Loops/For with a specified step: Difference between revisions
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<lang wart>for i 2 (i <= 8) (i <- i+2) |
<lang wart>for i 2 (i <= 8) (i <- i+2) |
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prn i</lang> |
prn i</lang> |
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=={{header|Wren}}== |
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There is currently no direct way to incorporate a step into a ''for'' loop but we can simulate it by declaring a second variable at the start of the loop which maps the loop variable to the value we want or we can simply using a ''while'' loop instead. |
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<lang ecmascript>// Print odd numbers under 20. |
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for (i in 1..10) { |
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var j = 2*i - 1 |
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System.write("%(j) ") |
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} |
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System.print("\n") |
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// Do the same using a 'while' loop. |
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var k = 1 |
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while (k < 20) { |
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System.write("%(k) ") |
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k = k + 2 |
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} |
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System.print() |
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</lang> |
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{{out}} |
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<pre> |
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1 3 5 7 9 11 13 15 17 19 |
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1 3 5 7 9 11 13 15 17 19 |
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</pre> |
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=={{header|XPL0}}== |
=={{header|XPL0}}== |
Revision as of 09:40, 13 April 2020
![Task](http://static.miraheze.org/rosettacodewiki/thumb/b/ba/Rcode-button-task-crushed.png/64px-Rcode-button-task-crushed.png)
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.
- Related tasks
- Loop over multiple arrays simultaneously
- Loops/Break
- Loops/Continue
- Loops/Do-while
- Loops/Downward for
- Loops/For
- Loops/For with a specified step
- Loops/Foreach
- Loops/Increment loop index within loop body
- Loops/Infinite
- Loops/N plus one half
- Loops/Nested
- Loops/While
- Loops/with multiple ranges
- Loops/Wrong ranges
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>
AArch64 Assembly
<lang AArch64 Assembly> /* ARM assembly AARCH64 Raspberry PI 3B */ /* program loopstep64.s */
/*******************************************/ /* Constantes file */ /*******************************************/ /* for this file see task include a file in language AArch64 assembly*/ .include "../includeConstantesARM64.inc"
.equ MAXI, 20 /*********************************/ /* Initialized data */ /*********************************/ .data szMessResult: .asciz "Counter = @ \n" // message result
/*********************************/ /* UnInitialized data */ /*********************************/ .bss sZoneConv: .skip 24 /*********************************/ /* code section */ /*********************************/ .text .global main main: // entry of program
mov x4,0 // init counter
1: // begin loop
mov x0,x4 ldr x1,qAdrsZoneConv // display value bl conversion10 // call function with 2 parameter (x0,x1) ldr x0,qAdrszMessResult ldr x1,qAdrsZoneConv bl strInsertAtCharInc // insert result at first @ character bl affichageMess // display message add x4,x4,2 // increment counter by 2 cmp x4,MAXI // ble 1b // loop
