Increment a numerical string

Task

Increment a numerical string.

ABAP

<lang ABAP>report zz_incstring perform test using: '0', '1', '-1', '10000000', '-10000000'.

form test using iv_string type string.

 data: lv_int  type i,
       lv_string type string.
 lv_int = iv_string + 1.
 lv_string = lv_int.
 concatenate '"' iv_string '" + 1 = "' lv_string '"' into lv_string.
 write / lv_string.

endform. </lang>

Output:

"0" + 1 = "1 "
"1" + 1 = "2 "
"-1" + 1 = "0 "
"10000000" + 1 = "10000001 "
"-10000000" + 1 = "9999999-"

ActionScript

<lang ActionScript>function incrementString(str:String):String { return String(Number(str)+1); }</lang>

The standard Ada package Ada.Strings.Fixed provides a function for trimming blanks from a string. <lang ada>S : String := "12345"; S := Ada.Strings.Fixed.Trim(Source => Integer'Image(Integer'Value(S) + 1), Side => Ada.Strings.Both);</lang>

Aime

ALGOL 68

Works with: ALGOL 68 version Revision 1 - no extensions to language used

Works with: ALGOL 68G version Any - tested with release 1.18.0-9h.tiny

<lang algol68>STRING s := "12345"; FILE f; INT i; associate(f, s); get(f,i); i+:=1; s:=""; reset(f); put(f,i); print((s, new line))</lang>

Output:

+12346

Apex

<lang apex> string count = '12345'; count = String.valueOf(integer.valueOf(count)+1); system.debug('Incremental Value : '+count); </lang>

Output:

AutoHotkey

<lang autohotkey>str = 12345 MsgBox % str += 1</lang>

Output:

AutoIt

<lang autoIt>Global $x = "12345" $x += 1 MsgBox(0,"",$x)</lang>

Output:

AWK

The example shows that the string s can be incremented, but after that still is a string of length 2. <lang awk>$ awk 'BEGIN{s="42"; s++; print s"("length(s)")" }' 43(2)</lang>

BASIC

Works with: BBC BASIC

Works with: QBasic

Works with: PowerBASIC

Works with: Visual Basic

Works with: Liberty BASIC

ZX Spectrum Basic

The ZX Spectrum needs line numbers and a let statement, but the same technique can be used:

Batch File

Since environment variables have no type distinction all numbers are simply numeric strings:

Works with: Windows NT version 4

BBC BASIC

This assumes the task is about incrementing an arbitrary-length decimal string. <lang bbcbasic> num$ = "567"

     REPEAT
       PRINT num$
       PROCinc$(num$)
     UNTIL FALSE
     END
     
     DEF PROCinc$(RETURN n$)
     LOCAL A$, I%
     I% = LEN(n$)
     REPEAT
       A$ = CHR$(ASCMID$(n$,I%) + 1)
       IF A$=":" A$ = "0"
       MID$(n$,I%,1) = A$
       I% -= 1
     UNTIL A$<>"0" OR I%=0
     IF A$="0" n$ = "1" + n$
     ENDPROC</lang>

Boo

<lang boo>s = "1234" s = (int.Parse(s) + 1).ToString()</lang>

Bracmat

Numbers are strings. Bracmat supports rational numbers, including integers, using arbitrary-precision arithmetic. Pure imaginary numbers are formed using a factor i (or -i). There is no support for floating point arithmetics. (Historically, because the ARM 2 processor in the Archimedes computer didn't sport an FPU.) <lang bracmat>(n=35664871829866234762187538073934873121878/6172839450617283945) &!n+1:?n &out$!n

  35664871829866234762193710913385490405823/6172839450617283945

</lang>

Brat

<lang brat>#Convert to integer, increment, then back to string p ("100".to_i + 1).to_s #Prints 101</lang>

Burlesque

C

Handling strings of arbitrary sizes:<lang c>#include <stdio.h>

include <string.h>
include <stdlib.h>

/* Constraints: input is in the form of (\+|-)?[0-9]+

*  and without leading zero (0 itself can be as "0" or "+0", but not "-0");
*  input pointer is realloc'able and may change;
*  if input has leading + sign, return may or may not keep it.
*  The constranits conform to sprintf("%+d") and this function's own output.
*/

char * incr(char *s) { int i, begin, tail, len; int neg = (*s == '-'); char tgt = neg ? '0' : '9';

/* special case: "-1" */ if (!strcmp(s, "-1")) { s[0] = '0', s[1] = '\0'; return s; }

len = strlen(s); begin = (*s == '-' || *s == '+') ? 1 : 0;

/* find out how many digits need to be changed */ for (tail = len - 1; tail >= begin && s[tail] == tgt; tail--);

if (tail < begin && !neg) { /* special case: all 9s, string will grow */ if (!begin) s = realloc(s, len + 2); s[0] = '1'; for (i = 1; i <= len - begin; i++) s[i] = '0'; s[len + 1] = '\0'; } else if (tail == begin && neg && s[1] == '1') { /* special case: -1000..., so string will shrink */ for (i = 1; i < len - begin; i++) s[i] = '9'; s[len - 1] = '\0'; } else { /* normal case; change tail to all 0 or 9, change prev digit by 1*/ for (i = len - 1; i > tail; i--) s[i] = neg ? '9' : '0'; s[tail] += neg ? -1 : 1; }

return s; }

void string_test(const char *s) { char *ret = malloc(strlen(s)); strcpy(ret, s);

printf("text: %s\n", ret); printf(" ->: %s\n", ret = incr(ret)); free(ret); }

int main() { string_test("+0"); string_test("-1"); string_test("-41"); string_test("+41"); string_test("999"); string_test("+999"); string_test("109999999999999999999999999999999999999999"); string_test("-100000000000000000000000000000000000000000000");

return 0; }</lang>

Output:

text: +0

 ->: +1

text: -1

 ->: 0

text: -41

 ->: -40

text: +41

 ->: +42

text: 999

 ->: 1000

text: +999

 ->: 1000

text: 109999999999999999999999999999999999999999

 ->: 110000000000000000000000000000000000000000

text: -100000000000000000000000000000000000000000000

->: -99999999999999999999999999999999999999999999

C++

<lang cpp>// standard C++ string stream operators

include <cstdlib>
include <string>
include <sstream>

// inside a function or method... std::string s = "12345";

int i; std::istringstream(s) >> i; i++; //or: //int i = std::atoi(s.c_str()) + 1;

std::ostringstream oss; if (oss << i) s = oss.str();</lang>

Works with: C++11

<lang cpp>#include <string>

std::string s = "12345"; s = std::to_string(1+std::stoi(s));</lang>

Library: Boost

<lang cpp>// Boost

include <cstdlib>
include <string>
include <boost/lexical_cast.hpp>

// inside a function or method... std::string s = "12345"; int i = boost::lexical_cast<int>(s) + 1; s = boost::lexical_cast<std::string>(i);</lang>

