Loops/For with a specified step

Demonstrate a for-loop where the step-value is greater than one.

360 Assembly

Basic - Algol style

The opcode BXH uses 3 registers, one for index one for step and one for limit. <lang 360asm>* Loops/For with a specified step 12/08/2015 LOOPFORS CSECT

        USING  LOOPFORS,R12
        LR     R12,R15

== Algol style ================ test at the beginning

        LA     R3,BUF             idx=0
        LA     R5,0               from 5 (from-step=0)
        LA     R6,5               step 5
        LA     R7,25              to 25

LOOPI BXH R5,R6,ELOOPI for i=5 to 25 step 5

        XDECO  R5,XDEC              edit i
        MVC    0(4,R3),XDEC+8       output i
        LA     R3,4(R3)             idx=idx+4
        B      LOOPI              next i

ELOOPI XPRNT BUF,80 print buffer

        BR     R14

BUF DC CL80' ' buffer XDEC DS CL12 temp for edit

        YREGS  
        END    LOOPFORS</lang>

Output:

   5  10  15  20  25

Basic - Fortran style

The opcode BXLE uses 3 registers, one for index one for step and one for limit. <lang 360asm>* == Fortran style ============== test at the end

        LA     R3,BUF             idx=0
        LA     R5,5               from 5
        LA     R6,5               step 5
        LA     R7,25              to 25

LOOPJ XDECO R5,XDEC for j=5 to 25 step 5; edit j

        MVC    0(4,R3),XDEC+8       output j
        LA     R3,4(R3)             idx=idx+4
        BXLE   R5,R6,LOOPJ        next j
        XPRNT  BUF,80             print buffer</lang>

Structured Macros

<lang 360asm>* == Algol style ================ test at the beginning

        LA     R3,BUF             idx=0
        LA     R5,5               from 5 
        LA     R6,5               step 5
        LA     R7,25              to 25
        DO WHILE=(CR,R5,LE,R7)    for i=5 to 25 step 5
          XDECO  R5,XDEC            edit i 
          MVC    0(4,R3),XDEC+8     output i
          LA     R3,4(R3)           idx=idx+4
          AR     R5,R6              i=i+step
        ENDDO  ,                  next i
        XPRNT  BUF,80             print buffer</lang>

Structured Macros HLASM

<lang 360asm>* == Fortran style ============== test at the end

        LA     R3,BUF             idx=0
        DO FROM=(R5,5),TO=(R7,25),BY=(R6,5)  for i=5 to 25 step 5
          XDECO  R5,XDEC            edit i 
          MVC    0(4,R3),XDEC+8     output i
          LA     R3,4(R3)           idx=idx+4
        ENDDO  ,                  next i
        XPRNT  BUF,80             print buffer</lang>

The FOR loop construct in Ada does not give the programmer the ability to directly modify the loop control variable during the execution of the loop. Instead, a valid range must always be provided before entering a loop. Because exact adherence to the task is impossible, we have three versions to approximate a solution. Looper_1 goes through a range of values which are even. Looper_2 multiples each value by two. Looper_3 most closely adheres to the requirements of this task, and achieves this by using a second range for the indices.

<lang ada>with Loopers; use Loopers;

procedure For_Main is begin

       Looper_1;
       Looper_2;
       Looper_3;

end For_Main;

package Loopers is

       procedure Looper_1;
       procedure Looper_2;
       procedure Looper_3;

end Loopers;

with Ada.Text_IO, Ada.Integer_Text_IO; use Ada.Text_IO, Ada.Integer_Text_IO;

package body Loopers is

       procedure Looper_1 is
               Values : array(1..5) of Integer := (2,4,6,8,10);
       begin
               for I in Values'Range loop
                       Put(Values(I),0);
                       if I = Values'Last then
                               Put_Line(".");
                       else
                               Put(",");
                       end if;
               end loop;
       end Looper_1;

       procedure Looper_2 is
               E : Integer := 5;
       begin
               for I in 1..E loop
                       Put(I*2,0);
                       if I = E then
                               Put_Line(".");
                       else
                               Put(",");
                       end if;
               end loop;
       end Looper_2;

       procedure Looper_3 is
               Values : array(1..10) of Integer := (1,2,3,4,5,6,7,8,9,10);
               Indices : array(1..5) of Integer := (2,4,6,8,10);
       begin
               for I in Indices'Range loop
                       Put(Values(Indices(I)),0);
                       if I = Indices'Last then
                               Put_Line(".");
                       else
                               Put(",");
                       end if;
               end loop;
       end Looper_3;

end Loopers;

</lang>

Agena

Tested with Agena 2.9.5 Win32 <lang agena>for i from 2 to 8 by 2 do

   print( i )

od</lang>

Aime

<lang aime>integer i;

i = 0; while (i < 10) {

   o_winteger(2, i);
   i += 2;

}

o_newline();</lang>

ALGOL 60

<lang algol60> FOR i:=5 UNTIL 25 STEP 5 DO

  OUTINTEGER(i)

</lang>

ALGOL 68

The ALGOL 68 "universal" for/while loop:

 [ for index ] [ from first ] [ by increment ] [ to last ] [ while condition ] do statements od
 The minimum form of a "loop clause" is thus: do statements od # an infinite loop #

The formal specification of ALGOL 68 states:

for i from u1 by u2 to u3 while condition do action od

"is thus equivalent to the following void-closed-clause:"

begin int f:= u1, int b = u2, t = u3;
   step2:
     if (b > 0 ∧ f ≤ t) ∨ (b < 0 ∧ f ≥ t) ∨ b = 0
     then int i = f;
         if condition
         then action; f +:= b; go to step2
         fi
     fi
end

Note: Highlighting is as per the formal specification, c.f. Category:ALGOL 68#Example of different program representations.

