Sum digits of an integer

This task is to take a Natural Number [[1]] in a given Base and return the sum of its digits:

1₁₀ sums to

1

;

1234₁₀ sums to

10

;

fe₁₆ sums to

29

;

f0e₁₆ sums to

29

.

Ada

Numeric constants in Ada are either decimal or written as B#Digits#. Here B is the base, written as a decimal number, and Digits is a base-B number. E.g., 30, 10#30# 2#11110#, and 16#1E# are the same number -- either written in decimal, binary or hexadecimal notation.

<lang Ada>with Ada.Integer_Text_IO;

procedure Sum_Digits is

  -- sums the digits of an integer (in whatever base)
  -- outputs the sum (in base 10)

  function Sum_Of_Digits(N: Natural; Base: Natural := 10) return Natural is
     Sum: Natural := 0;
     Val: Natural := N;
  begin
     while Val > 0 loop
        Sum := Sum + (Val mod Base);
        Val := Val / Base;
     end loop;
     return Sum;
  end Sum_Of_Digits;

  use Ada.Integer_Text_IO;

begin -- main procedure Sum_Digits

  Put(Sum_OF_Digits(1));            --   1
  Put(Sum_OF_Digits(12345));        --  15
  Put(Sum_OF_Digits(123045));       --  15
  Put(Sum_OF_Digits(123045,  50));  -- 104
  Put(Sum_OF_Digits(16#fe#,  10));  --  11
  Put(Sum_OF_Digits(16#fe#,  16));  --  29
  Put(Sum_OF_Digits(16#f0e#, 16));  --  29

end Sum_Digits;</lang>

Output:

          1         15         15        104         11         29         29

AutoHotkey

Translated from the C version.

<lang AutoHotkey>MsgBox % sprintf("%d %d %d %d %d`n" ,SumDigits(1, 10) ,SumDigits(12345, 10) ,SumDigits(123045, 10) ,SumDigits(0xfe, 16) ,SumDigits(0xf0e, 16) )

SumDigits(n,base) { sum := 0 while (n) { sum += Mod(n,base) n /= base } return sum }

sprintf(s,fmt*) { for each, f in fmt StringReplace,s,s,`%d, % f return s }</lang> Output:

1 15 15 29 29

AWK

MAWK only support base 10 numeric constants, so a conversion function is necessary.

Will sum digits in numbers from base 2 to base 16.

The output is in decimal. Output in other bases would require a function to do the conversion because MAWK's printf() does not support bases other than 10.

Other versions of AWK may not have these limitations.

<lang AWK>#!/usr/bin/awk -f

BEGIN {

   print sumDigits("1")
   print sumDigits("12")
   print sumDigits("fe")
   print sumDigits("f0e")

}

function sumDigits(num, nDigs, digits, sum, d, dig, val, sum) {

   nDigs = split(num, digits, "")
   sum = 0
   for (d = 1; d <= nDigs; d++) {
       dig = digits[d]
       val = digToDec(dig)
       sum += val
   }
   return sum

}

function digToDec(dig) {

   return index("0123456789abcdef", tolower(dig)) - 1

} </lang>

Example output:

BASIC

Works with: QBasic

Works with: PowerBASIC

Translation of: Visual Basic

Note that in order for this to work with the Windows versions of PowerBASIC, the test code needs to be with FUNCTION PBMAIN.

<lang qbasic>FUNCTION sumDigits(num AS STRING, bas AS LONG) AS LONG

   'can handle up to base 36
   DIM outp AS LONG
   DIM validNums AS STRING, tmp AS LONG, x AS LONG, lennum AS LONG, L0 AS LONG
   'ensure num contains only valid characters
   validNums = LEFT$("0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ", bas)
   lennum = LEN(num)
   FOR L0 = lennum TO 1 STEP -1
       x = INSTR(validNums, MID$(num, L0, 1)) - 1
       IF -1 = x THEN EXIT FUNCTION
       tmp = tmp + (x * (bas ^ (lennum - L0)))
   NEXT
   WHILE tmp
       outp = outp + (tmp MOD bas)
       tmp = tmp \ bas
   WEND
   sumDigits = outp

END FUNCTION

PRINT sumDigits(LTRIM$(STR$(1)), 10) PRINT sumDigits(LTRIM$(STR$(1234)), 10) PRINT sumDigits(LTRIM$(STR$(&HFE)), 16) PRINT sumDigits(LTRIM$(STR$(&HF0E)), 16) PRINT sumDigits("2", 2)</lang>

Output:

See also: BBC BASIC, Run BASIC, Visual Basic

Applesoft BASIC

<lang ApplesoftBasic>10 BASE = 10 20 N$ = "1" : GOSUB 100 : PRINT N 30 N$ = "1234" : GOSUB 100 : PRINT N 40 BASE = 16 50 N$ = "FE" : GOSUB 100 : PRINT N 60 N$ = "F0E" : GOSUB 100 : PRINT N 90 END

100 REM SUM DIGITS OF N$, BASE 110 IF BASE = 1 THEN N = LEN(N$) : RETURN 120 IF BASE < 2 THEN BASE = 10 130 N = 0 : V$ = LEFT$("0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ", BASE) 140 FOR I = 1 TO LEN(N$) : C$ = MID$(N$, I, 1) 150 FOR J = 1 TO LEN(V$) 160 IF C$ <> MID$(V$, J, 1) THEN NEXT J : N = SQR(-1) : STOP 170 N = N + J - 1 180 NEXT I 190 RETURN</lang>