Library: Qt

Uses: Qt (Components:{{#foreach: component$n$|{{{component$n$}}}Property "Uses Library" (as page type) with input value "Library/Qt/{{{component$n$}}}" contains invalid characters or is incomplete and therefore can cause unexpected results during a query or annotation process., }})

<lang cpp>// Qt QString num1 = "12345"; QString num2 = QString("%1").arg(v1.toInt()+1);</lang>

Library: MFC

Uses: Microsoft Foundation Classes (Components:{{#foreach: component$n$|{{{component$n$}}}Property "Uses Library" (as page type) with input value "Library/Microsoft Foundation Classes/{{{component$n$}}}" contains invalid characters or is incomplete and therefore can cause unexpected results during a query or annotation process., }})

Uses: C Runtime (Components:{{#foreach: component$n$|{{{component$n$}}}Property "Uses Library" (as page type) with input value "Library/C Runtime/{{{component$n$}}}" contains invalid characters or is incomplete and therefore can cause unexpected results during a query or annotation process., }})

<lang cpp>// MFC CString s = "12345"; int i = _ttoi(s) + 1; int i = _tcstoul(s, NULL, 10) + 1; s.Format("%d", i);</lang>

All of the above solutions only work for numbers <= INT_MAX. The following works for an (almost) arbitrary large number:

Works with: g++ version 4.0.2

<lang cpp>#include <string>

include <iostream>
include <ostream>

void increment_numerical_string(std::string& s) {

   std::string::reverse_iterator iter = s.rbegin(), end = s.rend();
   int carry = 1;
   while (carry && iter != end)
   {
       int value = (*iter - '0') + carry;
       carry = (value / 10);
       *iter = '0' + (value % 10);
       ++iter;
   }
   if (carry)
       s.insert(0, "1");

}

int main() {

   std::string big_number = "123456789012345678901234567899";
   std::cout << "before increment: " << big_number << "\n";
   increment_numerical_string(big_number);
   std::cout << "after increment:  " << big_number << "\n";

}</lang>

C#

<lang csharp>string s = "12345"; s = (int.Parse(s) + 1).ToString();</lang>

Ceylon

<lang ceylon>shared void run() {

"Increments a numeric string by 1. Returns a float or integer depending on the string. Returns null if the string isn't a number." function inc(String string) => if(exists integer = parseInteger(string)) then integer + 1 else if(exists float = parseFloat(string)) then float + 1.0 else null;

value a = "1"; print(a); value b = inc(a); print(b); value c = "1.0"; print(c); value d = inc(c); print(d); }</lang>

Clojure

<lang lisp>(str (inc (Integer/parseInt "1234")))</lang>

CMake

CMake performs all arithmetic with numeric strings, through its math() command.

<lang cmake>set(string "1599") math(EXPR string "${string} + 1") message(STATUS "${string}")</lang>

-- 1600

COBOL

<lang cobol> PROGRAM-ID. increment-num-str.

      DATA DIVISION.
      WORKING-STORAGE SECTION.
      01  str                    PIC X(5) VALUE "12345".
      01  num                    REDEFINES str PIC 9(5).
      
      PROCEDURE DIVISION.
          DISPLAY str
          ADD 1 TO num
          DISPLAY str

          GOBACK
          .</lang>

The following example also increments a numerical string, although it does not apear to. num-str is implicitly defined as USAGE DISPLAY which means its contents will be stored as characters. This means num-str is effectively a string of (numeric) characters. <lang cobol> PROGRAM-ID. increment-num-str.

      DATA DIVISION.
      WORKING-STORAGE SECTION.
      01  num-str                PIC 9(5) VALUE 12345.
      
      PROCEDURE DIVISION.
          DISPLAY num-str
          ADD 1 TO num-str
          DISPLAY num-str
      
          GOBACK
          .</lang>

Common Lisp

<lang lisp>(princ-to-string (1+ (parse-integer "1234")))</lang>

Component Pascal

BlackBox Component Builder <lang oberon2> MODULE Operations; IMPORT StdLog,Args,Strings;

PROCEDURE IncString(s: ARRAY OF CHAR): LONGINT; VAR resp: LONGINT; done: INTEGER; BEGIN Strings.StringToLInt(s,resp,done); INC(resp); RETURN resp END IncString;

PROCEDURE DoIncString*; VAR p: Args.Params; BEGIN Args.Get(p); IF p.argc > 0 THEN StdLog.String(p.args[0] + " + 1= ");StdLog.Int(IncString(p.args[0]));StdLog.Ln END END DoIncString;

END Operations. </lang> Execute: ^Q Operatiosn.DoIncString 124343~

Output:

124343 + 1=  124344

D

<lang d>void main() {

   import std.string;

   immutable s = "12349".succ;
   assert(s == "12350");

}</lang>

Delphi

<lang Delphi>program IncrementNumericalString;

{$APPTYPE CONSOLE}

uses SysUtils;

const

 STRING_VALUE = '12345';

begin

 WriteLn(Format('"%s" + 1 = %d', [STRING_VALUE, StrToInt(STRING_VALUE) + 1]));

 Readln;

end.</lang>

Output:

"12345" + 1 = 123456

Déjà Vu

Output:

"101"

DWScript

<lang Delphi>var value : String = "1234"; value := IntToStr(StrToInt(value) + 1); PrintLn(value);</lang>

E

<lang e>__makeInt("1234", 10).next().toString(10)</lang>

EchoLisp

<lang scheme> (number->string (1+ (string->number "665")))

   → "666"

</lang>

Eero

<lang objc>#import <Foundation/Foundation.h>

int main()

 i := (int)'123' + 1
 s := @(i).description

 Log( '%@', s )
 return 0</lang>

EGL

<lang EGL>s string = "12345"; s = 1 + s; // Note: s + 1 is a string concatenation.</lang>

Eiffel

<lang Eiffel> class APPLICATION

create make

feature {NONE}

make do io.put_string (increment_numerical_string ("7")) io.new_line io.put_string (increment_numerical_string ("99")) end

increment_numerical_string (s: STRING): STRING -- String 's' incremented by one. do Result := s.to_integer.plus (1).out end

end </lang> Output:

8

100

Elixir

Values can be converted to integers then converted back after incrementing <lang Elixir>increment1 = fn n -> to_string(String.to_integer(n) + 1) end

Or piped

increment2 = fn n -> n |> String.to_integer |> +1 |> to_string end

increment1.("1") increment2.("100")</lang>

Output:

"2"
"101"

Case of char list: <lang Elixir>iex(1)> (List.to_integer('12345') + 1) |> to_char_list '12346'</lang>

Emacs Lisp

<lang lisp>(1+ (string-to-number "12345"))</lang>

Erlang

<lang erlang>integer_to_list(list_to_integer("1336")+1).</lang>

ERRE

<lang>

  ....................
   s$="12345"
   s$=STR$(VAL(s$)+1)
  ....................