There are several unusual aspects of the construct:

- only the 'do ~ od' portion was compulsory, in which case the loop will iterate indefinitely.
- thus the clause 'to 100 do ~ od', will iterate only 100 times.
- the while "syntactic element" allowed a programmer to break from a for loop early. eg

int sum sq:=0;
for i while
  sum sq ≠ 70 × 70
do
  sum sq +:= i ↑ 2
od

Subsequent "extensions" to the standard Algol68 allowed the to syntactic element to be replaced with upto and downto to achieve a small optimisation. The same compilers also incorporated:

until^(C) - for late loop termination.
foreach^(S) - for working on arrays in parallel.

ALGOL-M

<lang algol>BEGIN

   INTEGER I;
   FOR I := 1 STEP 3 UNTIL 19 DO
       WRITE( I );

END</lang>

ALGOL W

<lang algolw>begin

   for i := 3 step 2 until 9 do write( i )

end.</lang>

AppleScript

<lang AppleScript>repeat with i from 2 to 10 by 2 log i end repeat</lang>

AutoHotkey

<lang AutoHotkey>SetBatchLines, -1 iterations := 5 step := 10 iterations *= step Loop, % iterations {

  If Mod(A_Index, step)
     Continue
  MsgBox, % A_Index

} ExitApp</lang>

AWK

<lang awk>BEGIN {

 for (i= 2; i <= 8; i = i + 2) {
   print i
 }
 print "Ain't never too late!"

}</lang>

Axe

Axe does not support a step size other than 1. However, one can modify the increment variable inside the loop to accomplish the same task.

This example increments by 2: <lang axe>For(I,0,10)

Disp I▶Dec,i
I++

End</lang>

BASIC

Applesoft BASIC

<lang qbasic>FOR I = 2 TO 8 STEP 2 : PRINT I; ", "; : NEXT I : PRINT "WHO DO WE APPRECIATE?"</lang>

Basic

Works with: QuickBasic version 4.5

<lang qbasic>for i = 2 to 8 step 2

  print i; ", ";

next i print "who do we appreciate?"</lang>

smart BASIC

-- Scott A. Rossell, 12-27-16

Notice how the ampersand (&) is used to concatenate the variable with the text instead of a semicolon.

<lang qbasic>FOR n = 2 TO 8 STEP 2

  PRINT n & "..";

NEXT n PRINT "who do we appreciate?" END</lang>

Commodore BASIC

<lang qbasic>10 FOR I = 1 TO 10 STEP 2 20 PRINT I 30 NEXT</lang>

Batch File

<lang dos>@echo off for /l %%A in (1,2,10) do (

    echo %%A

)</lang>

Output:

>Sample.BAT
1
3
5
7
9

>

BBC BASIC

<lang bbcbasic> FOR n = 2 TO 8 STEP 1.5

       PRINT n
     NEXT</lang>

Output:

bc

<lang bc>for (i = 2; i <= 10; i += 2) {

}</lang>

Befunge

Translation of: C

C

This prints all odd digits: <lang c>int i; for(i = 1; i < 10; i += 2)

 printf("%d\n", i);</lang>

ChucK

Chuck style <lang c> SinOsc s => dac;

for (0 => int i; i < 2000; 5 +=> i ) {

   i => s.freq;
   100::ms => now;

} </lang> General purpose style: <lang c> for (0 => int i; i < 2000; 5 +=> i ) {

   <<< i >>>;

} </lang>

C++

This prints all odd digits: <lang cpp>for (int i = 1; i < 10; i += 2)

 std::cout << i << std::endl;</lang>

C#

<lang csharp>using System;

class Program {

   static void Main(string[] args) {    
       for (int i = 2; i <= 8; i+= 2) {        
           Console.Write("{0}, ", i);
       }

       Console.WriteLine("who do we appreciate?");
   }

}</lang>

Ceylon

<lang ceylon>shared void run() {

for(i in (2..8).by(2)) { process.write("``i`` "); } print("who do we appreciate?"); }</lang>

Clojure

The first example here is following the literal specification, but is not idiomatic Clojure code. The second example achieves the same effect without explicit looping, and would (I think) be viewed as better code by the Clojure community. <lang Clojure>(loop [i 0]

 (println i)
 (when (< i 10)
   (recur (+ 2 i))))

(doseq [i (range 0 12 2)]

 (println i))</lang>

COBOL

<lang cobol> IDENTIFICATION DIVISION.

      PROGRAM-ID. Display-Odd-Nums.

      DATA DIVISION.
      WORKING-STORAGE SECTION.
      01  I PIC 99.

      PROCEDURE DIVISION.
          PERFORM VARYING I FROM 1 BY 2 UNTIL 10 < I
              DISPLAY I
          END-PERFORM

          GOBACK
          .</lang>

ColdFusion

 <Cfoutput>#i#</Cfoutput>

</cfloop> </lang>

Common Lisp

<lang lisp> (format t "~{~S, ~}who do we appreciate?~%" (loop for i from 2 to 8 by 2 collect i)) </lang>

Output:

2, 4, 6, 8, who do we appreciate?