BBC BASIC

Works with: BBC BASIC for Windows

This solution deliberately avoids MOD and DIV so it is not restricted to 32-bit integers. <lang bbcbasic> *FLOAT64

     PRINT "Digit sum of 1 (base 10) is "; FNdigitsum(1, 10)
     PRINT "Digit sum of 12345 (base 10) is "; FNdigitsum(12345, 10)
     PRINT "Digit sum of 9876543210 (base 10) is "; FNdigitsum(9876543210, 10)
     PRINT "Digit sum of FE (base 16) is "; ~FNdigitsum(&FE, 16) " (base 16)"
     PRINT "Digit sum of F0E (base 16) is "; ~FNdigitsum(&F0E, 16) " (base 16)"
     END
     
     DEF FNdigitsum(n, b)
     LOCAL q, s
     WHILE n <> 0
       q = INT(n / b)
       s += n - q * b
       n = q
     ENDWHILE
     = s</lang>

Output:

Digit sum of 1 (base 10) is 1
Digit sum of 12345 (base 10) is 15
Digit sum of 9876543210 (base 10) is 45
Digit sum of FE (base 16) is 1D (base 16)
Digit sum of F0E (base 16) is 1D (base 16)

C

<lang c>#include <stdio.h>

int SumDigits(unsigned long long n, const int base) {

   int sum = 0;
   for (; n; n /= base)
   	sum += n % base;
   return sum;

}

int main() {

   printf("%d %d %d %d %d\n",
       SumDigits(1, 10),
       SumDigits(12345, 10),
       SumDigits(123045, 10),
       SumDigits(0xfe, 16),
       SumDigits(0xf0e, 16) );
   return 0;

}</lang>

Output:

1 15 15 29 29

C#

<lang csharp>namespace RosettaCode.SumDigitsOfAnInteger {

   using System;
   using System.Collections.Generic;
   using System.Linq;

   internal static class Program
   {
       /// <summary>
       ///     Enumerates the digits of a number in a given base.
       /// </summary>
       /// <param name="number"> The number. </param>
       /// <param name="base"> The base. </param>
       /// <returns> The digits of the number in the given base. </returns>
       /// <remarks>
       ///     The digits are enumerated from least to most significant.
       /// </remarks>
       private static IEnumerable<int> Digits(this int number, int @base = 10)
       {
           while (number != 0)
           {
               int digit;
               number = Math.DivRem(number, @base, out digit);
               yield return digit;
           }
       }

       /// <summary>
       ///     Sums the digits of a number in a given base.
       /// </summary>
       /// <param name="number"> The number. </param>
       /// <param name="base"> The base. </param>
       /// <returns> The sum of the digits of the number in the given base. </returns>
       private static int SumOfDigits(this int number, int @base = 10)
       {
           return number.Digits(@base).Sum();
       }

       /// <summary>
       ///     Demonstrates <see cref="SumOfDigits" />.
       /// </summary>
       private static void Main()
       {
           foreach (var example in
               new[]
               {
                   new {Number = 1, Base = 10},
                   new {Number = 12345, Base = 10},
                   new {Number = 123045, Base = 10},
                   new {Number = 0xfe, Base = 0x10},
                   new {Number = 0xf0e, Base = 0x10}
               })
           {
               Console.WriteLine(example.Number.SumOfDigits(example.Base));
           }
       }
   }

}</lang> Output:

C++

<lang cpp>#include <iostream>

include <cmath>

int SumDigits(const unsigned long long int digits, const int BASE = 10) {

   int sum = 0;
   unsigned long long int x = digits;
   for (int i = log(digits)/log(BASE); i>0; i--){
       const double z = std::pow(BASE,i);

const unsigned long long int t = x/z; sum += t; x -= t*z;

   }
   return x+sum;

}

int main() {

       std::cout << SumDigits(1) << ' '
                 << SumDigits(12345) << ' '
                 << SumDigits(123045) << ' '
                 << SumDigits(0xfe, 16) << ' '
                 << SumDigits(0xf0e, 16) << std::endl;
       return 0;

}</lang>

Output:

1 15 15 29 29

C++ (template)

Template metaprogramming version. Tested with g++-4.6.3 (Ubuntu). <lang cpp> // Template Metaprogramming version by Martin Ettl

include <iostream>
include <cmath>

typedef unsigned long long int T; template <typename T, T i> void For(T &sum, T &x, const T &BASE) {

   const double z(std::pow(BASE,i));
   const T t = x/z;
   sum += t;
   x -= t*z; 
   For<T, i-1>(sum,x,BASE);

} template <> void For<T,0>(T &, T &, const T &){}

template <typename T, T digits, int BASE> T SumDigits()

{
   T sum(0);
   T x(digits);
   const T end(log(digits)/log(BASE));
   For<T,end>(sum,x,BASE);
   return x+sum;

}

int main() {

       std::cout << SumDigits<T, 1     , 10>()  << ' '
                 << SumDigits<T, 12345 , 10>()  << ' '
                 << SumDigits<T, 123045, 10>()  << ' '
                 << SumDigits<T, 0xfe  , 16>()  << ' '
                 << SumDigits<T, 0xf0e , 16>()  << std::endl;
       return 0;