</lang>

Euphoria

<lang euphoria>include get.e

function val(sequence s)

   sequence x
   x = value(s)
   return x[2]

end function

sequence s

s = "12345" s = sprintf("%d",{val(s)+1})</lang>

Factor

<lang factor>"1234" string>number 1 + number>string</lang>

Fantom

Within 'fansh':

<lang fantom> fansh> a := "123" 123 fansh> (a.toInt + 1).toStr 124 </lang>

Forth

This word causes the number whose string value is stored at the given location to be incremented. The address passed must contain enough space to hold the string representation of the new number. Error handling is rudimentary, and consists of aborting when the string does not contain a numerical value.

The word ">string" takes and integer and returns the string representation of that integer. I factored it out of the definitions below to keep the example simpler.

<lang forth>: >string ( d -- addr n )

 dup >r dabs <# #s r> sign #> ;

inc-string ( addr -- )

 dup count number? not abort" invalid number"
 1 s>d d+ >string rot place ;</lang>

Here is a version that can increment by any value

<lang forth>: inc-string ( addr n -- )

 over count number? not abort" invalid number"
 rot s>d d+  >string rot place ;</lang>

Test the first version like this:

<lang forth>s" 123" pad place pad inc-string pad count type</lang>

And the second one like this:

<lang forth>s" 123" pad place pad 1 inc-string pad count type</lang>

Fortran

Works with: Fortran version 90 and later

Using 'internal' files you can increment both integer and real strings <lang fortran>CHARACTER(10) :: intstr = "12345", realstr = "1234.5" INTEGER :: i REAL :: r

READ(intstr, "(I10)") i ! Read numeric string into integer i i = i + 1 ! increment i WRITE(intstr, "(I10)") i ! Write i back to string

READ(realstr, "(F10.1)") r r = r + 1.0 WRITE(realstr, "(F10.1)") r</lang>

F#

<lang fsharp>let next = string( int( "1234" ) + 1 )</lang>

Frink

The following works for integers, rational numbers, complex numbers, floating-point, etc. <lang frink> a = input["Enter number: "] toString[eval[a] + 1] </lang>

FutureBasic

<lang futurebasic> include "ConsoleWindow"

dim as Str255 numStr dim as long numeric, i

numStr = "12345" numeric = val(numStr) numeric++ numStr = str$(numeric) print numStr

print

for i = 0 to 10

  numeric++
  numStr = str$( numeric )
  print numStr

next </lang>

Output:

GAP

<lang gap># Using built-in functions Incr := s -> String(Int(s) + 1);

Implementing addition
(but here 9...9 + 1 = 0...0 since the string length is fixed)

Increment := function(s)

 local c, n, carry, digits;
 digits := "0123456789";
 n := Length(s);
 carry := true;
 while n > 0 and carry do
   c := Position(digits, s[n]) - 1; 
   if carry then
     c := c + 1;
   fi;
   if c > 9 then
     carry := true;
     c := c - 10;
   else
     carry := false;
   fi;
   s[n] := digits[c + 1];
   n := n - 1;
 od;

end;

s := "2399"; Increment(s); s;

"2400"</lang>

Go

Concise: <lang go>package main import "fmt" import "strconv" func main() {

 i, _ := strconv.Atoi("1234")
 fmt.Println(strconv.Itoa(i + 1))

}</lang> More: <lang go>package main

import (

   "math/big"
   "fmt"
   "strconv"

)

func main() {

   // integer
   is := "1234"
   fmt.Println("original:   ", is)
   i, err := strconv.Atoi(is)
   if err != nil {
       fmt.Println(err)
       return
   }
   // assignment back to original variable shows result is the same type.
   is = strconv.Itoa(i + 1)
   fmt.Println("incremented:", is)

   // error checking worthwhile
   fmt.Println()
   _, err = strconv.Atoi(" 1234") // whitespace not allowed
   fmt.Println(err)
   _, err = strconv.Atoi("12345678901")
   fmt.Println(err)
   _, err = strconv.Atoi("_1234")
   fmt.Println(err)
   _, err = strconv.ParseFloat("12.D34", 64)
   fmt.Println(err)

   // float
   fmt.Println()
   fs := "12.34"
   fmt.Println("original:   ", fs)
   f, err := strconv.ParseFloat(fs, 64)
   if err != nil {
       fmt.Println(err)
       return
   }
   // various options on FormatFloat produce different formats.  All are valid
   // input to ParseFloat, so result format does not have to match original
   // format.  (Matching original format would take more code.)
   fs = strconv.FormatFloat(f+1, 'g', -1, 64)
   fmt.Println("incremented:", fs)
   fs = strconv.FormatFloat(f+1, 'e', 4, 64)
   fmt.Println("what format?", fs)

   // complex
   // strconv package doesn't handle complex types, but fmt does.
   // (fmt can be used on ints and floats too, but strconv is more efficient.)
   fmt.Println()
   cs := "(12+34i)"
   fmt.Println("original:   ", cs)
   var c complex128
   _, err = fmt.Sscan(cs, &c)
   if err != nil {
       fmt.Println(err)
       return
   }
   cs = fmt.Sprint(c + 1)
   fmt.Println("incremented:", cs)

   // big integers have their own functions
   fmt.Println()
   bs := "170141183460469231731687303715884105728"
   fmt.Println("original:   ", bs)
   var b, one big.Int
   _, ok := b.SetString(bs, 10)
   if !ok {
       fmt.Println("big.SetString fail")
       return
   }
   one.SetInt64(1)
   bs = b.Add(&b, &one).String()
   fmt.Println("incremented:", bs)

}</lang>

Output:

original:    1234
incremented: 1235

strconv.ParseInt: parsing " 1234": invalid syntax
strconv.ParseInt: parsing "12345678901": value out of range
strconv.ParseInt: parsing "_1234": invalid syntax
strconv.ParseFloat: parsing "12.D34": invalid syntax

original:    12.34
incremented: 13.34
what format? 1.3340e+01

original:    (12+34i)
incremented: (13+34i)

original:    170141183460469231731687303715884105728
incremented: 170141183460469231731687303715884105729

Golfscript

<lang golfscript>~)`</lang> With a test framework to supply a number: <lang golfscript>"1234" ~)` p</lang>

Groovy

Solution: <lang groovy>println ((("23455" as BigDecimal) + 1) as String) println ((("23455.78" as BigDecimal) + 1) as String)</lang>

Output:

23456
23456.78

Haskell

HicEst

<lang hicest>CHARACTER string = "123 -4567.89"

  READ( Text=string) a,   b
  WRITE(Text=string) a+1, b+1 ! 124 -4566.89</lang>

HyperTalk

<lang hypertalk>put 0 into someVar add 1 to someVar -- without "into [field reference]" the value will appear -- in the message box put someVar -- into cd fld 1</lang>

Icon and Unicon

Icon and Unicon will automatically coerce type conversions where they make sense. Where a conversion can't be made to a required type a run time error is produced.