Chapel

<lang chapel> // Can be set on commandline via --N=x config const N = 3;

for i in 1 .. 10 by N {

 writeln(i);

} </lang>

Output:

$ ./loopby
1
4
7
10

$ ./loopby --N=4
1
5
9

D

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

void main() {

   // Print odd numbers up to 9.
   for (int i = 1; i < 10; i += 2)
       writeln(i);

   // Alternative way.
   foreach (i; iota(1, 10, 2))
       writeln(i);

}</lang>

Output:

Dao

<lang dao># first value: 1

max value: 9
step: 2

for( i = 1 : 2 : 9 ) io.writeln( i )</lang>

Delphi

Delphi's For loop doesn't support a step value. It would have to be simulated using something like a While loop.

<lang Delphi>program LoopWithStep;

{$APPTYPE CONSOLE}

var

 i: Integer;

begin

 i:=2;
 while i <= 8 do begin
   WriteLn(i);
   Inc(i, 2);
 end;

end.</lang>

Output:

DWScript

<lang Delphi>var i : Integer;

for i := 2 to 8 step 2 do

  PrintLn(i);</lang>

Output:

E

There is no step in the standard numeric range object (a..b and a..!b) in E, which is typically used for numeric iteration. An ordinary while loop can of course be used:

<lang e>var i := 2 while (i <= 8) {

   print(`$i, `)
   i += 2

} println("who do we appreciate?")</lang>

A programmer frequently in need of iteration with an arbitrary step should define an appropriate range object:

<lang e>def stepRange(low, high, step) {

   def range {
       to iterate(f) {
           var i := low
           while (i <= high) {
               f(null, i)
               i += step
           }
       }
   }
   return range

}

for i in stepRange(2, 9, 2) {

 print(`$i, `)

} println("who do we appreciate?")</lang>

The least efficient, but perhaps convenient, solution is to iterate over successive integers and discard undesired ones:

<lang e>for i ? (i %% 2 <=> 0) in 2..8 {

   print(`$i, `)

} println("who do we appreciate?")</lang>

EchoLisp

Steps may be integers, float, rationals. <lang lisp> (for ((i (in-range 0 15 2))) (write i))

   0 2 4 6 8 10 12 14

(for ((q (in-range 0 15 14/8))) (write q))

   0 7/4 7/2 21/4 7 35/4 21/2 49/4 14

(for ((x (in-range 0 15 PI))) (write x))

   0 3.141592653589793 6.283185307179586 9.42477796076938 12.566370614359172

</lang>

Ela

<lang ela>open monad io

for m s n | n > m = do return ()

         | else = do
             putStrLn (show n) 
             for m s (n+s)

_ = for 10 2 0 ::: IO</lang>

Output:

Elena

<lang elena>#import extensions.

program = [

   2 to:8 &by:2 &doEach:i
   [
       console writeLine:i.
   ].

].</lang>

Elixir

<lang elixir>defmodule Loops do

 def for_step(n, step) do
   IO.inspect Enum.take_every(1..n, step)
 end

end

Loops.for_step(20, 3)</lang>

Output:

[1, 4, 7, 10, 13, 16, 19]

or <lang elixir>iex(1)> Stream.iterate(1, &(&1+2)) |> Enum.take(10) [1, 3, 5, 7, 9, 11, 13, 15, 17, 19]</lang>

ERRE

     FOR N=2 TO 8 STEP 1.5 DO
       PRINT(N)
     END FOR

</lang>

Output:

Erlang

<lang erlang>%% Implemented by Arjun Sunel %% for_loop/4 by Bengt Kleberg. -module(loop_step). -export([main/0, for_loop/1, for_loop/4]).

% This Erlang code for "For Loop" is equivalent to: " for (i=start;  i<end ; i=i+2){ printf("* ");} " in C language.

main() -> for_loop(1).

for_loop( N ) ->

for_loop( N, 4, 2, fun() -> io:fwrite("* ") end ).

for_loop( I, End, Step, Do ) when N < End ->

   Do(),
   for_loop( I+Step, End, Step, Do );

for_loop( _I, _End, _Step, _Do ) -> ok. </lang>

Output:

* * ok

Euphoria

<lang Euphoria> for i = 1 to 10 by 2 do

? i

end for </lang> As a note, ? something is shorthand for: <lang Euphoria> print(1, something) puts(1, "\n") </lang>

print() differs from puts() in that print() will print out the actual sequence it is given. If it is given an integer, or an atom (Any number that is not an integer), it will print those out as-is.

Factor

Prints odd digits. <lang factor>1 10 2 <range> [ . ] each</lang>

FALSE

<lang false>2[$9\>][$.", "2+]#"who do we appreciate!"</lang>

Fantom

<lang fantom> class Main {

 public static Void main ()
 {
   Int step := 5
   for (Int i := 0; i < 100; i += step)
   {
     echo (i)
   }
 }

} </lang>

FBSL

<lang qbasic>#APPTYPE CONSOLE

DIM n AS INTEGER FOR n = 2 TO 8 STEP 2

   PRINT n;
   IF n < 8 THEN PRINT " ";

NEXT PRINT ", who will we obliterate?" PAUSE </lang>

FOCAL

If a FOR statement has three parameters, they are (in order) the start, the step, and the end; if only two parameters are supplied, they are taken to be the start and the end. The step is then set to 1. <lang focal>FOR I = 1,3,10; TYPE I, !</lang>

Forth

<lang forth>: test

 9 2 do
   i .
 2 +loop
 ." who do we appreciate?" cr ;</lang>

Fortran

Works with: Fortran version 90 and later

<lang fortran>do i = 1,10,2

  print *, i

end do</lang>

Works with: Fortran version 77 and later

<lang fortran> PROGRAM STEPFOR

       INTEGER I

C This will print all even numbers from -10 to +10, inclusive.