} </lang>

Output:

1 15 15 29 29

Common Lisp

<lang lisp>(defun sum-digits (number base)

 (loop for n = number then q
       for (q r) = (multiple-value-list (truncate n base))
       sum r until (zerop q)))</lang>

Example: <lang lisp>(loop for (number base) in '((1 10) (1234 10) (#xfe 16) (#xf0e 16))

     do (format t "(~a)_~a = ~a~%" number base (sum-digits number base)))</lang>

Output:

(1)_10 = 1
(1234)_10 = 10
(254)_16 = 29
(3854)_16 = 29

D

<lang d>import std.stdio, std.bigint;

uint sumDigits(T)(T n, in uint base=10) pure /*nothrow*/ in {

   assert(base > 1);

} body {

   typeof(return) total = 0;
   for ( ; n; n /= base)
       total += n % base;
   return total;

}

void main() {

   1.sumDigits.writeln;
   1_234.sumDigits.writeln;
   sumDigits(0xfe, 16).writeln;
   sumDigits(0xf0e, 16).writeln;
   1_234.BigInt.sumDigits.writeln;

}</lang>

Output:

Erlang

<lang erlang> -module(sum_digits). -export([sum_digits/2, sum_digits/1]).

sum_digits(N) ->

   sum_digits(N,10).

sum_digits(N,B) ->

   sum_digits(N,B,0).

sum_digits(0,_,Acc) ->

   Acc;

sum_digits(N,B,Acc) when N < B ->

   Acc+N;

sum_digits(N,B,Acc) ->

   sum_digits(N div B, B, Acc + (N rem B)).

</lang>

Example usage:

2> sum_digits:sum_digits(1).
1
3> sum_digits:sum_digits(1234).
10
4> sum_digits:sum_digits(16#fe,16).
29
5> sum_digits:sum_digits(16#f0e,16).
29

Ezhil

இது ஒரு எழில் தமிழ் நிரலாக்க மொழி உதாரணம்

sum of digits of a number
எண்ணிக்கையிலான இலக்கங்களின் தொகை

நிரல்பாகம் எண்_கூட்டல்( எண் )

 தொகை = 0
 @( எண் > 0 ) வரை
    d = எண்%10;
    பதிப்பி "digit = ",d
    எண் = (எண்-d)/10;
    தொகை  = தொகை  + d
 முடி
 பின்கொடு தொகை

முடி

பதிப்பி எண்_கூட்டல்( 1289)#20 பதிப்பி எண்_கூட்டல்( 123456789)# 45 </lang>

F#

<lang fsharp>open System

let digsum b n =

   let rec loop acc = function
       | n when n > 0 ->
           let m, r = Math.DivRem(n, b)
           loop (acc + r) m
       | _ -> acc
   loop 0 n

[<EntryPoint>] let main argv =

   let rec show = function 
       | n :: b :: r -> printf " %d" (digsum b n); show r
       | _ -> ()

   show [1; 10; 1234; 10; 0xFE; 16; 0xF0E; 16]     // ->  1 10 29 29
   0</lang>

Forth

This is an easy task for Forth, that has built in support for radices up to 36. You set the radix by storing the value in variable BASE. <lang forth>: sum_int 0 begin over while swap base @ /mod swap rot + repeat nip ;

2 base ! 11110 sum_int decimal  . cr

10 base ! 12345 sum_int decimal . cr 16 base ! f0e sum_int decimal . cr</lang>

Fortran

Please find GNU/linux compilation instructions along with the sample output within the comments at the start of this FORTRAN 2008 source. Thank you. Review of this page shows a solution to this task with the number input as text. The solution is the sum of index positions in an ordered list of digit characters. (awk). Other solutions ignore the representations of the input, encode digits using the base, then sum the encoding. Both methods appear in this implementation. <lang FORTRAN> !-*- mode: compilation; default-directory: "/tmp/" -*- !Compilation started at Fri Jun 7 21:00:12 ! !a=./f && make $a && $a !gfortran -std=f2008 -Wall -fopenmp -ffree-form -fall-intrinsics -fimplicit-none f.f08 -o f !f.f08:57.29: ! ! subroutine process1(fmt,s,b) ! 1 !Warning: Unused dummy argument 'b' at (1) !digit sum n ! 1 1 ! 10 1234 ! 29 fe ! 29 f0e ! sum of digits of n expressed in base is... ! n base sum ! 1 10 1 ! 1234 10 10 ! 254 16 29 ! 3854 16 29 ! !Compilation finished at Fri Jun 7 21:00:12

module base_mod

 private :: reverse

contains

 subroutine reverse(a)
   integer, dimension(:), intent(inout) :: a
   integer :: i, j, t
   do i=1,size(a)/2
      j = size(a) - i + 1
      t = a(i)
      a(i) = a(j)
      a(j) = t
   end do
 end subroutine reverse

 function antibase(b, n) result(a)
   integer, intent(in) :: b,n
   integer, dimension(32) :: a
   integer :: m, i
   a = 0
   m = n
   i = 1
   do while (m .ne. 0)
      a(i) = mod(m, b)
      m = m/b
      i = i+1
   end do
   call reverse(a)
 end function antibase