<lang Icon>s := "123" # s is a string s +:= 1 # s is now an integer</lang>

IDL

<lang idl>str = '1234' print, string(fix(str)+1)

==> 1235</lang>

In fact, IDL tries to convert types cleverly. That works, too:

<lang idl>print, '1234' + 1

==> 1235</lang>

Inform 7

This solution works for numbers that fit into a single word (16-bit signed int for Z-machine, 32-bit signed int for Glulx virtual machine). <lang inform7>Home is a room.

To decide which indexed text is incremented (T - indexed text): let temp be indexed text; let temp be the player's command; change the text of the player's command to T; let N be a number; if the player's command matches "[number]": let N be the number understood; change the text of the player's command to temp; decide on "[N + 1]".

When play begins: say incremented "12345"; end the story.</lang>

J

<lang j>incrTextNum=: >:&.".</lang>

Note that in addition to working for a single numeric value, this will increment multiple values provided within the same string, on a variety of number types and formats including rational and complex numbers. <lang j> incrTextNum '34.5' 35.5

  incrTextNum '7 0.2 3r5 2j4 5.7e_4'

8 1.2 1.6 3j4 1.00057</lang>

Note also that the result here is a list of characters, and not a list of integers, which becomes obvious when you manipulate the result. For example, consider the effect of reversing the contents of the list:

<lang j> |.incrTextNum'123 456' 754 421

  |.1+123 456

457 124</lang>

Java

When using Integer.parseInt in other places, it may be beneficial to call trim on the String, since parseInt will throw an Exception if there are spaces in the String. <lang java>String s = "12345"; s = String.valueOf(Integer.parseInt(s) + 1);</lang>

Another solution that works with big decimal numbers: <lang java>String s = "123456789012345678901234567890.12345"; s = new BigDecimal(s).add(BigDecimal.ONE).toString();</lang>

JavaScript

<lang javascript>var s = "12345"; s = (Number(s) + 1).toString(); </lang>

jq

tonumber

jq's string-to-number filter is called tonumber. For example, if we have a file named input.txt consisting of string representations of numbers, one per line, we could compute the sum as follows: <lang sh>$ jq -n -M -s 'map(tonumber) | add' input.txt</lang>

More precisely, tonumber will convert string representations of JSON numbers (integers and decimals) to numbers, but very large integers will be converted to decimals with possible loss of precision, and similar problems will be noticeable for very small and very large decimals.

tostring can be used to convert numbers to strings.

long_add

<lang jq># This function assumes its string arguments represent non-negative decimal integers. def long_add(num1; num2):

 if (num1|length) < (num2|length) then long_add(num2; num1)
 else  (num1 | explode | map(.-48) | reverse) as $a1
     | (num2 | explode | map(.-48) | reverse) as $a2
     | reduce range(0; num1|length) as $ix
         ($a2;  # result
          ( $a1[$ix] + .[$ix] ) as $r
          | if $r > 9 # carrying
            then
              .[$ix + 1] = ($r / 10 | floor) +
                (if $ix + 1 >= length then 0 else .[$ix + 1] end )
              | .[$ix] = $r - ( $r / 10 | floor ) * 10
            else
              .[$ix] = $r
            end )
     | reverse | map(.+48) | implode
 end ;</lang>

Example <lang jq> long_add("9" * 100; "1")</lang>

Output:

"10000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000"

Julia

We add a method to the + function so as to be able to deal with strings, including support for arbitrary-precision arithmetic if a "big" integer is used. <lang Julia>+(s::String, n::Integer) = string(parse(Int, s) + n)</lang>

Output:

julia> "123" + 1
"124"
julia> "1234567890987654321" + big(1)
"1234567890987654322"

K

"." is a built-in function that evaluates a valid K expression.

<lang K> 1 + ."1234" 1235

  1 + ."1234.56"

1235.56

  / As a function
  inc:{1 + . x}
  inc "1234"

1235</lang>

Some other examples. <lang K> 1 + .:' ("1";"2";"3";"4") 2 3 4 5

  1 + . "123 456"

124 457

  . "1","+","-10"

-9</lang>

LabVIEW

This image is a VI Snippet, an executable image of LabVIEW code. The LabVIEW version is shown on the top-right hand corner. You can download it, then drag-and-drop it onto the LabVIEW block diagram from a file browser, and it will appear as runnable, editable code.

Lasso

<lang Lasso>(integer('123') + 1) -> asstring</lang> -> 124

$L a T e X$

<lang latex>\documentclass{article}

% numbers are stored in counters \newcounter{tmpnum}

% macro to increment a string (given as argument) \newcommand{\stringinc}[1]{% \setcounter{tmpnum}{#1}% setcounter effectively converts the string to a number \stepcounter{tmpnum}% increment the counter; alternatively: \addtocounter{tmpnum}{1} \arabic{tmpnum}% convert counter value to arabic (i.e. decimal) number string }

%example usage \begin{document} The number 12345 is followed by \stringinc{12345}. \end{document}</lang>

Liberty BASIC

<lang lb>' [RC] Increment a numerical string.

o$ ="12345" print o$

v =val( o$) o$ =str$( v +1) print o$

end</lang>

LiveCode

LiveCode casts types transparently. When storing a number in a variable, the internal representation is numeric (a double, I think), and if the variable is used as a number, there is no type conversion. If the variable is used as a string, the conversion is automatic; likewise if a string variable containing a number is used as a number: <lang LiveCode>put "0" & "1234" into myString -- I think this will result in an internal string representation add 1 to myString -- automatically converts to a number put "The number is:" && myString -- outputs "The number is: 1235"</lang>

Logo

Logo is weakly typed, so numeric strings can be treated as numbers and numbers can be treated as strings. <lang logo>show "123 + 1 ; 124 show word? ("123 + 1) ; true</lang>

Logtalk

<lang logtalk>number_chars(Number, "123"), Number2 is Number+1, number_chars(Number2, String2)</lang>

LOLCODE

LOLCODE is weakly typed, so the arithmetic operators work "as expected" on strings.