       DO 10 I = -10, 10, 2
         WRITE (*,*) I
  10   CONTINUE

       STOP
     END</lang>

FreeBASIC

<lang freebasic>' FB 1.05.0 Win64

For i As Integer = 1 To 21 Step 2

 Print i; " ";

Next Print Sleep</lang>

Output:

 1  3  5  7  9  11  13  15  17  19  21

F#

<lang fsharp>for i in 2..2..8 do

  printf "%d, " i

printfn "done"</lang>

Output:

2, 4, 6, 8, done

FutureBasic

<lang futurebasic> include "ConsoleWindow"

dim as Str15 s(11) dim as long i

s(0) = "Somewhere" s(2) = " over" s(4) = " the" s(6) = " rainbow" + chr$(13) s(8) = "Bluebirds" s(10) = " fly."

for i = 0 to 10 step 2 print s(i); next </lang> Output:

Somewhere over the rainbow
Bluebirds fly.

Gambas

<lang gambas>Public Sub Main() Dim siCount As Short

For siCount = 1 To 50 Step 5

 Print "Gambas is great!"

GAP

Use a range [a, b .. c], where the step is b-a (b is the value following a), and c-a must be a multiple of the step.

<lang gap>for i in [1, 3 .. 11] do

   Print(i, "\n");

od;

1 3 5 7 9 11 </lang>

GML

<lang GML>for(i = 0; i < 10; i += 2)

   show_message(string(i))</lang>

Go

This prints all odd digits: <lang go>for i := 1; i < 10; i += 2 {

 fmt.Printf("%d\n", i)

}</lang>

Groovy

"for" loop: <lang groovy>for(i in (2..9).step(2)) {

   print "${i} "

} println "Who do we appreciate?"</lang>

"each() method: Though technically not a loop, most Groovy programmers would use the slightly more terse "each()" method on the collection itself, instead of a "for" loop. <lang groovy>(2..9).step(2).each {

   print "${it} "

} println "Who do we appreciate?"</lang>

Output:

2 4 6 8 Who do we appreciate?

Go Team!

Haskell

<lang haskell>import Control.Monad (forM_) main = do forM_ [2,4..8] (\x -> putStr (show x ++ ", "))

         putStrLn "who do we appreciate?"</lang>

HicEst

<lang hicest>DO i = 1, 6, 1.25 ! from 1 to 6 step 1.25

  WRITE() i

ENDDO</lang>

Icon and Unicon

Icon and Unicon accomplish loop stepping through the use of a generator, the ternary operator to-by, and the every clause which forces a generator to consume all of its results. Because to-by is an operator it has precedence (just higher than assignments) and associativity (left) and can be combined with other operators. <lang Icon>

  every 1 to 10 by 2                       # the simplest case that satisfies the task, step by 2

  every 1 to 10                            # no to, step is by 1 by default
  every EXPR1 to EXPR2 by EXPR3 do EXPR4   # general case - EXPRn can be complete expressions including other generators such as to-by, every's do is optional
  steps := [2,3,5,7]                       # a list
  every i := 1 to 100 by !steps            # . more complex, several passes with each step in the list steps, also we might want to know what value we are at
  every L[1 to 100 by 2]                   # as a list index
  every i := 1 to 100 by (k := !steps)     # . need () otherwise := generates an error
  every 1 to 5 to 10                       # simple case of combined to-by - 1,..,10, 2,..10, ..., 5,..,10
  every 1 to 15 by 2 to 5                  # combined to-by
  every (1 to 15 by 2) to 5                # . made explicit

  every writes( (TO_BY_EXPR) | "\n", " " ) # if you want to see how any of these work

</lang> The ability to combine to-by arbitrarily is quite powerful. Yet it can lead to unexpected results. In cases of combined to-by operators the left associativity seems natural where the by is omitted. In cases where the by is used it might seem more natural to be right associative. If in doubt parenthesize.

Io

<lang Io>for(i,2,8,2,

   write(i,", ")

) write("who do we appreciate?")</lang>

J

<lang J> ' who do we appreciate?' ,~ ": 2 * >: i.4 2 4 6 8 who do we appreciate?</lang>

Or, using an actual for loop:

<lang J> 3 :0

 r=.$0
 for_n. 2 * >: i.4 do.
   r=.r,n
 end.
 ' who do we appreciate?' ,~ ":n

) 2 4 6 8 who do we appreciate?</lang>

That said, note also that J's steps verb lets us specify how many steps to take:

<lang J> i:8 _8 _7 _6 _5 _4 _3 _2 _1 0 1 2 3 4 5 6 7 8

  i:8j8

_8 _6 _4 _2 0 2 4 6 8</lang>

Or, if we prefer, we could borrow the definition of thru from the Downward for task and then filter for the desired values:

Example use:

<lang J> (#~ 0 = 2&|) 1 thru 20 2 4 6 8 10 12 14 16 18 20

  (#~ 0 = 3&|) 1 thru 20

3 6 9 12 15 18

  (#~ 1 = 3&|) 1 thru 20

1 4 7 10 13 16 19</lang>

And, of course, like filtering in any language, this approach supports non-constant step sizes:

Java

<lang java>for(int i = 2; i <= 8;i += 2){

  System.out.print(i + ", ");

} System.out.println("who do we appreciate?");</lang>

JavaScript

<lang javascript>var output = ,

i;

for (i = 2; i <= 8; i += 2) {

  output += i + ', ';

} output += 'who do we appreciate?'; document.write(output);</lang>

In a functional idiom of JavaScript, however, we will only be able to compose this computation within the superordinate expressions of our program if it has the the form of an expression returning a value, rather than that of a statement which fires off side-effects but returns no value.