end module base_mod

program digit_sum

 use base_mod
 call still
 call confused

contains

 subroutine still
   character(len=6),parameter :: fmt = '(i9,a)'
   print'(a9,a8)','digit sum','n'
   call process1(fmt,'1',10)
   call process1(fmt,'1234',10)
   call process1(fmt,'fe',16)
   call process1(fmt,'f0e',16)
 end subroutine still

 subroutine process1(fmt,s,b)
   character(len=*), intent(in) :: fmt, s
   integer, intent(in), optional :: b
   integer :: i
   print fmt,sum((/(index('123456789abcdef',s(i:i)),i=1,len(s))/)),' '//s
 end subroutine process1

 subroutine confused
   character(len=5),parameter :: fmt = '(3i7)'
   print*,'sum of digits of n expressed in base is...'
   print'(3a7)','n','base','sum'
   call process0(10,1,fmt)
   call process0(10,1234,fmt)
   call process0(16,254,fmt)
   call process0(16,3854,fmt)
 end subroutine confused

 subroutine process0(b,n,fmt)
   integer, intent(in) :: b, n
   character(len=*), intent(in) :: fmt
   print fmt,n,b,sum(antibase(b, n))
 end subroutine process0

end program digit_sum </lang>

Go

Handling numbers up to 2^63 and bases from 2 to 36 is pretty easy. <lang go>package digit

import (

   "errors"
   "strconv"

)

func Sum(n string, base int) (int64, error) {

   if base < 2 || base > 36 {
       return 0, errors.New("base must be from 2 to 36")
   }
   i, err := strconv.ParseInt(n, base, 64)
   if err != nil {
       return 0, err
   }
   if i < 0 {
       return 0, errors.New("number must be non-negative")
   }
   b64 := int64(base)
   var sum int64
   for i > 0 {
       sum += i % b64
       i /= b64
   }
   return sum, nil

}</lang> <lang go>package digit_test

import (

   "testing"

   "digit"

)

type testCase struct {

   n    string
   base int
   dSum int64

}

var testData = []testCase{

   {"1", 10, 1},
   {"1234", 10, 10},
   {"fe", 16, 29},
   {"f0e", 16, 29},

}

func testSum(t *testing.T) {

   for _, tc := range testData {
       ds, err := digit.Sum(tc.n, tc.base)
       if err != nil {
           t.Fatal("test case", tc, err)
       }
       if ds != tc.dSum {
           t.Fatal("test case", tc, "got", ds, "expected", tc.dSum)
       }
   }

}</lang>

Groovy

Solution: <lang groovy>def digitsum = { number, radix = 10 ->

   Integer.toString(number, radix).collect { Integer.parseInt(it, radix) }.sum()

}</lang>

Test: <lang groovy>[[30, 2], [30, 10], [1, 10], [12345, 10], [123405, 10], [0xfe, 16], [0xf0e, 16]].each {

   println """
   Decimal value:     ${it[0]}
   Radix:             ${it[1]}
   Radix value:       ${Integer.toString(it[0], it[1])}
   Decimal Digit Sum: ${digitsum(it[0], it[1])}
   Radix Digit Sum:   ${Integer.toString(digitsum(it[0], it[1]), it[1])}
   """

}</lang>

Output:

    Decimal value:     30
    Radix:             2
    Radix value:       11110
    Decimal Digit Sum: 4
    Radix Digit Sum:   100
    

    Decimal value:     30
    Radix:             10
    Radix value:       30
    Decimal Digit Sum: 3
    Radix Digit Sum:   3
    

    Decimal value:     1
    Radix:             10
    Radix value:       1
    Decimal Digit Sum: 1
    Radix Digit Sum:   1
    

    Decimal value:     12345
    Radix:             10
    Radix value:       12345
    Decimal Digit Sum: 15
    Radix Digit Sum:   15
    

    Decimal value:     123405
    Radix:             10
    Radix value:       123405
    Decimal Digit Sum: 15
    Radix Digit Sum:   15
    

    Decimal value:     254
    Radix:             16
    Radix value:       fe
    Decimal Digit Sum: 29
    Radix Digit Sum:   1d
    

    Decimal value:     3854
    Radix:             16
    Radix value:       f0e
    Decimal Digit Sum: 29
    Radix Digit Sum:   1d

Haskell

<lang haskell>digsum base = f 0 where f a 0 = a f a n = f (a+r) q where (q,r) = n `divMod` base

main = print $ digsum 16 255 -- "FF": 15 + 15 = 30</lang>

Icon and Unicon

This solution works in both languages. This solution assumes the input number is expressed in the indicated base. This assumption differs from that made in some of the other solutions.

<lang unicon>procedure main(a)

   write(dsum(a[1]|1234,a[2]|10))

end

procedure dsum(n,b)

   n := integer((\b|10)||"r"||n)
   sum := 0
   while sum +:= (0 < n) % b do n /:= b
   return sum

end</lang>

Sample runs:

->sdi 1
1
->sdi 1234
10
->sdi fe 16
29
->sdi f0e 16
29
->sdi ff 16
30
->sdi 255 16
12
->sdi fffff 16
75
->sdi 254 16
11
->

J

<lang j>digsum=: 10&$: : (+/@(#.inv))</lang>

Example use:

<lang J> digsum 1234 10

  10 digsum 254

11

  16 digsum 254

29</lang>

Illustration of mechanics:

<lang j> 10 #. 1 2 3 4 1234

 10 #.inv 1234

1 2 3 4

 10 +/ 1 2 3 4

10

 10 +/@(#.inv) 1234

10</lang>

So #.inv gives us the digits, +/ gives us the sum, and @ glues them together with +/ being a "post processor" for #.inv or, as we say in the expression: (#.inv). We need the parenthesis or inv will try to look up the inverse of +/@#. and that's not well defined.