<lang LOLCODE>HAI 1.3

I HAS A foo ITZ "1234" foo R SUM OF foo AN 1 VISIBLE foo BTW, prints 1235

KTHXBYE</lang>

LSL

To test it yourself; rez a box on the ground, and add the following as a New Script. <lang LSL>default { state_entry() { llListen(PUBLIC_CHANNEL, "", llGetOwner(), ""); llOwnerSay("Say a Number and I'll Increment it."); } listen(integer iChannel, string sName, key kId, string sMessage) { llOwnerSay("You said '"+sMessage+"' + 1 = "+(string)(((integer)sMessage)+1)); } }</lang>

Output:

Increment_a_Numerical_String: You said '99999999' + 1 = 100000000
Increment_a_Numerical_String: You said '-100000000' + 1 = -99999999

Lua

<lang lua>print(tonumber("2345")+1)</lang>

M4

M4 can handle only integer signed 32 bit numbers, and they can be only written as strings <lang m4>define(`V',`123')dnl define(`VN',`-123')dnl eval(V+1) eval(VN+1)</lang>

If the expansion of any macro in the argument of eval gives something that can't be interpreted as an expression, an error is raised (but the interpretation of the whole file is not stopped)

Maple

<lang maple>s := "12345"; s := convert(parse(s)+1, string);</lang>

Mathematica / Wolfram Language

<lang Mathematica>Print[FromDigits["1234"] + 1]</lang>

MATLAB

<lang MATLAB>function numStr = incrementNumStr(numStr)

   numStr = num2str(str2double(numStr) + 1);

end</lang>

MAXScript

<lang maxscript>str = "12345" str = ((str as integer) + 1) as string</lang>

Metafont

<lang metafont>string s; s := "1234"; s := decimal(scantokens(s)+1); message s;</lang>

mIRC Scripting Language

ML

mLite

Standard ML

<lang sml>Int.toString (1 + valOf (Int.fromString "1234"))</lang>

Modula-2

<lang modula2>MODULE addstr;

IMPORT InOut, NumConv, Strings;

VAR str1, str2 : Strings.String;

       num             : CARDINAL;
       ok              : BOOLEAN;

BEGIN

 str1 := "12345";
 InOut.Write ('"');    InOut.WriteString (str1);       InOut.WriteString ('" + 1 = ');
 NumConv.Str2Num (num, 10, str1, ok);
 INC (num);
 NumConv.Num2Str (num, 10, str2,  ok);
 InOut.WriteString (str2);
 InOut.WriteLn

END addstr.</lang>

"12345" + 1 = 12346

Modula-3

Modula-3 provides the module Scan for lexing. <lang modula3>MODULE StringInt EXPORTS Main;

IMPORT IO, Fmt, Scan;

VAR string: TEXT := "1234";

   num: INTEGER := 0;

BEGIN

 num := Scan.Int(string);
 IO.Put(string & " + 1 = " & Fmt.Int(num + 1) & "\n");

END StringInt.</lang>

Output:

1234 + 1 = 1235

MUMPS

Just add. MUMPS has strings of characters as its native datatype. The "+" (plus) binary operator interprets its two arguments as numbers, so the MUMPS system does incrementing a string naturally. MUMPS portability standards require that the result must have at least 15 significant digits. Some implementations use Binary Coded Digits (BCD) and long fixed point (64 bit) integers to accomplish this. <lang MUMPS>

SET STR="123"
WRITE STR+1

</lang>

Neko

<lang Neko>var str = "123"; var str = $string($int(str) + 1);

$print(str);</lang>

Nemerle

<lang Nemerle>mutable str = "12345"; str = $"$(Int32.Parse(str)+1)";</lang>

NetRexx

In concert with Rexx, NetRexx can use typeless variables. Typeless variable support is provided through the default NetRexx Rexx object. Values are stored as variable length character strings and can be treated as either a string or a numeric value, depending on the context in which they are used. <lang NetRexx>/* NetRexx */

options replace format comments java crossref savelog symbols nobinary

numbers = '12345' say numbers numbers = numbers + 1 say numbers

return </lang>

Output:

12345
12346

NewLISP

<lang NewLISP>(string (++ (int "123")))</lang>

Nim

<lang nim>import strutils let next = $(parseInt("123") + 1)</lang>

Oberon-2

<lang oberon2>MODULE addstr;

IMPORT Out, Strings;

VAR str1, str2 : ARRAY 9 OF CHAR;

       num, pos        : INTEGER;
       carry           : BOOLEAN;
       ch              : CHAR;

BEGIN

 str1 := "9999";
 Out.Char ('"');       Out.String (str1);      Out.String ('" + 1 = ');
 num := Strings.Length (str1) - 1;
 pos := num;
 IF  str1 [0] = '9'  THEN  INC (pos)  END;
 str2 [pos + 1] := 0X;
 carry := TRUE;
 REPEAT
   ch := str1 [num];
   IF  carry  THEN
     ch := CHR (ORD (ch) + 1)
   END;
   IF  ch > '9'  THEN
     carry := TRUE;
     ch := '0'
   ELSE
     carry := FALSE
   END;
   str2 [pos] := ch;
   DEC (num);
   DEC (pos)
 UNTIL num < 0;
 IF  carry  THEN  str2 [0] := '1'  END;
 Out.String (str2);
 Out.Ln

END addstr.</lang> Producing:

jan@Beryllium:~/Oberon/obc$ Add
"12345" + 1 = 12346
"9999" + 1 = 10000

Objeck

<lang objeck> s := "12345"; i := int->ToInt(s) + 1; s := i->ToString(); </lang>

Objective-C

<lang objc>NSString *s = @"12345"; int i = [s intValue] + 1; s = [NSString stringWithFormat:@"%i", i]</lang>

OCaml

<lang ocaml>string_of_int (succ (int_of_string "1234"))</lang>

Octave

We convert the string to a number, increment it, and convert it back to a string.

<lang octave>nstring = "123"; nstring = sprintf("%d", str2num(nstring) + 1); disp(nstring);</lang>

Oforth

+ on strings is a concatenation, not an addition. To increment, the string is converted as integer then as string again.

<lang Oforth>"999" 1 + println "999" asInteger 1 + asString println</lang>

Output:

9991
1000

OoRexx

ooRexx supports the += etc. operators: <lang oorexx>i=1 i+=1 Say i</lang>

Output:

OpenEdge/Progress

<lang progress>DEFINE VARIABLE cc AS CHARACTER INITIAL "12345".

MESSAGE

  INTEGER( cc ) + 1

VIEW-AS ALERT-BOX.</lang>

Oz

<lang oz>{Int.toString {String.toInt "12345"} + 1}</lang>

PARI/GP

Pascal

Works with: Free Pascal

not like Delphi doing two conversion, but increment a string by 1 is much faster.