Following the example of languages like Haskell and J on this page, we can begin by generating the stepped series as an expression. In functional JavaScript we will typically replace a state-changing loop with a non-mutating map or fold, writing, for example, something like:

<lang JavaScript>// range(iMax) // range(iMin, iMax) // range(iMin, iMax, dI) function range() {

 var lngArgs = arguments.length,
   lngMore = lngArgs - 1;

 iMin = lngMore ? arguments[0] : 1;
 iMax = arguments[lngMore ? 1 : 0];
 dI = lngMore > 1 ? arguments[2] : 1;

 return lngArgs ? Array.apply(null, Array(
   Math.floor((iMax - iMin) / dI) + 1
 )).map(function (_, i) {
   return iMin + (dI * i);
 }) : [];

}

console.log(

 range(2, 8, 2).join(', ') + ', who do we appreciate ?'

);</lang>

Output:

2, 4, 6, 8, who do we appreciate ?

jq

To generate the stream: 2,4,6,8:<lang jq># If your version of jq does not have range/3, use this: def range(m;n;step): range(0; ((n-m)/step) ) | m + (. * step);

range(2;9;2)</lang> Example: <lang jq>reduce range(2;9;2) as $i

  (""; . + "\($i), ") +
  "whom do we appreciate?"</lang>

Julia

<lang julia>for i in 2:2:8

   print(i, ", ")

end println("whom do we appreciate?")</lang>

Kotlin

<lang scala>// version 1.0.6

fun main(args: Array<String>) {

   for (i in 1 .. 21 step 2) print("$i ")

}</lang>

Output:

1 3 5 7 9 11 13 15 17 19 21

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.

Lang5

<lang lang5>: <range> over iota swap * rot + tuck swap <= select ; : tuck swap over ;

>>say.(*) . ;

1 10 2 <range> >>say.</lang>

Lasso

<lang Lasso>loop(-to=100, -from=1, -by=2) => {^

   loop_count
   '\r' // for formatting

^}</lang>

Liberty BASIC

<lang lb> for i = 2 to 8 step 2

  print i; ", ";

next i print "who do we appreciate?" end </lang>

Lingo

Lingo loops don't support a "step" parameter, so it has to be implemented manually: <lang lingo>step = 3 repeat with i = 0 to 10

 put i
 i = i + (step-1)

end repeat</lang>

Output:

-- 0
-- 3
-- 6
-- 9

Lisaac

<lang Lisaac>1.to 9 by 2 do { i : INTEGER;

 i.print;
 '\n'.print;

};</lang>

LiveCode

<lang LiveCode>repeat with n = 0 to 10 step 2

   put n after loopn
   if n is not 10 then put comma after loopn

end repeat put loopn</lang> Output<lang LiveCode>0,2,4,6,8,10</lang>

Logo

<lang logo>for [i 2 8 2] [type :i type "|, |] print [who do we appreciate?]</lang>

Lua

<lang lua> for i=2,9,2 do

 print(i)

end </lang>

Output:

M4

<lang M4>define(`for',

  `ifelse($#,0,``$0,
  `ifelse(eval($2<=$3),1,
  `pushdef(`$1',$2)$5`'popdef(`$1')$0(`$1',eval($2+$4),$3,$4,`$5')')')')dnl

for(`x',`1',`5',`3',`x ')</lang>

Output:

1
4

Maple

<lang Maple>for i from 2 to 8 by 2 do

i;

end do;</lang>

Output:

Mathematica

<lang Mathematica>Do[

Print@i,
{i, 1, 20, 4}]</lang>

Output:

MATLAB / Octave

<lang Matlab> for k = 0:10:100,

       printf('%d\n',k)
   end; </lang>

A vectorized version of the code is

<lang Matlab> printf('%d\n',0:10:100); </lang>

Maxima

<lang maxima>for i: 1 step 2 thru 10 do print(i); /* 1

  3
  5
  7 */</lang>

MAXScript

<lang MAXScript>for i = 0 to 10 by 2 do format "%\n" i</lang> Output: <lang MAXScript> 0 2 4 6 8 10 OK </lang>

МК-61/52

In this example, the step is 3, the lowest value is 1 and the upper limit is 10.

Modula-2

<lang modula2>MODULE ForBy;

 IMPORT InOut;

 VAR
   i: INTEGER;

BEGIN

 FOR i := 0 TO 100 BY 2 DO
   InOut.WriteInt(i, 3);
   InOut.WriteLn
 END

END ForBy.</lang>

Modula-3

<lang modula3>FOR i := 1 TO 100 BY 2 DO

 IO.Put(Fmt.Int(i) & " ");

END;</lang>

MUMPS

<lang MUMPS>FOR I=65:3:122 DO

.WRITE $CHAR(I)," "</lang>

Output:

A D G J M P S V Y \ _ b e h k n q t w z

NewLISP

<lang NewLISP>(for (i 0 10 2)

 (println i))</lang>

Nim

<lang nim>for x in countdown(10,0,3): echo(x)</lang>

Output:

Nemerle

<lang Nemerle>for (i = 2; i <= 8; i +=2)</lang> <lang Nemerle>foreach (i in [2, 4 .. 8])</lang>

NetRexx

<lang NetRexx>/* NetRexx */ options replace format comments java crossref savelog symbols nobinary

say

 say 'Loops/For with a specified step'

 loop i_ = -1.4 to 10.6 by 1.7
   say i_.format(3, 1) || '\0'
   end i_
 say</lang>

