling up the string heap with a lot of useless strings. Based on the way the GC works, however, I think that this also is not a problem in practice. — Jeffrey L Whitledge
– Jeffrey L Whitledge, Commented Mar 24, 2010 at 16:42

stmax · Accepted Answer · 2010-03-24 17:14:44Z

38

you can use logartihms to calculate the length of the int:

public static int IntLength(int i) {
  if (i <= 0) throw new ArgumentOutOfRangeException();

  return (int)Math.Floor(Math.Log10(i)) + 1;
}

the test passes:

[Test]
public void TestIntLength() {
  Assert.AreEqual(1, IntLength(1));
  Assert.AreEqual(1, IntLength(9));
  Assert.AreEqual(2, IntLength(10));
  Assert.AreEqual(2, IntLength(99));
  Assert.AreEqual(3, IntLength(100));
  Assert.AreEqual(3, IntLength(999));
  Assert.AreEqual(4, IntLength(1000));
  Assert.AreEqual(10, IntLength(int.MaxValue));
}

a quick test has shown that the log-method is 4 times faster than the int.ToString().Length method..

the method shown by GvS below (using if-statements) is another 6 times (!) faster than the log method:

public static int IntLengthIf(int i) {
  if (i < 10) return 1;
  if (i < 100) return 2;
  if (i < 1000) return 3;
  if (i < 10000) return 4;
  if (i < 100000) return 5;
  if (i < 1000000) return 6;
  if (i < 10000000) return 7;
  if (i < 100000000) return 8;
  if (i < 1000000000) return 9;
  throw new ArgumentOutOfRangeException();
}

here are the exact timings for the numbers 1 to 10000000:

IntLengthToString: 4205ms
IntLengthLog10: 1122ms
IntLengthIf: 201ms

edited Mar 24, 2010 at 17:14

answered Mar 24, 2010 at 9:34

stmax

6,6154 gold badges31 silver badges47 bronze badges

Sign up to request clarification or add additional context in comments.

4 Comments

Matt Ellen Over a year ago

To cope with negative numbers, you could get the absolute value of i and then add 1 to the result of the function.

Jeffrey L Whitledge Over a year ago

If you have a quick test for this already, you should compare it to GvS's solution. That should be interesting.

Tobias Knauss Over a year ago

you could add support for negative numbers: public static int IntLengthIf(int i, bool WithMinus) { int iCount = 0; if (i_iValue < 0) { i = -i; if (WithMinus) iCount++; } if (i_iValue < 10) iCount += 1; ...; return iCount;

Tobias Knauss Over a year ago

By the way, there is a bug in IntLengthIf: if you have values > 1e9 like MAXINT, then you have 10 digits, but this case is not handled. So you should add if (i_iValue <= int.MaxValue) return 10;

GvS · Accepted Answer · 2010-03-24 17:20:52Z

12

If input is in range 0-10000

if (i < 10) return 1;
if (i < 100) return 2;
if (i < 1000) return 3;
if (i < 10000) return 4;
// etc

edited Mar 24, 2010 at 17:20

answered Mar 24, 2010 at 10:02

GvS

52.5k16 gold badges104 silver badges139 bronze badges

2 Comments

President James K. Polk Over a year ago

This is the simplest solution, and it reads well. It probably can't compete with an optimized floating point logarithm.

Jeffrey L Whitledge Over a year ago

I am not going to bother doing the test, but I bet this solution is faster than the logarithm one. Logs take quite a few clock cycles to compute.

Andrew Bezzub · Accepted Answer · 2010-03-24 09:25:59Z

4

You could use something like this:

        int integer = 100;

        int charachtersCount = 0;
        while (integer > 0)
        {
            integer = integer/10;
            charachtersCount++;
        }

But do you really need to optimize this? I would actually prefer using string (looks much better):

integer.ToString().Length

answered Mar 24, 2010 at 9:25

Andrew Bezzub

16.1k7 gold badges54 silver badges73 bronze badges

2 Comments

Simon Linder Over a year ago

Nice solution, but as you said I would doubt that the performance is good. I guess that string conversion is best.

Enigmativity Over a year ago

I would suspect that the performance here is quite good. It's using integer maths afterall.

Community · Accepted Answer · 2017-05-23 11:54:07Z

1

If you need to deal with negative numbers also, you can take stmax solution with a spin:

public static int IntLength(int i) { 
  if (i == 0) return 1; // no log10(0)
  int n = (i < 0) ? 2 : 1;
  i = (i < 0) ? -i : i;

  return (int)Math.Floor(Math.Log10(i)) + n; 
}

edited May 23, 2017 at 11:54

CommunityBot

11 silver badge

answered Mar 24, 2010 at 9:56

Paulo Santos

11.7k5 gold badges45 silver badges67 bronze badges

Comments

Mike Dunlavey · Accepted Answer · 2010-03-24 17:00:52Z

1

You can do:

int ndig = 1;
if (n < 0){n = -n; ndig++;}
if (n >= 100000000){n /= 100000000; ndig += 8;}
if (n >=     10000){n /=     10000; ndig += 4;}
if (n >=       100){n /=       100; ndig += 2;}
if (n >=        10){n /=        10; ndig += 1;}

or something along those lines. It takes 4 comparisons and 0-4 divisions.

(On 64 bits you have to add a fifth level.)

edited Mar 24, 2010 at 17:00

answered Mar 24, 2010 at 16:35

Mike Dunlavey

40.8k15 gold badges95 silver badges140 bronze badges

1 Comment

JeffC Over a year ago

This one gets clever points. It's slightly slower than the straight if but I like it.

roxton · Accepted Answer · 2024-11-19 08:57:01Z

0

Handles all ints, including 0, negative values and int.MinValue:

int GetFormattedLength(int i)
{
    return i == 0 ? 1 : (int)Math.Floor(Math.Log10(Math.Abs((double)i))) + (i > 0 ? 1 : 2);
}

And here's a version optimized for performance:

int GetFormattedLength(int i)
{
    if (i == 0)
        return 1;

    uint absI;
    int signLength = 0;
    if (i < 0)
    {
        absI = ((uint)~i) + 1;
        signLength = 1;
    }
    else
    {
        absI = (uint)i;
    }

    return absI switch
    {
        < 10 => 1,
        < 100 => 2,
        < 1000 => 3,
        < 10000 => 4,
        < 100000 => 5,
        < 1000000 => 6,
        < 10000000 => 7,
        < 100000000 => 8,
        < 1000000000 => 9,
        _ => 10,
    } + signLength;
}

edited Nov 19, 2024 at 8:57

answered Nov 8, 2024 at 13:05

roxton

2,06612 silver badges10 bronze badges

Comments

Rik · Accepted Answer · 2010-03-24 09:36:51Z

-1

If you want to do it by maths you could try this:

int integer = 100
int charCount = (int) Math.Ceiling(Math.Log10(integer+1));

I doubt if this is very much faster than converting to a string though

answered Mar 24, 2010 at 9:36

Rik

29.3k14 gold badges52 silver badges69 bronze badges

Collectives™ on Stack Overflow

Finding the string length of a integer in .NET

7 Answers 7

4 Comments

2 Comments

2 Comments

Comments

1 Comment

Comments

Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

7 Answers 7

4 Comments

2 Comments

2 Comments

Comments

1 Comment

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related