Finding the number of digits of an integer

Question

What is the best method to find the number of digits of a positive integer?

I have found this 3 basic methods:

• conversion to string

  String s = new Integer(t).toString(); 
  int len = s.length();
  

• for loop

  for(long long int temp = number; temp >= 1;)
  {
      temp/=10;
      decimalPlaces++;
  } 
  

• logaritmic calculation

  digits = floor( log10( number ) ) + 1;
  

where you can calculate log10(x) = ln(x) / ln(10) in most languages.

First I thought the string method is the dirtiest one but the more I think about it the more I think it's the fastest way. Or is it?

Answer 1

Well the correct answer would be to measure it - but you should be able to make a guess about the number of CPU steps involved in converting strings and going through them looking for an end marker

Then think how many FPU operations/s your processor can do and how easy it is to calculate a single log.

edit: wasting some more time on a monday morning :-)

String s = new Integer(t).toString(); 
int len = s.length();

One of the problems with high level languages is guessing how much work the system is doing behind the scenes of an apparently simple statement. Mandatory Joel link

This statement involves allocating memory for a string, and possibly a couple of temporary copies of a string. It must parse the integer and copy the digits of it into a string, possibly having to reallocate and move the existing memory if the number is large. It might have to check a bunch of locale settings to decide if your country uses "," or ".", it might have to do a bunch of unicode conversions.
Then finding the length has to scan the entire string, again considering unicode and any local specific settings such as - are you in a right->left language?.

Alternatively:

digits = floor( log10( number ) ) + 1;

Just because this would be harder for you to do on paper doesn't mean it's hard for a computer! In fact a good rule in high performance computing seems to have been - if something is hard for a human (fluid dynamics, 3d rendering) it's easy for a computer, and if it's easy for a human (face recognition, detecting a voice in a noisy room) it's hard for a computer!

You can generally assume that the builtin maths functions log/sin/cos etc - have been an important part of computer design for 50years. So even if they don't map directly into a hardware function in the FPU you can bet that the alternative implementation is pretty efficient.

Answer 2

log(x,n)-mod(log(x,n),1)+1

Where x is a the base and n is the number.

Answer 3

Here is the measurement in Swift 4.

Algorithms code:

extension Int {
    var numberOfDigits0: Int {
        var currentNumber = self
        var n = 1
        if (currentNumber >= 100000000) {
            n += 8
            currentNumber /= 100000000
        }
        if (currentNumber >= 10000) {
            n += 4
            currentNumber /= 10000
        }
        if (currentNumber >= 100) {
            n += 2
            currentNumber /= 100
        }
        if (currentNumber >= 10) {
            n += 1
        }
        return n
    }

    var numberOfDigits1: Int {
        return String(self).count
    }

    var numberOfDigits2: Int {
        var n = 1
        var currentNumber = self
        while currentNumber > 9 {
            n += 1
            currentNumber /= 10
        }
        return n
    }

}

Measurement code:

var timeInterval0 = Date()
for i in 0...10000 {
    i.numberOfDigits0
}
print("timeInterval0: \(Date().timeIntervalSince(timeInterval0))")

var timeInterval1 = Date()
for i in 0...10000 {
    i.numberOfDigits1
}
print("timeInterval1: \(Date().timeIntervalSince(timeInterval1))")

var timeInterval2 = Date()
for i in 0...10000 {
    i.numberOfDigits2
}
print("timeInterval2: \(Date().timeIntervalSince(timeInterval2))")

Output

timeInterval0: 1.92149806022644

timeInterval1: 0.557608008384705

timeInterval2: 2.83262193202972

On this measurement basis String conversion is the best option for the Swift language.

