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

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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  gt  2     n   Math abs n        int len   1      long min   radix - 1       while  n  gt  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  amp le r  amp 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  gt  min            len            if  min    overflowpt radix   break          n -  min          min    radix              return len

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With design  based on problem    This is an alternate of divide-and-conquer   We ll first define an enum  considering it s only for an unsigned int    public enum IntegerLength       One  byte 1 10       Two  byte 2 100       Three  byte 3 1000       Four  byte 4 10000       Five  byte 5 100000       Six  byte 6 1000000       Seven  byte 7 10000000       Eight  byte 8 100000000       Nine  byte 9 1000000000        byte length      int value       IntegerLength byte len int value            this length   len          this value   value             public byte getLenght             return length             public int getValue             return value            Now we ll define a class that goes through the values of the enum and compare and return the appropriate length   public class IntegerLenght       public static byte calculateIntLenght int num                for IntegerLength v   IntegerLength values                  if num  lt  v getValue                     return v getLenght                                    return 0            The run time of this solution is the same as the divide-and-conquer approach

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Enter the number and create an Arraylist  and the while loop will record all the digits into the Arraylist   Then we can take out the size of array  which will be the length of the integer value you entered   ArrayList lt Integer gt  a new ArrayList lt  gt      while number  gt  0          remainder   num   10       a add remainder       number   number   10       int m a size

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One wants to do this mostly because he she wants to  present  it  which mostly mean it finally needs to be  toString-ed   or transformed in another way  explicitly or implicitly anyway  before it can be presented  printed for example   If that is the case then just try to make the necessary  toString  explicit and count the bits

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Using Java  int nDigits   Math floor Math log10 Math abs the integer      1    use import java lang Math    in the beginning  Using C  int nDigits   floor log10 abs the integer      1    use inclue math h in the beginning

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I see people using String libraries or even using the Integer class  Nothing wrong with that but the algorithm for getting the number of digits is not that complicated  I am using a long in this example but it works just as fine with an int     private static int getLength long num         int count   1       while  num  gt   10            num   num   10          count               return count

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Marian s Solution  now with Ternary    public int len int n           return  n lt 100000    n lt 100    n lt 10  1 2   n lt 1000  3   n lt 10000  4 5     n lt 10000000    n lt 1000000  6 7    n lt 100000000  8   n lt 1000000000  9 10             Because we can

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Can I try      based on Dirk s solution  final int digits   number  0 1  1    int Math floor Math log10 Math abs number

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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  lt  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  lt 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    lt  0 0000001                condition   false            else               digitLength                numberCounter    10                      return digitLength - 1

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How about plain old Mathematics  Divide by 10 until you reach 0   public static int getSize long number            int count   0          while  number  gt  0                count    1              number    number   10                     return count

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Your String-based solution is perfectly OK  there is nothing  un-neat  about it  You have to realize that mathematically  numbers don t have a length  nor do they have digits  Length and digits are both properties of a physical representation of a number in a specific base  i e  a String   A logarithm-based solution does  some of  the same things the String-based one does internally  and probably does so  insignificantly  faster because it only produces the length and ignores the digits  But I wouldn t actually consider it clearer in intent - and that s the most important factor

