Loops/For with a specified step: Difference between revisions

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=={{header|Scilab}}==
<lang>
for i=1:2:10
printf("%d\n",i)
end
</lang>
{{out}}
<pre style="height:20ex">
1
3
5
7
8
</pre>



=={{header|Seed7}}==
=={{header|Seed7}}==

Revision as of 15:00, 18 May 2015

Task
Loops/For with a specified step
You are encouraged to solve this task according to the task description, using any language you may know.

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

Ada

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>

Aime

<lang aime>integer i;

i = 0; while (i < 10) {

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

}

o_newline();</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 W

<lang algolw>begin

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

end.</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>

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>

Applesoft BASIC

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

BBC BASIC

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

       PRINT n
     NEXT</lang>
Output:
         2
       3.5
         5
       6.5
         8

bc

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

   i

}</lang>

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>

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

<lang cfm> <cfloop from="0" to="99" step="3" index="i">

 <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:
1
3
5
7
9
1
3
5
7
9

Dao

<lang dao># first value: 1

  1. max value: 9
  2. 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:
2
4
6
8

DWScript

<lang Delphi>var i : Integer;

for i := 2 to 8 step 2 do

  PrintLn(i);</lang>
Output:
2
4
6
8

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>

Ela

<lang ela>open console

for m s n | n > m = ()

         | else = writen n $ for m s (n+s)

for 10 2 0</lang>

Output:
0
2
4
6
8
10


ERRE

<lang ERRE>

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

</lang>

Output:
         2
       3.5
         5
       6.5
         8

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>

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>

F#

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

  printf "%d, " i

printfn "done"</lang>

Output:
2, 4, 6, 8, done

GAP

  1. 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.

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:

<lang J> thru=: <./ + i.@(+*)@-~</lang>

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:

<lang J> (#~ 1&p:) 1 thru 20 2 3 5 7 11 13 17 19</lang>

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>

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>

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>

Lisaac

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

 i.print;
 '\n'.print;

};</lang>

<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:
2
4
6
8

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:
                               2
                               4
                               6
                               8

Mathematica

<lang Mathematica>Do[

Print@i,
{i, 1, 20, 4}]</lang>
Output:
1
5
9
13
17

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

<lang>1 П0 ИП0 3 + П0 1 0 - x#0 02 С/П</lang>

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:
10
7
4
1

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>

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:

0
3
6
9
10
7
4
1

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>

  1. for_step 0 8 2 (fun i -> Printf.printf " %d\n" i) ;;
0
2
4
6
8

- : 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>

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>

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>

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>

Racket

<lang racket>

  1. 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.

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>

Salmon

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

   print(x, ", ");;

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

SAS

<lang sas>data _null_; do i=1 to 10 by 2; put i; end; run;</lang>

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

<lang scala> for (i <- 2 to 8 by 2) {

 println(i)

} </lang>

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

Scheme

<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>

Output:
2
4
6
8

Scilab

<lang> for i=1:2:10

   printf("%d\n",i)

end </lang>

Output:
1
3
5
7
8


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

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

   say i;

}</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>

Swift

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

 println(i)

}</lang> Alternately: <lang swift>for var i = 1; i < 10; i += 2 {

 println(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

<lang TorqueScript>for(%i = 0; %i < 201; %i += 2) { echo(%i); }</lang>

TUSCRIPT

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

Output:
2
4
6
8

UNIX Shell

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

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

<lang bash>for x in `jot - 2 8 2`; do echo $x; done</lang>

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>

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>

Vim Script

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

   echo i

endfor</lang>

Output:
2                                                                                                                 
4                                                                                                                 
6                                                                                                                 
8                                                                                                                 
10

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:
1
5
9

1
4
7
10

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>