The rest of it is about using 10 as the default left argument when no left argument is defined. A J verb has a monadic definition (for use with one argument) and a dyadic definition (for use with two arguments) and : derives a new verb where the monadic definition is used from the verb on the left and the dyadic definition is used from the verb on the right. $: is a self reference to the top-level defined verb.

Full examples:

<lang j> digsum 1 1

  digsum 1234

10

  16 digsum 16bfe

29

  16 digsum 16bf0e

29</lang>

Note that J implements numeric types -- J tries to ensure that the semantics of numbers match their mathematical properties. So it doesn't matter how we originally obtained a number.

<lang j> 200+54 254

254

  2.54e2

254

  16bfe

254

  254b10 , 1r254b0.1  NB. 10 in base 254 , 0.1 in base 1/254

254 254</lang>

Java

<lang java>import java.math.BigInteger; public class SumDigits {

   public static int sumDigits(long num) {

return sumDigits(num, 10);

   }
   public static int sumDigits(long num, int base) {

String s = Long.toString(num, base); int result = 0; for (int i = 0; i < s.length(); i++) result += Character.digit(s.charAt(i), base); return result;

   }
   public static int sumDigits(BigInteger num) {

return sumDigits(num, 10);

   }
   public static int sumDigits(BigInteger num, int base) {

String s = num.toString(base); int result = 0; for (int i = 0; i < s.length(); i++) result += Character.digit(s.charAt(i), base); return result;

   public static void main(String[] args) {

System.out.println(sumDigits(1)); System.out.println(sumDigits(12345)); System.out.println(sumDigits(123045)); System.out.println(sumDigits(0xfe, 16)); System.out.println(sumDigits(0xf0e, 16)); System.out.println(sumDigits(new BigInteger("12345678901234567890")));

}</lang>

Output:

Julia

Using the built-in digits function: <lang julia>sumdigits(n, base=10) = sum(digits(n, base))</lang>

Lasso

<lang Lasso>define br => '
\n'

define sumdigits(int, base = 10) => { fail_if(#base < 2, -1, 'Base need to be at least 2') local( out = integer, divmod ) while(#int) => { #divmod = #int -> div(#base) #int = #divmod -> first #out += #divmod -> second } return #out }

sumdigits(1) br sumdigits(12345) br sumdigits(123045) br sumdigits(0xfe, 16) br sumdigits(0xf0e, 16)</lang> Output:

Mathematica

<lang Mathematica>Total[IntegerDigits[1234]] Total[IntegerDigits[16^^FE, 16]]</lang>

Output:

10
29

МК-61/52

NetRexx

Strings

Processes data as text from the command line. Provides a representative sample if no input is supplied: <lang NetRexx>/* NetRexx */ options replace format comments java crossref symbols nobinary

parse arg input inputs = ['1234', '01234', '0xfe', '0xf0e', '0', '00', '0,2' '1', '070', '77, 8' '0xf0e, 10', '070, 16', '0xf0e, 36', '000999ABCXYZ, 36', 'ff, 16', 'f, 10', 'z, 37'] -- test data if input.length() > 0 then inputs = [input] -- replace test data with user input loop i_ = 0 to inputs.length - 1

 in = inputs[i_]
 parse in val . ',' base .
 dSum = sumDigits(val, base)
 say 'Sum of digits for integer "'val'" for a given base of "'base'":' dSum'\-'
 -- Carry the exercise to it's logical conclusion and sum the results to give a single digit in range 0-9
 loop while dSum.length() > 1 & dSum.datatype('n')
   dSum = sumDigits(dSum, 10)
   say ',' dSum'\-'
   end
 say
 end i_

-- Sum digits of an integer method sumDigits(val = Rexx, base = Rexx ) public static returns Rexx

 rVal = 0
 parse normalizeValue(val, base) val base .
 loop label digs for val.length()
   -- loop to extract digits from input and sum them
   parse val dv +1 val
   do
     rVal = rVal + Integer.valueOf(dv.toString(), base).intValue()
   catch ex = NumberFormatException
     rVal = 'NumberFormatException:' ex.getMessage()
     leave digs
   end
   end digs
 return rVal

-- Clean up the input, normalize the data and determine which base to use method normalizeValue(inV = Rexx, base = Rexx ) private static returns Rexx

 inV = inV.strip('l')
 base = base.strip()
 parse inV xpref +2 . -
        =0 opref +1 . -
        =0 . '0x' xval . ',' . -
        =0 . '0'  oval . ',' . -
        =0 dval .