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The fastest approach  divide and conquer  Assuming your range is 0 to MAX INT  then you have 1 to 10 digits  You can approach this interval using divide and conquer  with up to 4 comparisons per each input  First  you divide  1  10  into  1  5  and  6  10  with one comparison  and then each length 5 interval you divide using one comparison into one length 3 and one length 2 interval  The length 2 interval requires one more comparison  total 3 comparisons   the length 3 interval can be divided into length 1 interval  solution  and a length 2 interval  So  you need 3 or 4 comparisons  No divisions  no floating point operations  no expensive logarithms  only integer comparisons  Code  long but fast   if  n  lt  100000           5 or less     if  n  lt  100              1 or 2         if  n  lt  10              return 1          else             return 2        else              3 or 4 or 5         if  n  lt  1000              return 3          else                  4 or 5             if  n  lt  10000                  return 4              else                 return 5                    else          6 or more     if  n  lt  10000000               6 or 7         if  n  lt  1000000              return 6          else             return 7        else              8 to 10         if  n  lt  100000000              return 8          else                  9 or 10             if  n  lt  1000000000                  return 9              else                 return 10                     Benchmark  after JVM warm-up  - see code below to see how the benchmark was run   baseline method  with String length   2145ms log10 method  711ms   3 02 times as fast as baseline repeated divide  2797ms   0 77 times as fast as baseline divide-and-conquer  74ms   28 99 times as fast as baseline  Full code  public static void main String   args  throws Exception               validate methods      for  int i   0  i  lt  1000  i            if  method1 i     method2 i               System out println i       for  int i   0  i  lt  1000  i            if  method1 i     method3 i               System out println i    quot   quot    method1 i     quot   quot    method3 i        for  int i   333  i  lt  2000000000  i    1000          if  method1 i     method3 i               System out println i    quot   quot    method1 i     quot   quot    method3 i        for  int i   0  i  lt  1000  i            if  method1 i     method4 i               System out println i    quot   quot    method1 i     quot   quot    method4 i        for  int i   333  i  lt  2000000000  i    1000          if  method1 i     method4 i               System out println i    quot   quot    method1 i     quot   quot    method4 i                work-up the JVM - make sure everything will be run in hot-spot mode     allMethod1        allMethod2        allMethod3        allMethod4                run benchmark     Chronometer c           c   new Chronometer true       allMethod1        c stop        long baseline   c getValue        System out println c            c   new Chronometer true       allMethod2        c stop        System out println c    quot     quot    StringTools formatDouble  double baseline   c getValue      quot 0 00 quot      quot  times as fast as baseline quot             c   new Chronometer true       allMethod3        c stop        System out println c    quot     quot    StringTools formatDouble  double baseline   c getValue      quot 0 00 quot      quot  times as fast as baseline quot             c   new Chronometer true       allMethod4        c stop        System out println c    quot     quot    StringTools formatDouble  double baseline   c getValue      quot 0 00 quot      quot  times as fast as baseline quot        private static int method1 int n        return Integer toString n  length       private static int method2 int n        if  n    0          return 1      return  int  Math log10 n    1      private static int method3 int n        if  n    0          return 1      int l      for  l   0   n  gt  0    l          n    10      return l     private static int method4 int n        if  n  lt  100000               5 or less         if  n  lt  100                   1 or 2             if  n  lt  10                  return 1              else                 return 2            else                  3 or 4 or 5             if  n  lt  1000                  return 3              else                      4 or 5                 if  n  lt  10000                      return 4                  else                     return 5                                else              6 or more         if  n  lt  10000000                   6 or 7             if  n  lt  1000000                  return 6              else                 return 7            else                  8 to 10             if  n  lt  100000000                  return 8              else                      9 or 10                 if  n  lt  1000000000                      return 9                  else                     return 10                                    private static int allMethod1         int x   0      for  int i   0  i  lt  1000  i            x   method1 i       for  int i   1000  i  lt  100000  i    10          x   method1 i       for  int i   100000  i  lt  1000000  i    100          x   method1 i       for  int i   1000000  i  lt  2000000000  i    200          x   method1 i            return x     private static int allMethod2         int x   0      for  int i   0  i  lt  1000  i            x   method2 i       for  int i   1000  i  lt  100000  i    10          x   method2 i       for  int i   100000  i  lt  1000000  i    100          x   method2 i       for  int i   1000000  i  lt  2000000000  i    200          x   method2 i            return x     private static int allMethod3         int x   0      for  int i   0  i  lt  1000  i            x   method3 i       for  int i   1000  i  lt  100000  i    10          x   method3 i       for  int i   100000  i  lt  1000000  i    100          x   method3 i       for  int i   1000000  i  lt  2000000000  i    200          x   method3 i            return x     private static int allMethod4         int x   0      for  int i   0  i  lt  1000  i            x   method4 i       for  int i   1000  i  lt  100000  i    10          x   method4 i       for  int i   100000  i  lt  1000000  i    100          x   method4 i       for  int i   1000000  i  lt  2000000000  i    200          x   method4 i            return x     Again  benchmark   baseline method  with String length   2145ms log10 method  711ms   3 02 times as fast as baseline repeated divide  2797ms   0 77 times as fast as baseline divide-and-conquer  74ms   28 99 times as fast as baseline   Edit After I wrote the benchmark  I took a sneak peak into Integer toString from Java 6  and I found that it uses  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  lt   sizeTable i               return i 1     I benchmarked it against my divide-and-conquer solution   divide-and-conquer  104ms Java 6 solution - iterate and compare  406ms  Mine is about 4x as fast as the Java 6 solution

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I wrote this function after looking Integer java source code   private static int stringSize int x        final int   sizeTable    9  99  999  9 999  99 999  999 999  9 999 999              99 999 999  999 999 999  Integer MAX VALUE       for  int i   0      i            if  x  lt   sizeTable i                 return i   1

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Another string approach  Short and sweet - for any integer n   int length         n  length

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You could could the digits using successive division by ten   int a 0   if  no  lt  0        no   -no    else if  no    0        no   1     while  no  gt  0        no   no   10      a       System out println  Number of digits in given number is    a

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The logarithm is your friend   int n   1000  int length    int  Math log10 n  1     NB  only valid for n   0

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A really simple solution   public int numLength int n      for  int length   1  n   Math pow 10  length     n  length         return length