Here with different bases upto 10.After this there must be a correction to convert values > '9' to 'A'... Only for positive integer strings as high speed counter. <lang pascal>program StrInc; {$IFDEF FPC}

 {$Mode Delphi}
 {$Optimization ON}{$Align 16}{$Codealign proc=16,loop=4}

{$ENDIF}

uses

 sysutils;

type

 myString =  AnsiString; // string[32];//

function InCLoop(ps: pChar;le,Base: NativeInt):NativeInt; //Add 1 and correct carry //returns 0, if no overflow, else -1 var

 dg: nativeInt;

Begin

 dec(le);//ps is 0-based
 repeat
   dg := ord(ps[le])+(-ord('0')+1);
   result  := -ord(dg>=base);// -1 or 0 -> $FF...FF or $00...00
   ps[le] := chr(-(result AND base)+dg+ord('0'));
   dec(le);
 until (result = 0) or (le<0);

end;

procedure IncIntStr(base:NativeInt;var s:myString); var

 le :nativeInt;

begin

 le := length(s);
 IF le > 0 then
 Begin
   if (InCLoop(pChar(@s[1]),le,base) <>0) then
   begin
     setlength(s,le+1);
     move(s[1],s[2],le);
     s[1] := '1';
   end
 end
 else
 Begin
   setlength(s,1);
   s[1] := '1';
 end;

end;

const

 strLen = 26;
 MAX = 1 shl strLen -1;

var

 s  : myString;
 i,base : nativeInt;
 T1,T0: TDateTime;

Begin

 For base := 2 to 10 do
 Begin
   s:= ;

{

   //Zero pad string
   //s:= '0';// doesn't work for AnsiString for FPC 3.0 but for 2.6.4?
   //This works for all Ansi-string
   setlength(s,strLen);fillchar(s[1],strLen,'0');

}

   T0 := time;
   For i := 1 to MAX do
     IncIntStr(Base,s);
   T0 := (time-T0)*86400;
   writeln(s:strLen,' base ',base:2,T0:8:3,' s');

//One Billion Digits

 setlength(s,1000*1000*1000+1);
 s[1]:= '0';//don't measure setlength in IncIntStr
 fillchar(s[2],length(s)-1,'9');
 writeln('first 5 digits ',s[1],s[2],s[3],s[4],s[5]);
 T0 := time;
 IncIntStr(s,10);
 T0 := (time-T0)*86400;
 writeln(length(s):10,T0:8:3,' s');
 writeln('first 5 digits ',s[1],s[2],s[3],s[4],s[5]);
 end;

end.</lang> Output:

 ppc386 aka 32-Bit fpc 3.0.1
11111111111111111111111111 base  2   0.423 s
         11200021111001110 base  3   0.364 s
             3333333333333 base  4   0.331 s
              114134440423 base  5   0.345 s
               10354213103 base  6   0.325 s
                1443262443 base  7   0.318 s
                 377777777 base  8   0.312 s
                 150244043 base  9   0.308 s
                  67108863 base 10   0.305 s
first 5 digits 09999
1000000001   1.299 s 
//4.55 cpu-cycles per digit == 3.5 Ghz [cycle/s]*1.3[s]/1e9[digits]  (IPC = 17/4.55 = 3.73 wow)
first 5 digits 10000

Perl

Perl 6

Works with: Rakudo version #22 "Thousand Oaks"

Phix

<lang Phix>integer Template:N = scanf("2047","%d") printf(1,"%d\n",{n+1})</lang>

Output:

PHP

PicoLisp

<lang PicoLisp>(format (inc (format "123456")))</lang>

Pike

<lang Pike>string number = "0"; number = (string)((int)number+1); Result: "1"</lang>

PL/I

<lang pli>declare s picture '999999999'; s = '123456789'; s = s + 1; put skip list (s);</lang>

Warning:
With s='999999999'
the result shown would be 
000000000
It is advisable to enable the SIZE condition 
(size):
 s = s + 1; 
which will diagnose this problem: 
IBM0342I  ONCODE=0340  The SIZE condition was raised.
   At offset +000000B9 in procedure with entry IB1

Plain $T e X$

<lang tex>\newcount\acounter \def\stringinc#1{\acounter=#1\relax% \advance\acounter by 1\relax% \number\acounter} The number 12345 is followed by \stringinc{12345}. \bye</lang>

The generated page will contain the text:

The number 12345 is followed by 12346.

Pop11

<lang pop11>lvars s = '123456789012123456789999999999'; (strnumber(s) + 1) >< -> s;</lang>

PowerShell

The easiest way is to cast the string to int, incrementing it and casting back to string: <lang powershell>$s = "12345" $t = [string] ([int] $s + 1)</lang> One can also take advantage of the fact that PowerShell casts automatically according to the left-most operand to save one cast: <lang powershell>$t = [string] (1 + $s)</lang>

Prolog

Works with SWI-Prolog. <lang Prolog>incr_numerical_string(S1, S2) :- string_to_atom(S1, A1), atom_number(A1, N1), N2 is N1+1, atom_number(A2, N2), string_to_atom(S2, A2). </lang>

Output:

<lang Prolog> ?- incr_numerical_string("123", S2). S2 = "124". </lang>

PureBasic

<lang PureBasic>string$="12345" string$=Str(Val(string$)+1) Debug string$</lang>

Python

Works with: Python version 2.3 through 3.4

R

Racket

lang racket

(define next (compose number->string add1 string->number)) </lang>

Rascal

<lang rascal> import String;

public str IncrNumStr(str s) = "<toInt(s) + 1>"; </lang>

Output:

rascal>IncrNumStr("123")
str: "124"

REBOL

<lang REBOL>REBOL [ Title: "Increment Numerical String" Author: oofoe Date: 2009-12-23 URL: http://rosettacode.org/wiki/Increment_numerical_string ]

Note the use of unusual characters in function name. Also note that
because REBOL collects terms from right to left, I convert the
string argument (s) to integer first, then add that result to one.

s++: func [s][to-string 1 + to-integer s]

Examples. Because the 'print' word actually evaluates the block
(it's effectively a 'reduce' that gets printed space separated),
it's possible for me to assign the test string to 'x' and have it
printed as a side effect. At the end, 'x' is available to submit to
the 's++' function. I 'mold' the return value of s++ to make it
obvious that it's still a string.

print [x: "-99" "plus one equals" mold s++ x] print [x: "42" "plus one equals" mold s++ x] print [x: "12345" "plus one equals" mold s++ x]</lang>

Output:

-99 plus one equals "-98"
42 plus one equals "43"
12345 plus one equals "12346"

Retro

<lang Retro>"123" toNumber 1+ toString</lang>

REXX

REXX, like many other scripting languages, uses typeless variables.
Typeless variables are stored as variable length character strings and can be treated as
either a string or a numeric value, depending on the context in which they are used.

version 1

<lang rexx>/*REXX program demonstrates a method how to increment a numerical string*/ count = "3" /*REXX variables (and constants) are character strings.*/ count = 3 /*(identical to the above statement.) */ count = count+1 /*strings in a numerical context are treated as numbers*/ say 'sum=' count /*display the value of COUNT to the terminal (screen)*/

/*────────────────── The default numeric digits for REXX is 9 digits. */ /*────────────────── However, that can be increased with NUMERIC DIGITS.*/

numeric digits 15000 /*let's go ka-razy with fifteen thousand digits. */

count=copies(2,15000) /*stressing REXX's brains with lots of two's, */

                      /*the above is considered a number in REXX.      */

count=count+3 /*make that last digit of COUNT a "5". */

if 1==0 then /*let's not display this gihugeic number to term,*/ say 'count=' count /*ya most likely don't want to display this thing*/