 select
   when xpref = '0x' & base.length() = 0 then do
     -- value starts with '0x' and no base supplied.  Assign hex as base
     inval = xval
     base = 16
     end
   when opref = '0'  & base.length() = 0 then do
     -- value starts with '0' and no base supplied.  Assign octal as base
     inval = oval
     base = 8
     end
   otherwise do
     inval = dval
     end
   end
 if base.length() = 0 then base = 10 -- base not set.  Assign decimal as base
 if inval.length() <= 0 then inval = 0 -- boundary condition.  Invalid input or a single zero
 rVal = inval base

 return rVal

</lang> Output

Sum of digits for integer "1234" for a given base of "": 10, 1
Sum of digits for integer "01234" for a given base of "": 10, 1
Sum of digits for integer "0xfe" for a given base of "": 29, 11, 2
Sum of digits for integer "0xf0e" for a given base of "": 29, 11, 2
Sum of digits for integer "0" for a given base of "": 0
Sum of digits for integer "00" for a given base of "": 0
Sum of digits for integer "0" for a given base of "2": 0
Sum of digits for integer "070" for a given base of "": 7
Sum of digits for integer "77" for a given base of "8": 14, 5
Sum of digits for integer "070" for a given base of "16": 7
Sum of digits for integer "0xf0e" for a given base of "36": 62, 8
Sum of digits for integer "000999ABCXYZ" for a given base of "36": 162, 9
Sum of digits for integer "ff" for a given base of "16": 30, 3
Sum of digits for integer "f" for a given base of "10": NumberFormatException: For input string: "f"
Sum of digits for integer "z" for a given base of "37": NumberFormatException: radix 37 greater than Character.MAX_RADIX

Type `int`

Processes sample data as int arrays: <lang NetRexx>/* NetRexx */ options replace format comments java crossref symbols binary

inputs = [[int 1234, 10], [octal('01234'), 8], [0xfe, 16], [0xf0e,16], [8b0, 2], [16b10101100, 2], [octal('077'), 8]] -- test data loop i_ = 0 to inputs.length - 1

 in = inputs[i_, 0]
 ib = inputs[i_, 1]
 dSum = sumDigits(in, ib)
 say 'Sum of digits for integer "'Integer.toString(in, ib)'" for a given base of "'ib'":' dSum'\-'
 -- Carry the exercise to it's logical conclusion and sum the results to give a single digit in range 0-9
 loop while dSum.length() > 1 & dSum.datatype('n')
   dSum = sumDigits(dSum, 10)
   say ',' dSum'\-'
   end
 say
 end i_

-- Sum digits of an integer method sumDigits(val = int, base = int 10) public static returns Rexx

 rVal = Rexx 0
 sVal = Rexx(Integer.toString(val, base))
 loop label digs for sVal.length()
   -- loop to extract digits from input and sum them
   parse sVal dv +1 sVal
   do
     rVal = rVal + Integer.valueOf(dv.toString(), base).intValue()
   catch ex = NumberFormatException
     rVal = 'NumberFormatException:' ex.getMessage()
     leave digs
   end
   end digs
 return rVal

-- if there's a way to insert octal constants into an int in NetRexx I don't remember it method octal(oVal = String) private constant returns int signals NumberFormatException

 iVal = Integer.valueOf(oVal, 8).intValue()
 return iVal

</lang> Output

Sum of digits for integer "1234" for a given base of "10": 10, 1
Sum of digits for integer "1234" for a given base of "8": 10, 1
Sum of digits for integer "fe" for a given base of "16": 29, 11, 2
Sum of digits for integer "f0e" for a given base of "16": 29, 11, 2
Sum of digits for integer "0" for a given base of "2": 0
Sum of digits for integer "10101100" for a given base of "2": 4
Sum of digits for integer "77" for a given base of "8": 14, 5

Oberon-2

<lang oberon2> MODULE SumDigits; IMPORT Out; PROCEDURE Sum(n: LONGINT;base: INTEGER): LONGINT; VAR sum: LONGINT; BEGIN sum := 0; WHILE (n > 0) DO INC(sum,(n MOD base)); n := n DIV base END; RETURN sum END Sum; BEGIN Out.String("1 : ");Out.LongInt(Sum(1,10),10);Out.Ln; Out.String("1234 : ");Out.LongInt(Sum(1234,10),10);Out.Ln; Out.String("0FEH : ");Out.LongInt(Sum(0FEH,16),10);Out.Ln; Out.String("OF0EH : ");Out.LongInt(Sum(0F0EH,16),10);Out.Ln END SumDigits. </lang> Output:

1     :          1
1234  :         10
0FEH  :         29
OF0EH :         29

OCaml

<lang ocaml>let sum_digits ~digits ~base =

 let rec aux sum x =
   if x <= 0 then sum else
   aux (sum + x mod base) (x / base)
 in
 aux 0 digits

let () =

 Printf.printf "%d %d %d %d %d\n"
   (sum_digits 1 10)
   (sum_digits 12345 10)
   (sum_digits 123045 10)
   (sum_digits 0xfe 16)
   (sum_digits 0xf0e 16)</lang>

Output:

1 15 15 29 29

PARI/GP

<lang parigp>dsum(n,base)=my(s); while(n, s += n%base; n \= base); s</lang>

Also the built-in sumdigits can be used for base 10.

Perl

<lang Perl>#!/usr/bin/perl use strict ; use warnings ;

whatever the number base, a number stands for itself, and the letters start
at number 10 !

sub sumdigits {

  my $number = shift ;
  my $hashref = shift ;
  my $sum = 0 ;
  map { if ( /\d/ ) { $sum += $_ } else { $sum += ${$hashref}{ $_ } } } 
     split( // , $number ) ;
  return $sum ;

}

my %lettervals ; my $base = 10 ; for my $letter ( 'a'..'z' ) {

  $lettervals{ $letter } = $base++ ;

} map { print "$_ sums to " . sumdigits( $_ , \%lettervals) . " !\n" }

  ( 1 , 1234 , 'fe' , 'f0e' ) ;

</lang> Output:

1 sums to 1 !
1234 sums to 10 !
fe sums to 29 !
f0e sums to 29 !