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Two comments on your benchmark  Java is a complex environment  what with just-in-time compiling and garbage collection and so forth  so to get a fair comparison  whenever I run a benchmark  I always   a  enclose the two tests in a loop that runs them in sequence 5 or 10 times  Quite often the runtime on the second pass through the loop is quite different from the first  And  b  After each  approach   I do a System gc   to try to trigger a garbage collection  Otherwise  the first approach might generate a bunch of objects  but not quite enough to force a garbage collection  then the second approach creates a few objects  the heap is exhausted  and garbage collection runs  Then the second approach is  charged  for picking up the garbage left by the first approach  Very unfair   That said  neither of the above made a significant difference in this example   With or without those modifications  I got very different results than you did  When I ran this  yes  the toString approach gave run times of 6400 to 6600 millis  while the log approach topok 20 000 to 20 400 millis  Instead of being slightly faster  the log approach was 3 times slower for me   Note that the two approaches involve very different costs  so this isn t totally shocking  The toString approach will create a lot of temporary objects that have to be cleaned up  while the log approach takes more intense computation  So maybe the difference is that on a machine with less memory  toString requires more garbage collection rounds  while on a machine with a slower processor  the extra computation of log would be more painful   I also tried a third approach  I wrote this little function   static int numlength int n        if  n    0  return 1      int l      n Math abs n       for  l 0 n gt 0   l          n  10      return l                 That ran in 1600 to 1900 millis -- less than 1 3 of the toString approach  and 1 10 the log approach on my machine   If you had a broad range of numbers  you could speed it up further by starting out dividing by 1 000 or 1 000 000 to reduce the number of times through the loop  I haven t played with that

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simple solution   public class long length       long x l 1 n      for  n 10 n lt x n  10           if  x n  0               l                        System out print l

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int num   02300      int count   0      while num gt 0            if num    0  break           num num 10           count              System out println count

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We can achieve this using a recursive loop      public static int digitCount int numberInput  int i            while  numberInput  gt  0            i            numberInput   numberInput   10          digitCount numberInput  i                     return i             public static void printString             int numberInput   1234567          int digitCount   digitCount numberInput  0            System out println  Count of digit in    numberInput    is    digitCount

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Marian s solution adapted for long type numbers  up to 9 223 372 036 854 775 807   in case someone want s to Copy amp Paste it  In the program I wrote this for numbers up to 10000 were much more probable  so I made a specific branch for them  Anyway it won t make a significative difference   public static int numberOfDigits  long n                Guessing 4 digit numbers will be more probable         They are set in the first branch      if  n  lt  10000L       from 1 to 4         if  n  lt  100L       1 or 2             if  n  lt  10L                    return 1                else                   return 2                          else      3 or 4             if  n  lt  1000L                    return 3                else                   return 4                                           else       from 5 a 20  albeit longs can t have more than 18 or 19          if  n  lt  1000000000000L       from 5 to 12             if  n  lt  100000000L       from 5 to 8                 if  n  lt  1000000L       5 or 6                     if  n  lt  100000L                            return 5                        else                           return 6                                          else      7 u 8                     if  n  lt  10000000L                            return 7                        else                           return 8                                                        else      from 9 to 12                 if  n  lt  10000000000L       9 or 10                     if  n  lt  1000000000L                            return 9                        else                           return 10                                          else      11 or 12                     if  n  lt  100000000000L                            return 11                        else                           return 12                                                                  else      from 13 to      18 or 20              if  n  lt  10000000000000000L       from 13 to 16                 if  n  lt  100000000000000L       13 or 14                     if  n  lt  10000000000000L                             return 13                        else                           return 14                                          else      15 or 16                     if  n  lt  1000000000000000L                            return 15                        else                           return 16                                                        else      from 17 to      20                  if  n  lt  1000000000000000000L       17 or 18                     if  n  lt  100000000000000000L                            return 17                        else                           return 18                                          else      19  Can it be                         10000000000000000000L is nt a valid long                      return 19

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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  lt  95000000           cardDao createCard cardNumber  cardStatus  customerId

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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 lt  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 lt  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

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Since the number of digits in base 10 of an integer is just 1   truncate log10 number    you can do   public class Test        public static void main String   args             final int number   1234          final int digits   1    int Math floor Math log10 number             System out println digits             Edited because my last edit fixed the code example  but not the description

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Easy recursive way  int    get int lenght current lenght  value     if  value   10  lt  10      return  current lenght   1   return  get int lenght current lenght   1  value       not tested

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What about this recursive method       private static int length   0       public static int length int n        length        if  n   10   lt  10            length          else           length n   10             return length

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no String API  no utils  no type conversion  just pure java iteration -   public static int getNumberOfDigits int input        int numOfDigits   1      int base   1      while  input  gt   base   10            base   base   10          numOfDigits              return numOfDigits       You can go long for bigger values if you please

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Can t leave a comment yet  so I ll post as a separate answer    The logarithm-based solution doesn t calculate the correct number of digits for very big long integers  for example   long n   99999999999999999L      correct answer  17 int numberOfDigits   String valueOf n  length        incorrect answer  18 int wrongNumberOfDigits    int   Math log10 n    1      Logarithm-based solution calculates incorrect number of digits in large integers

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Try converting the int to a string and then get the length of the string  That should get the length of the int   public static int intLength int num       String n   Integer toString num       int newNum   n length        return newNum

[java] Way to get number of digits in an int?

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