                      /* [↓]  show the six leftmost and rightmost digs.*/

say 'count=' left(count,6)'···'right(count,6)

                                      /*stick a fork in it, we're done.*/</lang>

Output:

sum= 4
count= 222222···222225

version 2

Looking at the PL/I code I started investigating this situation in Rexx. These are my findings: <lang rexx>/* REXX ************************************************************

11.12.2012 Walter Pachl
There is no equivalent to PL/I's SIZE condition in REXX.
The result of an arithmetic expression is rounded
according to the current setting of Numeric Digits.
ooRexx introduced, however, a LOSTDIGITS condition that checks
if any of the OPERANDS exceeds this number of digits.
Unfortunately this check is currently a little too weak
and will not recognise a 10-digit number to be too large.
This little bug will be fixed in the next release of ooRexx.
- - - - /

Parse Version v . Say v z=999999998 Do i=1 To 3

 z=z+1
 Say z
 End

Numeric Digits 20 z=999999998 Do i=1 To 3

 z=z+1
 Say z
 End

Numeric Digits 9 If left(v,11)='REXX-ooRexx' Then

 Signal On Lostdigits

z=100000000012 Say z z=z+1 Say z Exit lostdigits:

 Say 'LOSTDIGITS condition raised in line' sigl
 Say 'sourceline='sourceline(sigl)
 Say "condition('D')="condition('D')</lang>

Output:

REXX-ooRexx_4.1.2(MT)
999999999
1.00000000E+9
1.00000000E+9
999999999
1000000000
1000000001
100000000012
LOSTDIGITS condition raised in line 30
sourceline=z=z+1
condition('D')= 100000000012

Ring

<lang ring> x = "1234" See 1+x # print 1235 </lang>

Ruby

If a string represents a number, the succ method will increment the number: <lang ruby>'1234'.succ #=> '1235' '99'.succ #=> '100'</lang>

Rust

<lang rust>fn next_string(input: &str) -> String {

   (input.parse::<i64>().unwrap() + 1).to_string()

}

fn main() {

   let s = "-1";
   let s2 = next_string(s);
   println!("{:?}", s2);

}</lang>

Output:

"0"

Run BASIC

Run BASIC has trim command for left and right <lang runbasic>string$ = "12345" numeric = val(string$) numeric = numeric + 1 string$ = str$(numeric) print string$ </lang>

Scala

The string needs to be converted to a numeric type. BigDecimal should handle most numeric strings. We define a method to do it.

<lang scala>implicit def toSucc(s: String) = new { def succ = BigDecimal(s) + 1 toString }</lang>

Usage:

scala> "123".succ
res5: String = 124

Scheme

<lang scheme>(number->string (+ 1 (string->number "1234")))</lang>

Sed

Reads a decimal integer from stdin and outputs the same with the magnitude incremented by one.

(TODO: Since it deals only with the magnitude, the result is incorrect for negative numbers—though adding this support is definitely possible.)

The routine starts by suffixing the input number with a carry mark (a : in this case) indicating that the digit to its left still needs to be incremented. In a loop, the following happens:

If there is a carry mark on the far left, replace it with a 1.
If there are no more carry marks, exit the loop.
Hold the current number. (h)
Extract the digit to the left of the first carry mark. (s)
Replace the digit with the same digit incremented by one, with 9 incrementing to a carry mark (i.e. 10). (y)
If the result of such replacement was a carry mark, suffix the mark with a 0, indicating that the digit has rolled over and the digit to the left must be incremented. (s)
Retrieve the held number (G) and replace the first carry mark and the digit to its left with the result of the computation. (s)
Repeat. (b)

<lang sed>s/^.*$/&:/

bubble

s/^:/1/ /.:/ {

   h
   s/^.*\(.\):.*$/\1/
   y/0123456789/123456789:/
   s/:/:0/
   G
   s/\(.*\)\n\(.*\).:\(.*\)$/\2\1\3/
   b bubble

}</lang>

Seed7

<lang seed7>var string: s is "12345";

s := str(succ(integer parse s));</lang>

Sidef

Slate

<lang slate>((Integer readFrom: '123') + 1) printString</lang>

Smalltalk

<lang smalltalk>('123' asInteger + 1) printString</lang> (a note to non-smalltalkers: "printString" does not print, but return the "printString")

SNOBOL4

    output = trim(input) + 1
    output = "123" + 1

end</lang>

Input:
 123

Output:
 124
 124

Sparkling

<lang sparkling>function numStrIncmt(s) {

   return fmtstr("%d", toint(s) + 1);

}

spn:1> numStrIncmt("12345") = 12346</lang>

SuperTalk

<lang supertalk>put 0 into someVar add 1 to someVar -- without "into [field reference]" the value will appear -- in the message box put someVar -- into cd fld 1</lang>

Swift

Works with: Swift version 2.x+

<lang swift>let s = "1234" if let x = Int(s) {

 print("\(x + 1)")

}</lang>

Works with: Swift version 1.x

<lang swift>let s = "1234" if let x = s.toInt() {

 println("\(x + 1)")

}</lang>

Tcl

In the end, all variables are strings in Tcl. A "number" is merely a particular interpretation of a string of bytes. <lang tcl>set str 1234 incr str</lang>

TI-89 BASIC

<lang ti89b>string(expr(str)+1)</lang>

TI-83 BASIC

There is no single command to convert a number to a string; you have to store it to one of the Function variables which acts as both a number and a string. <lang ti83b>:"1"→Str1

expr(Str1)+1→A

{0,1}→L₁

{0,A}→L₂

LinReg(ax+b) Y₁

Equ►String(Y₁,Str1)

sub(Str1,1,length(Str1)-3)→Str1</lang>

Toka

<lang toka>" 100" >number drop 1 + >string</lang>

TorqueScript

To increment by 1:

 $string = "12345";
 $string++;

$string is now 12346.

To increment by more than 1:

 $string = "12345";
 $string += 10;

$string is now 12355.

TUSCRIPT

<lang tuscript> $$ MODE TUSCRIPT teststring="0'1'-1'12345'10000000'-10000000" LOOP/CLEAR n=teststring n=n+1 PRINT n ENDLOOP </lang>

Output:

TXR

Two implementations of what the task says: incrementing a numerical string. (Not: converting a string to a number, then incrementing the number, then converting back to string.)