Output:

D:\>java lst

Loops/For with a specified step
 -1.4  0.3  2.0  3.7  5.4  7.1  8.8 10.5

Oberon-2

Works with oo2c Version 2 <lang oberon2> MODULE LoopForStep; IMPORT

 Out;

VAR

 i: INTEGER;

BEGIN

 FOR i := 0 TO 10 BY 3 DO
   Out.LongInt(i,0);Out.Ln
 END;
 FOR i := 10 TO 0 BY -3 DO
   Out.LongInt(i,0);Out.Ln
 END

END LoopForStep. </lang> Output:

Objeck

<lang objeck> for(i := 0; i < 10; i += 2;) {

 i->PrintLine();

}; </lang>

OCaml

<lang ocaml># let for_step a b step fn =

   let rec aux i =
     if i <= b then begin
       fn i;
       aux (i+step)
     end
   in
   aux a
 ;;

val for_step : int -> int -> int -> (int -> 'a) -> unit = <fun>

for_step 0 8 2 (fun i -> Printf.printf " %d\n" i) ;;

- : unit = ()</lang>

Octave

<lang octave>for i = 1:2:10

 disp(i)

endfor</lang>

Oforth

<lang Oforth> 1 100 2 step: i [ i println ]</lang>

Openscad

<lang openscad>/* Loop from 3 to 9 in steps of 2 */

for ( l = [3:2:9] ) {

 echo (l);

} echo ("on a double white line.");</lang>

Oz

<lang oz>for I in 2..8;2 do

  {System.show I}

end {System.show done} </lang>

PARI/GP

<lang parigp>forstep(n=1,10,2,print(n))</lang>

The forstep construct is actually more powerful. For example, to print numbers with last digit relatively prime to 10: <lang parigp>forstep(n=1,100,[2,4,2,2],print(n))</lang>

Panda

Panda doesn't nativly have a number generator with steps, so let's add it. <lang panda>fun for(from,to,step) type integer,integer,integer->integer

 t=to.minus(from).divide(step)
 0..t.times(step).plus(from)
 /test it for(1 6 2) -> 1 3 5

for(1 3 5)</lang>

Pascal

See Delphi

Perl

<lang perl>for($i=2; $i <= 8; $i += 2) {

 print "$i, ";

} print "who do we appreciate?\n";</lang>

Perl 6

Works with: Rakudo version 2010.07

With at least two values on the left-hand side, the sequence operator (...) can infer an arithmetic series. (With at least three values, it can infer a geometric sequence, too.)

<lang perl6>for 2, 4 ... 8 {

   print "$_, ";

}

say 'whom do we appreciate?';</lang>

Phix

<lang Phix>for i=2 to 8 by 2 do

   printf(1,"%d, ",i)

end for printf(1,"who do we appreciate?\n")</lang>

PHP

<lang php><?php foreach (range(2, 8, 2) as $i)

   echo "$i, ";

echo "who do we appreciate?\n"; ?></lang>

Output:

2, 4, 6, 8, who do we appreciate?

PicoLisp

<lang PicoLisp>(for (N 1 (> 10 N) (+ N 2))

  (printsp N) )</lang>

Pike

<lang pike>int main() {

  for(int i = 2; i <= 16; i=i+2) {
     write(i + "\n");
  }

}</lang>

PILOT

One of the advantages of needing to create loops manually by using conditional jumps is that a step of any integer is just as easy as a step of one. <lang pilot>R : Prints the odd numbers less than 10. C :i = 1

Loop

T :#i C :i = i + 2 J ( i < 10 ) :*Loop END:</lang>

PL/I

<lang PL/I> declare (n, i) fixed binary;

get list (n); do i = 1 to n by 4;

  put skip list (i);

end; </lang>

PowerShell

<lang powershell>for ($i = 0; $i -lt 10; $i += 2) {

$i

}</lang>

PureBasic

<lang PureBasic>For i=-15 To 25 Step 5

 Debug i

Next i</lang>

Python

Works with: Python version 2.x

<lang python>for i in xrange(2, 9, 2):

   print "%d," % i,

print "who do we appreciate?"</lang>

Works with: Python version 3.x

<lang python>for i in range(2, 9, 2):

   print("%d, " % i, end="")

print("who do we appreciate?")</lang>

Output:

2, 4, 6, 8, who do we appreciate?

R

<lang R>for(a in seq(2,8,2)) {

 cat(a, ", ")

} cat("who do we appreciate?\n")</lang>

Here the loop may be done implicitly by first concatenating the string and then printing:

<lang R>cat(paste(c(seq(2, 8, by=2), "who do we appreciate?\n"), collapse=", "))</lang>

Racket

lang racket

(for ([i (in-range 2 9 2)])

 (printf "~a, " i))

(printf "who do we appreciate?~n") </lang>

Raven

List of numbers: <lang Raven>[ 2 4 6 8 ] each "%d, " print "who do we appreciate?\n" print</lang>

Range: <lang Raven>2 10 2 range each "%d, " print "who do we appreciate?\n" print</lang>

Output:

2, 4, 6, 8, who do we appreciate?

REBOL

<lang REBOL>for i 2 8 2 [ prin rejoin [i ", "]] print "who do we appreciate?"</lang>

Output:

2, 4, 6, 8, who do we appreciate?

REXX

version 1

<lang rexx> do x=1 to 10 by 1.5

 say x
 end</lang>

Output:

1
2.5
4.0
5.5
7.0
8.5
10.0

version 2

<lang rexx> do thing=1 by 3/2 to 10

 say  thing
 end</lang>

output is the same as above.