Answer 4

You can use a recursive solution instead of a loop, but somehow similar:

@tailrec
def digits (i: Long, carry: Int=1) : Int =  if (i < 10) carry else digits (i/10, carry+1)

digits (8345012978643L)

With longs, the picture might change - measure small and long numbers independently against different algorithms, and pick the appropriate one, depending on your typical input. :)

Of course nothing beats a switch:

switch (x) {
  case 0:  case 1:  case 2:  case 3:  case 4:  case 5:  case 6:  case 7:  case 8:  case 9: return 1;
  case 10: case 11: // ...
  case 99: return 2;
  case 100: // you get the point :) 
  default: return 10; // switch only over int
}

except a plain-o-array:

   int [] size = {1,1,1,1,1,1,1,1,1,2,2,2,2,2,... };
   int x = 234561798;
   return size [x];

Some people will tell you to optimize the code-size, but yaknow, premature optimization ...

Answer 5

You can obviously eliminate the method 1 from the competition, because the atoi/toString algorithm it uses would be similar to method 2.

Method 3's speed depends on whether the code is being compiled for a system whose instruction set includes log base 10.

Answer 6

Test conditions

• Decimal numeral system
• Positive integers
• Up to 10 digits
• Language: ActionScript 3

Results

digits: [1,10],

no. of runs: 1,000,000

random sample: 8777509,40442298,477894,329950,513,91751410,313,3159,131309,2

result: 7,8,6,6,3,8,3,4,6,1

CONVERSION TO STRING: 724ms

LOGARITMIC CALCULATION: 349ms

DIV 10 ITERATION: 229ms

MANUAL CONDITIONING: 136ms

Note: Author refrains from making any conclusions for numbers with more than 10 digits.

---

Script

package {
    import flash.display.MovieClip;
    import flash.utils.getTimer;
    /**
     * @author Daniel
     */
    public class Digits extends MovieClip {
        private const NUMBERS : uint = 1000000;
        private const DIGITS : uint = 10;

        private var numbers : Array;
        private var digits : Array;

        public function Digits() {
            // ************* NUMBERS *************
            numbers = [];
            for (var i : int = 0; i < NUMBERS; i++) {
                var number : Number = Math.floor(Math.pow(10, Math.random()*DIGITS));
                numbers.push(number);
            }   
            trace('Max digits: ' + DIGITS + ', count of numbers: ' + NUMBERS);
            trace('sample: ' + numbers.slice(0, 10));

            // ************* CONVERSION TO STRING *************
            digits = [];
            var time : Number = getTimer();

            for (var i : int = 0; i < numbers.length; i++) {
                digits.push(String(numbers[i]).length);
            }

            trace('\nCONVERSION TO STRING - time: ' + (getTimer() - time));
            trace('sample: ' + digits.slice(0, 10));

            // ************* LOGARITMIC CALCULATION *************
            digits = [];
            time = getTimer();

            for (var i : int = 0; i < numbers.length; i++) {
                digits.push(Math.floor( Math.log( numbers[i] ) / Math.log(10) ) + 1);
            }

            trace('\nLOGARITMIC CALCULATION - time: ' + (getTimer() - time));
            trace('sample: ' + digits.slice(0, 10));

            // ************* DIV 10 ITERATION *************
            digits = [];
            time = getTimer();

            var digit : uint = 0;
            for (var i : int = 0; i < numbers.length; i++) {
                digit = 0;
                for(var temp : Number = numbers[i]; temp >= 1;)
                {
                    temp/=10;
                    digit++;
                } 
                digits.push(digit);
            }

            trace('\nDIV 10 ITERATION - time: ' + (getTimer() - time));
            trace('sample: ' + digits.slice(0, 10));

            // ************* MANUAL CONDITIONING *************
            digits = [];
            time = getTimer();

            var digit : uint;
            for (var i : int = 0; i < numbers.length; i++) {
                var number : Number = numbers[i];
                if (number < 10) digit = 1;
                else if (number < 100) digit = 2;  
                else if (number < 1000) digit = 3;  
                else if (number < 10000) digit = 4;  
                else if (number < 100000) digit = 5;  
                else if (number < 1000000) digit = 6;  
                else if (number < 10000000) digit = 7;  
                else if (number < 100000000) digit = 8;  
                else if (number < 1000000000) digit = 9;  
                else if (number < 10000000000) digit = 10;  
                digits.push(digit);
            }

            trace('\nMANUAL CONDITIONING: ' + (getTimer() - time));
            trace('sample: ' + digits.slice(0, 10));
        }
    }
}