TXR Lisp

<lang txr>@(do (defun inc-num-str (str-in)

      (let ((len (length str-in))
            (str (copy-str str-in)))
        (for ((i (- len 1)))
             ((>= i 0) `1@str`)
             ((dec i))
          (if (<= (inc [str i]) #\9)
            (return str)
            (set [str i] #\0))))))

@(bind a @(inc-num-str "9999")) @(bind b @(inc-num-str "1234"))</lang>

$ ./txr -B incnum.txr 
a="10000"
b="1235"

No TXR Lisp

<lang txr>@(deffilter incdig ("0" "1") ("1" "2") ("2" "3") ("3" "4") ("4" "5")

                  ("5" "6") ("6" "7") ("7" "8") ("8" "9"))

@(define increment (num out)) @ (local prefix dig junk) @ (next :string num) @ (cases) 9 @ (bind out "10") @ (or) @*{prefix}@{dig /[0-8]/} @ (bind out `@prefix@{dig :filter incdig}`) @ (or) @*{prefix}9 @ (bind out `@{prefix :filter (:fun increment)}0`) @ (or) @junk @ (throw error `bad input: @junk`) @ (end) @(end) @in @(increment in out)</lang>

$ echo 1 | ./txr -B incnum.txr -
input="1"
result="2"
$ echo 123 | ./txr -B incnum.txr -
input="123"
result="124"
$ echo 899999 | ./txr -B incnum.txr -
input="899999"
result="900000"
$ echo 999998 | ./txr -B incnum.txr -
input="999998"
result="999999"
$ echo 999999 | ./txr -B incnum.txr -
input="999999"
result="1000000"

UNIX Shell

Traditional Unix shell does not directly support arithmetic operations, so external tools, such as expr are used to perform arithmetic calculations when required. The following example demonstrates how a variable can be incremented by using the expr function:

Works with: Bourne Shell

<lang bash># All variables are strings within the shell

Although num look like a number, it is in fact a numerical string

num=5 num=`expr $num + 1` # Increment the number</lang>

The Korn Shell and some newer shells do support arithmetic operations directly, and several syntax options are available:

Works with: bash

Works with: ksh93

Works with: pdksh

Works with: zsh

<lang bash>num=5 let num=num+1 # Increment the number let "num = num + 1" # Increment again. (We can use spaces inside quotes) ((num = num + 1)) # This time we use doublebrackets let num+=1 # This time we use += let "num += 1" ((num += 1))</lang>

Works with: ksh93

Works with: pdksh

Works with: zsh

<lang bash>integer num=5 # Declare an integer... num=$num+1 # ...then increment without the let keyword.</lang>

C Shell

The @ assignment command uses strings as integers. <lang csh>@ num = 5 @ num += 1</lang>

Ursa

<lang ursa>decl string num set num "123" set num (int (+ (int num) 1))</lang>

Ursala

<lang Ursala>#import nat

instring = ~&h+ %nP+ successor+ %np@iNC # convert, do the math, convert back</lang> test program: <lang Ursala>#cast %sL

tests = instring* <'22435','4','125','77','325'></lang>

Output:

<'22436','5','126','78','326'>

VBA

The easy method assumes that the number can be represented as a Long integer: <lang VBA> Public Function incr(astring As String) As String 'simple function to increment a number string

  incr = CStr(CLng(astring) + 1)

End Function </lang> Examples:

print incr("345343434")
345343435
print incr("-10000000")
-9999999

The long version handles arbitrary-length strings: <lang VBA> Public Function Lincr(astring As String) As String 'increment a number string, of whatever length 'calls function "increment" or "decrement" Dim result As String

'see if it is a negative number If left$(astring, 1) = "-" Then

 'negative x: decrease |x| by 1, then add "-"
 '(except if the result is zero)
 result = decrement(Mid$(astring, 2))
 If result <> "0" Then result = "-" & result

Else

 '0 or positive x: increase x by 1
 If left$(astring, 1) = "+" Then  'allow a + before the number
   result = increment(Mid$(astring, 2))
 Else
   result = increment(astring)
 End If

End If Lincr = result End Function

Public Function increment(astring) As String Dim result As String 'increment a string representing a positive number 'does not work with negative numbers carry = 1 L = Len(astring) result = "" For j = L To 1 Step -1

 digit = Val(Mid$(astring, j, 1)) + carry
 If digit > 9 Then
   digit = digit - 10
   carry = 1
 Else
   carry = 0
 End If
 result = CStr(digit) & result

Next If carry = 1 Then result = CStr(carry) & result increment = result End Function

Public Function decrement(astring) As String Dim result As String 'decrement a string representing a positive number 'does not work with zero or negative numbers borrow = 1 L = Len(astring) result = "" For j = L To 1 Step -1

 digit = Val(Mid$(astring, j, 1)) - borrow
 If digit < 0 Then
   digit = digit + 10
   borrow = 1
 Else
   borrow = 0
 End If
 result = CStr(digit) & result

Next 'remove leading zero, if necessary If (Len(result) > 1) And (left$(result, 1) = "0") Then result = Mid$(result, 2) decrement = result End Function </lang> Examples:

print Lincr("99999999999999999")
100000000000000000
print Lincr("-10000000000000000")
-9999999999999999
print Lincr("-1")
0
print Lincr("0")
1
print Lincr("+1234567890987654321009")
1234567890987654321010

Vedit macro language

This example increments numeric string by converting it into numeric value, as most other language examples do. The string is located in text register 10. <lang vedit>itoa(atoi(10)+1, 10)</lang>

The following example increments unsigned numeric string of unlimited length. The current line in the edit buffer contains the string. <lang vedit>EOL do {

   if (At_BOL) {

Ins_Char('1') // add new digit Break

   }
   Char(-1)
   #1 = Cur_Char+1		// digit
   #2 = 0			// carry bit
   if (#1 > '9') {

#1 = '0' #2 = 1

   }
   Ins_Char(#1, OVERWRITE)
   Char(-1)

} while (#2) // repeat until no carry</lang>

Visual Basic .NET

<lang vbnet> Dim s As String = "123"

   s = CStr(CInt("123") + 1)
   ' or
   s = (CInt("123") + 1).ToString</lang>

XLISP

<lang lisp>(DEFUN INCREMENT-STRING (X)

   (NUMBER->STRING (+ (STRING->NUMBER X) 1)))</lang>

XPL0

<lang XPL0>string 0; \use zero-terminated string convention code Text=12;

func StrLen(A); \Return number of characters in an ASCIIZ string char A; int I; for I:= 0 to -1>>1-1 do

       if A(I) = 0 then return I;

proc IncStr(S); \Increment a numeric string char S; int I; [for I:= StrLen(S)-1 downto 0 do

       [S(I):= S(I)+1;
       if S(I) > ^9 then S(I):= S(I)-10 else return;
       ];

];

char Str; [Str:= "0123999999999"; \MSD first (big endian) IncStr(Str); IncStr(Str); Text(0, Str); ]</lang>

Output:

0124000000001

zkl

In zkl, the first operand "wins" and transforms the second. So 5+"1"-->6 <lang zkl>fcn numStringPlusOne(s){1+s} numStringPlusOne("123") //-->124</lang>