Ring

we use step keyword to define step length in this example we print Even numbers between 0 and 10 <lang ring> for i = 0 to 10 step 2 see i + nl next </lang>

Output:

we can use step with double values as well: <lang ring> for i = 0 to 10 step 0.5 see i + nl next </lang>

Output:

Ruby

<lang ruby>2.step(8,2) {|n| print "#{n}, "} puts "who do we appreciate?"</lang> or: <lang ruby>(2..8).step(2) {|n| print "#{n}, "} puts "who do we appreciate?"</lang> or: <lang ruby>for n in (2..8).step(2)

 print "#{n}, "

end puts "who do we appreciate?"</lang>

Output:

2, 4, 6, 8, who do we appreciate?

Run BASIC

<lang runbasic>for i = 2 to 8 step 2

  print i; ", ";

next i print "who do we appreciate?"</lang>

Output:

2, 4, 6, 8, who do we appreciate?

Rust

<lang rust>fn main() {

   let mut i = 2;
   while i <= 8 {
       print!("{}, ", i);
       i += 2;
   }
   println!("who do we appreciate?!");

}</lang>

Prettier, but currently feature gated version (Rust 1.0)

Does not work in stable Rust <lang rust>#![feature(step_by)]

fn main() {

 for i in (2..8+1).step_by(2) {
   print!("{}", i);
 }
 println!("who do we appreciate?!");

}</lang> Additionally, there is a crate on crates.io called cfor which allows for the use of a C-like for loop. <lang rust> #[macro_use] extern crate cfor;

fn main() {

   cfor!(let mut i = 2; i <= 8; i+=2; {
       println!("{}", i);
   });
   println!("Who do we appreciate?");

}</lang>

Salmon

<lang Salmon>for (x; 2; x <= 8; 2)

   print(x, ", ");;

print("who do we appreciate?\n");</lang>

SAS

Sather

See Loops/For#Sather: the implementation for for! allows to specify a step, even though the built-in stepto! can be used; an example of usage could be simply: <lang sather> i :INT;

   loop
     i := for!(1, 50, 2);
     -- OR
     -- i := 1.stepto!(50, 2);
     #OUT + i + "\n";
   end;</lang>

(Print all odd numbers from 1 to 50)

Scala

Alternatively: <lang scala>(2 to 8 by 2) foreach println</lang>

Scheme

The built-in for-like form in Scheme is the do form:

<lang scheme>(do ((i 2 (+ i 2))) ; list of variables, initials and steps -- you can iterate over several at once

 ((>= i 9))         ; exit condition
 (display i)        ; body
 (newline))</lang>

Some people prefer to use the recursive-style and more flexible _named let_ form:

<lang scheme>(let loop ((i 2)) ; function name, parameters and starting values

 (cond ((< i 9)
        (display i)
        (newline)
        (loop (+ i 2))))))  ; tail-recursive call, won't create a new stack frame</lang>

You can add to the language by wrapping the loop in a function:

<lang scheme>(define (for-loop start end step func)

 (let loop ((i start))
   (cond ((< i end)

(func i) (loop (+ i step))))))

(for-loop 2 9 2

 (lambda (i)
   (display i)
   (newline)))</lang>

... or in a macro, which allows for making the (lambda) implicit:

<lang scheme>(define-syntax for-loop

 (syntax-rules () 
   ((for-loop index start end step body ...)
    (let ((evaluated-end end) (evaluated-step step))
      (let loop ((i start))
        (if (< i evaluated-end)
          ((lambda (index) body ... (loop (+ i evaluated-step))) i)))))))

(for-loop i 2 9 2

 (display i)
 (newline))</lang>

Output:

Scilab

Works with: Scilab version 5.5.1

<lang>for i=1:2:10

   printf("%d\n",i)

end</lang>

Output:

Seed7

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

const proc: main is func

 local
   var integer: number is 0;
 begin
   for number range 1 to 10 step 2 do
     writeln(number);
   end for;
 end func;</lang>

Sidef

for(;;) loop: <lang ruby>for (var i = 2; i <= 8; i += 2) {

   say i

}</lang>

for-in loop: <lang ruby>for i in (2 .. (8, 2)) {

   say i

}</lang>

.each method: <lang ruby>2.to(8).by(2).each { |i|

   say i

}</lang>

Simula

<lang simula>begin

    integer i;
    for i:=5 step 5 until 25 do outint(i,5)
end</lang>

Slate

<lang slate>2 to: 8 by: 2 do: [| :i | Console ; i printString ; ', ']. inform: 'enough with the cheering already!'.</lang>

Smalltalk

<lang smalltalk>2 to: 8 by: 2 do: [ :i |

 Transcript show: i; show ', '