Perl 6

This will handle input numbers in any base from 2 to 36. The results are in base 10. <lang perl6>say Σ $_ for <1 1234 1020304 fe f0e DEADBEEF>;

sub Σ { [+] $^n.comb.map: { :36($_) } }</lang>

Output:

PHP

<lang php><?php function sumDigits($num, $base = 10) {

   $s = base_convert($num, 10, $base);
   foreach (str_split($s) as $c)
       $result += intval($c, $base);
   return $result;

} echo sumDigits(1), "\n"; echo sumDigits(12345), "\n"; echo sumDigits(123045), "\n"; echo sumDigits(0xfe, 16), "\n"; echo sumDigits(0xf0e, 16), "\n"; ?></lang>

Output:

PicoLisp

<lang PicoLisp>(de sumDigits (N Base)

  (or
     (=0 N)
     (+ (% N Base) (sumDigits (/ N Base) Base)) ) )</lang>

Test: <lang PicoLisp>: (sumDigits 1 10) -> 1

(sumDigits 1234 10)

-> 10

(sumDigits (hex "fe") 16)

-> 29

(sumDigits (hex "f0e") 16)

-> 29</lang>

PL/I

<lang PL/I> sum_digits: procedure options (main); /* 4/9/2012 */

  declare ch character (1);
  declare (k, sd) fixed;

  on endfile (sysin) begin; put skip data (sd); stop; end;
  sd = 0;
  do forever;
     get edit (ch) (a(1)); put edit (ch) (a);
     k = index('abcdef', ch);
     if k > 0 then /* we have a base above 10 */
        sd = sd + 9 + k;
     else
        sd = sd + ch;
  end;

end sum_digits; </lang> results:

5c7e
SD=      38;
10111000001
SD=       5;

Powershell

<lang Powershell>function Get-DigitalSum ($n) {

   if ($n -lt 10) {$n}
   else {
       ($n % 10) + (Get-DigitalSum ([math]::Floor($n / 10)))
   }

}</lang>

Python

<lang python>def toBaseX(num, base):

   output = []
   while num:
       num, rem = divmod(num, base)
       output.append(rem)
   return output

def sumDigits(num, base=10):

   if base < 2:
       print "Error: Base must be at least 2"
       return
   return sum(toBaseX(num, base))

print sumDigits(1) print sumDigits(12345) print sumDigits(123045) print sumDigits(0xfe, 16) print sumDigits(0xf0e, 16)</lang> Output

Racket

<lang Racket>#lang racket (define (sum-of-digits n base (sum 0))

 (if (= n 0)
     sum
     (sum-of-digits (quotient n base)
                    base
                    (+ (remainder n base) sum))))

(for-each

(lambda (number-base-pair)
  (define number (car number-base-pair))
  (define base (cadr number-base-pair))
  (displayln (format "(~a)_~a = ~a" number base (sum-of-digits number base))))
'((1 10) (1234 10) (#xfe 16) (#xf0e 16)))

outputs
(1)_10 = 1
(1234)_10 = 10
(254)_16 = 29
(3854)_16 = 29</lang>

REXX

version 1

<lang rexx> /* REXX **************************************************************

04.12.2012 Walter Pachl
- - - - /

digits='0123456789ABCDEF' Do i=1 To length(digits)

 d=substr(digits,i,1)                                                  
 value.d=i-1                                                           
 End

Call test '1' Call test '1234' Call test 'FE' Call test 'F0E' Exit test:

 Parse Arg number                                                      
 res=right(number,4)                                                   
 dsum=0                                                                
 Do While number<>                                                   
   Parse Var number d +1 number                                        
   dsum=dsum+value.d                                                   
   End                                                                 
 Say res '->' right(dsum,2)                                            
 Return</lang>

Output:

   1 ->  1
1234 -> 10
  FE -> 29
 F0E -> 29

version 2

This REXX version allows:

leading signs (+ -)
decimal points
leading and/or trailing whitespace
numbers may be in mixed case
numbers may include commas (,)
numbers may be expressed up to base 36
numbers may be any length (size)

<lang rexx>/*REXX pgm sums the digits of natural numbers in any base up to base 36.*/ parse arg z /*get optional #s or use default.*/ if z= then z='1 1234 fe f0e +F0E -666.00 11111112222222333333344444449'

    do j=1  for words(z);     _=word(z,j)
    say right(sumDigs(_),9) ' is the sum of the digits for the number ' _
    end   /*j*/

exit /*stick a fork in it, we're done.*/ /*──────────────────────────────────SUMDIGS subroutine──────────────────*/ sumDigs: procedure; arg x; @=123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ s=0; do k=1 for length(x); s=s+pos(substr(x,k,1),@)

                   end   /*k*/

return s</lang> output when using the default input

        1  is the sum of the digits for the number  1
       10  is the sum of the digits for the number  1234
       29  is the sum of the digits for the number  fe
       29  is the sum of the digits for the number  f0e
       29  is the sum of the digits for the number  +F0E
       18  is the sum of the digits for the number  -666.00
       79  is the sum of the digits for the number  11111112222222333333344444449

Ruby

<lang ruby>>> def sumDigits(num, base = 10) >> num.to_s(base).split(//).inject(0) {|z, x| z + x.to_i(base)} >> end => nil >> sumDigits(1) => 1 >> sumDigits(12345) => 15 >> sumDigits(123045) => 15 >> sumDigits(0xfe, 16) => 29 >> sumDigits(0xf0e, 16) => 29 </lang>

Run BASIC

This example is incorrect. Please fix the code and remove this message.