]. Transcript showCr: 'enough with the cheering already!'</lang>

SSEM

Implementing loops with a step other than one is precisely as easy (or as fiddly) as implementing loops with a step equal to one. This example program uses a loop to perform integer division. It should be run with the dividend in storage location 21 and the divisor in storage location 22. To show that it works, we shall ask the machine to count from 387 in steps of -5 and to halt with the accumulator showing the number of times it has done so before producing a negative result. <lang ssem>10101000000000100000000000000000 0. -21 to c 00101000000001100000000000000000 1. c to 20 00101000000000100000000000000000 2. -20 to c 01101000000000010000000000000000 3. Sub. 22 10101000000001100000000000000000 4. c to 21 00000000000000110000000000000000 5. Test 01001000000001000000000000000000 6. Add 18 to CI 00011000000000000000000000000000 7. 24 to CI 11101000000000100000000000000000 8. -23 to CI 01001000000000010000000000000000 9. Sub. 18 00101000000001100000000000000000 10. c to 20 00101000000000100000000000000000 11. -20 to c 11101000000001100000000000000000 12. c to 23 11001000000000000000000000000000 13. 19 to CI 11101000000000100000000000000000 14. -23 to c 00101000000001100000000000000000 15. c to 20 00101000000000100000000000000000 16. -20 to c 00000000000001110000000000000000 17. Stop 10000000000000000000000000000000 18. 1 11111111111111111111111111111111 19. -1 00000000000000000000000000000000 20. 0 11000001100000000000000000000000 21. 387 10100000000000000000000000000000 22. 5 00000000000000000000000000000000 23. 0 10110000000000000000000000000000 24. 13</lang> After executing 1,012 instructions, the computer halts with the correct quotient—77—in the accumulator.

Stata

<lang stata>forvalues i=1(2)10 { display "`i'" }

1 3 5 7 9</lang>

Swift

This prints all odd digits: <lang swift>for i in 1.stride(to: 10, by: 2) {

 print(i)

}</lang> Alternately (removed in Swift 3): <lang swift>for var i = 1; i < 10; i += 2 {

 print(i)

}</lang>

Tcl

<lang tcl>for {set i 2} {$i <= 8} {incr i 2} {

   puts -nonewline "$i, "

} puts "enough with the cheering already!"</lang>

TI-83 BASIC

Prints numbers from 0 to 100 stepping by 5. <lang ti83b>:For(I,0,100,5

Disp I

End</lang>

TI-89 BASIC

<lang ti89b>Local i For i, 0, 100, 5

   Disp i

EndFor</lang>

TorqueScript

TUSCRIPT

<lang tuscript> $$ MODE TUSCRIPT LOOP i=2,9,2 PRINT i ENDLOOP </lang>

Output:

UNIX Shell

All these loops iterate 2, 4, 6, 8.

Bourne Shell

Works with: Bourne Shell

<lang bash>x=2 while test $x -le 8; do echo $x x=`expr $x + 2` || exit $? done</lang>

Works with: Bourne Shell

Library: jot

Korn Shell

Works with: Korn Shell

<lang bash>x=2 while $x -le 8; do echo $x ((x=x+2)) done</lang>

Works with: Korn Shell

<lang bash>x=2 while ((x<=8)); do echo $x ((x+=2)) done</lang>

Bourne Again Shell

Works with: Bourne Again SHell version 3

<lang bash>for (( x=2; $x<=8; x=$x+2 )); do

 printf "%d, " $x

done</lang>

Works with: Bourne Again SHell version 4

Bash v4.0+ has inbuilt support for setting up a step value <lang bash>for x in {2..8..2} do

 echo $x

done</lang>

C Shell

Library: jot

<lang csh>foreach x (`jot - 2 8 2`) echo $x end</lang>

Ursa

Translation of: Python

Vedit macro language

This prints all odd digits in range 1 to 9: <lang vedit>for (#1 = 1; #1 < 10; #1 += 2) {

   Num_Type(#1)

}</lang>

VBA

<lang vb>Sub MyLoop()

   For i = 2 To 8 Step 2
       Debug.Print i;
   Next i
   Debug.Print

End Sub</lang>

Output:

 2  4  6  8

VBScript

<lang vb>buffer = "" For i = 2 To 8 Step 2

   buffer = buffer & i & " "

Next WScript.Echo buffer</lang>

Output:

2 4 6 8

Vim Script

<lang vim>for i in range(2, 10, 2)

   echo i

endfor</lang>

Output:

Visual Basic

Works with: Visual Basic version VB6 Standard

<lang vb>Sub MyLoop()

   For i = 2 To 8 Step 2
       Debug.Print i;
   Next i
   Debug.Print

End Sub</lang>

Output:

 2  4  6  8

Visual Basic .NET

Works with: Visual Basic .NET version 2011

<lang vbnet>Public Class FormPG

   Private Sub FormPG_Load(sender As Object, e As EventArgs) Handles MyBase.Load
       Dim i As Integer, buffer As String
       buffer = ""
       For i = 2 To 8 Step 2
           buffer = buffer & i & " "
       Next i
       Debug.Print(buffer)
   End Sub

End Class</lang>

Output:

2 4 6 8

Vorpal

<lang vorpal>for(i = 2, i <= 8, i = i + 2){

  i.print()

}</lang>

Wart

<lang wart>for i 2 (i <= 8) (i <- i+2)

 prn i</lang>

XPL0

The 'for' loop always steps by 1 (or -1 for 'downto'). However there is no restriction on how the control variable can be used or manipulated, thus a step by 2 can be implemented like this:

<lang XPL0>include c:\cxpl\codes; int I; [for I:= 2 to 8 do

       [IntOut(0, I);  Text(0, ", ");
       I:= I+1;
       ];

Text(0, "who do we appreciate?"); ]</lang>

Output:

2, 4, 6, 8, who do we appreciate?

zkl

<lang zkl>foreach n in ([1..10,4]) {println(n)} [1..10,3].pump(Console.println)</lang>

Output:

A few others: <lang zkl>fcn loop(i=0){println(i); if(i<10)return(self.fcn(i+2))} (0).pump(10,Console.println,fcn(n){if(n%2)return(Void.Skip); n})</lang>

ZX Spectrum Basic

<lang basic>10 FOR l = 2 TO 8 STEP 2 20 PRINT l; ", "; 30 NEXT l 40 PRINT "Who do we appreciate?"</lang>