Details: It only handles base 10, but should be able to handle multiple and/or arbitrary bases.

<lang runbasic>input "Gimme a number:";n

print "Sum of digits :";n;" is ";sum(n) end function sum(n) n$ = str$(n) for i = 1 to len(n$)

 sum = sum + val(mid$(n$,i,1))

next i

end function</lang>

Gimme a number:?123456789
Sum of digits :123456789 is 45

Scala

<lang scala>def sumDigits(x:BigInt, base:Int=10):BigInt=sumDigits(x.toString(base), base) def sumDigits(x:String, base:Int):BigInt = x map(_.asDigit) sum</lang> Test: <lang scala>sumDigits(0) // => 0 sumDigits(0, 2) // => 0 sumDigits(0, 16) // => 0 sumDigits("00", 2) // => 0 sumDigits("00", 10) // => 0 sumDigits("00", 16) // => 0 sumDigits(1234) // => 10 sumDigits(0xfe) // => 11 sumDigits(0xfe, 16) // => 29 sumDigits(0xf0e, 16) // => 29 sumDigits(077) // => 9 sumDigits(077, 8) // => 14 sumDigits("077", 8) // => 14 sumDigits("077", 10) // => 14 sumDigits("077", 16) // => 14 sumDigits("0xf0e", 36) // => 62 sumDigits("000999ABCXYZ", 36) // => 162 sumDigits(BigInt("12345678901234567890")) // => 90 sumDigits("12345678901234567890", 10) // => 90</lang>

Seed7

<lang seed7>$ include "seed7_05.s7i";

const func integer: sumDigits (in var integer: num, in integer: base) is func

 result
    var integer: sum is 0;
 begin
   while num > 0 do
     sum +:= num rem base;
     num := num div base;
   end while;
 end func;

const proc: main is func

 begin
   writeln(sumDigits(1,      10));
   writeln(sumDigits(12345,  10));
   writeln(sumDigits(123045, 10));
   writeln(sumDigits(123045, 50));
   writeln(sumDigits(16#fe,  10));
   writeln(sumDigits(16#fe,  16));
   writeln(sumDigits(16#f0e, 16));
 end func;</lang>

Output:

Tcl

Supporting arbitrary bases makes this primarily a string operation. <lang tcl>proc sumDigits {num {base 10}} {

   set total 0
   foreach d [split $num ""] {

if {[string is alpha $d]} { set d [expr {[scan [string tolower $d] %c] - 87}] } elseif {![string is digit $d]} { error "bad digit: $d" } if {$d >= $base} { error "bad digit: $d" } incr total $d

   }
   return $total

}</lang> Demonstrating: <lang tcl>puts [sumDigits 1] puts [sumDigits 12345] puts [sumDigits 123045] puts [sumDigits fe 16] puts [sumDigits f0e 16] puts [sumDigits 000999ABCXYZ 36]</lang>

Output:

Visual Basic

This version checks that only valid digits for the indicated base are passed in, exiting otherwise.

<lang vb>Function sumDigits(num As Variant, base As Long) As Long

   'can handle up to base 36
   Dim outp As Long
   Dim validNums As String, tmp As Variant, x As Long, lennum As Long
   'ensure num contains only valid characters
   validNums = Left$("0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ", base)
   lennum = Len(num)
   For L0 = lennum To 1 Step -1
       x = InStr(validNums, Mid$(num, L0, 1)) - 1
       If -1 = x Then Exit Function
       tmp = tmp + (x * (base ^ (lennum - L0)))
   Next
   While tmp
       outp = outp + (tmp Mod base)
       tmp = tmp \ base
   Wend
   sumDigits = outp

End Function

Sub tester()

   Debug.Print sumDigits(1, 10)
   Debug.Print sumDigits(1234, 10)
   Debug.Print sumDigits(&HFE, 16)
   Debug.Print sumDigits(&HF0E, 16)
   Debug.Print sumDigits("2", 2)

End Sub</lang>

Output (in the debug window):

XPL0

<lang XPL0>code ChOut=8, CrLf=9, IntOut=11;

func SumDigits(N, Base); int N, Base, Sum; [Sum:= 0; repeat N:= N/Base;

       Sum:= Sum + rem(0);

until N=0; return Sum; ];

[IntOut(0, SumDigits(1, 10)); ChOut(0, ^ );

IntOut(0, SumDigits(12345,  10));  ChOut(0, ^ );
IntOut(0, SumDigits(123045, 10));  ChOut(0, ^ );
IntOut(0, SumDigits($FE,    10));  ChOut(0, ^ );
IntOut(0, SumDigits($FE,    16));  ChOut(0, ^ );
IntOut(0, SumDigits($F0E,   16));  CrLf(0);

]</lang>

Output:

1 15 15 11 29 29

Icon and Unicon

Strings

Type int

version 1

version 2

Type `int`