100: // standard end of the program
mov x0,0 // return code mov x8,EXIT // request to exit program svc 0 // perform the system call
qAdrsZoneConv: .quad sZoneConv qAdrszMessResult: .quad szMessResult /********************************************************/ /* File Include fonctions */ /********************************************************/ /* for this file see task include a file in language AArch64 assembly */ .include "../includeARM64.inc" </lang>
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>
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
for i:=5 step 5 until 25 do OUTINTEGER(i)
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>
ALGOL-M
<lang algol>BEGIN
INTEGER I; FOR I := 1 STEP 3 UNTIL 19 DO WRITE( I );
END</lang>
AppleScript
<lang AppleScript>repeat with i from 2 to 10 by 2 log i end repeat</lang>
ARM Assembly
<lang ARM Assembly>
/* ARM assembly Raspberry PI */ /* program loopstep2.s */
/* Constantes */ .equ STDOUT, 1 @ Linux output console .equ EXIT, 1 @ Linux syscall .equ WRITE, 4 @ Linux syscall .equ MAXI, 20 /*********************************/ /* Initialized data */ /*********************************/ .data szMessResult: .ascii "Counter = " @ message result sMessValeur: .fill 12, 1, ' '
.asciz "\n"
/*********************************/ /* UnInitialized data */ /*********************************/ .bss /*********************************/ /* code section */ /*********************************/ .text .global main main: @ entry of program
push {fp,lr} @ saves 2 registers mov r4,#0
1: @ begin loop
mov r0,r4 ldr r1,iAdrsMessValeur @ display value bl conversion10 @ call function with 2 parameter (r0,r1) ldr r0,iAdrszMessResult bl affichageMess @ display message add r4,#2 @ increment counter by 2 cmp r4,#MAXI @ ble 1b @ loop
100: @ standard end of the program
mov r0, #0 @ return code pop {fp,lr} @restaur 2 registers mov r7, #EXIT @ request to exit program svc #0 @ perform the system call
iAdrsMessValeur: .int sMessValeur iAdrszMessResult: .int szMessResult /******************************************************************/ /* display text with size calculation */ /******************************************************************/ /* r0 contains the address of the message */ affichageMess:
push {r0,r1,r2,r7,lr} @ save registres mov r2,#0 @ counter length
1: @ loop length calculation
ldrb r1,[r0,r2] @ read octet start position + index cmp r1,#0 @ if 0 its over addne r2,r2,#1 @ else add 1 in the length bne 1b @ and loop @ so here r2 contains the length of the message mov r1,r0 @ address message in r1 mov r0,#STDOUT @ code to write to the standard output Linux mov r7, #WRITE @ code call system "write" svc #0 @ call systeme pop {r0,r1,r2,r7,lr} @ restaur des 2 registres */ bx lr @ return
/******************************************************************/ /* Converting a register to a decimal */ /******************************************************************/ /* r0 contains value and r1 address area */ conversion10:
push {r1-r4,lr} @ save registers mov r3,r1 mov r2,#10
1: @ start loop
bl divisionpar10 @ r0 <- dividende. quotient ->r0 reste -> r1 add r1,#48 @ digit strb r1,[r3,r2] @ store digit on area sub r2,#1 @ previous position cmp r0,#0 @ stop if quotient = 0 */ bne 1b @ else loop @ and move spaces in first on area mov r1,#' ' @ space
2:
strb r1,[r3,r2] @ store space in area subs r2,#1 @ @ previous position bge 2b @ loop if r2 >= zéro
100:
pop {r1-r4,lr} @ restaur registres bx lr @return
/***************************************************/ /* division par 10 signé */ /* Thanks to http://thinkingeek.com/arm-assembler-raspberry-pi/* /* and http://www.hackersdelight.org/ */ /***************************************************/ /* r0 dividende */ /* r0 quotient */ /* r1 remainder */ divisionpar10:
/* r0 contains the argument to be divided by 10 */ push {r2-r4} /* save registers */ mov r4,r0 mov r3,#0x6667 @ r3 <- magic_number lower movt r3,#0x6666 @ r3 <- magic_number upper smull r1, r2, r3, r0 @ r1 <- Lower32Bits(r1*r0). r2 <- Upper32Bits(r1*r0) mov r2, r2, ASR #2 /* r2 <- r2 >> 2 */ mov r1, r0, LSR #31 /* r1 <- r0 >> 31 */ add r0, r2, r1 /* r0 <- r2 + r1 */ add r2,r0,r0, lsl #2 /* r2 <- r0 * 5 */ sub r1,r4,r2, lsl #1 /* r1 <- r4 - (r2 * 2) = r4 - (r0 * 10) */ pop {r2-r4} bx lr /* leave function */
/***************************************************/ /* integer division unsigned */ /***************************************************/ division:
/* r0 contains dividend */ /* r1 contains divisor */ /* r2 returns quotient */ /* r3 returns remainder */ push {r4, lr} mov r2, #0 @ init quotient mov r3, #0 @ init remainder mov r4, #32 @ init counter bits b 2f
1: @ loop
movs r0, r0, LSL #1 @ r0 <- r0 << 1 updating cpsr (sets C if 31st bit of r0 was 1) adc r3, r3, r3 @ r3 <- r3 + r3 + C. This is equivalent to r3 ? (r3 << 1) + C cmp r3, r1 @ compute r3 - r1 and update cpsr subhs r3, r3, r1 @ if r3 >= r1 (C=1) then r3 ? r3 - r1 adc r2, r2, r2 @ r2 <- r2 + r2 + C. This is equivalent to r2 <- (r2 << 1) + C
2:
subs r4, r4, #1 @ r4 <- r4 - 1 bpl 1b @ if r4 >= 0 (N=0) then loop pop {r4, lr} bx lr
</lang>
Arturo
<lang arturo>loop [rangeBy 0 10 2] { print & }</lang>
- Output:
0 2 4 6 8 10
AutoHotkey
<lang AutoHotkey>SetBatchLines, -1 iterations := 5 step := 10 iterations *= step Loop, % iterations {
If Mod(A_Index, step) Continue MsgBox, % A_Index
} ExitApp</lang>
Avail
<lang Avail>For each i from 0 to 100 by 7 do [Print: “i” ++ " is a multiple of 7!\n";];</lang>
Note the 0 to 100 by 7
segment isn't a fixed part of the loop syntax, but a call to the _to_by_
method, which returns a tuple of integers in a range separated by a particular step size.
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>
BaCon
This prints all odd digits: <lang freebasic> FOR i = 1 TO 10 STEP 2
PRINT i
NEXT</lang>
Basic
<lang qbasic>for i = 2 to 8 step 2
print i; ", ";
next i print "who do we appreciate?"</lang>
IS-BASIC
<lang IS-BASIC>100 FOR I=1 TO 10 STEP 2 110 PRINT I 120 NEXT</lang>
smart BASIC
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>
BBC BASIC
<lang bbcbasic> FOR n = 2 TO 8 STEP 1.5
PRINT n NEXT</lang>
- Output:
2 3.5 5 6.5 8
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>
XBasic
<lang xbasic> PROGRAM "forby"
DECLARE FUNCTION Entry()
FUNCTION Entry()
FOR i% = 0 TO 100 STEP 2 PRINT FORMAT$("###", i%) NEXT i%
END FUNCTION END PROGRAM </lang>
TI-83 BASIC
Prints numbers from 0 to 100 stepping by 5. <lang ti83b>:For(I,0,100,5
- Disp I
- End</lang>
TI-89 BASIC
<lang ti89b>Local i For i, 0, 100, 5
Disp i
EndFor</lang>
PureBasic
<lang PureBasic>For i=-15 To 25 Step 5
Debug i
Next i</lang>
Run BASIC
<lang runbasic>for i = 2 to 8 step 2
print i; ", ";
next i print "who do we appreciate?"</lang>
- Output:
2, 4, 6, 8, who do we appreciate?
FreeBASIC
<lang freebasic>' FB 1.05.0 Win64
For i As Integer = 1 To 21 Step 2
Print i; " ";
Next Print Sleep</lang>
FutureBasic
<lang futurebasic> include "ConsoleWindow"
dim as Str15 s(11) dim as long i
s(0) = "Somewhere" s(2) = " over" s(4) = " the" s(6) = " rainbow" + chr$(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
Click this link to run this code <lang gambas>Public Sub Main() Dim siCount As Short
For siCount = 1 To 50 Step 5
Print "Gambas is great!"
Next
End</lang>
Gambas is great! Gambas is great! Gambas is great! Gambas is great! Gambas is great! Gambas is great! Gambas is great! Gambas is great! Gambas is great! Gambas is great!
Liberty BASIC
<lang lb> for i = 2 to 8 step 2
print i; ", ";
next i print "who do we appreciate?" end </lang>
Microsoft Small Basic
<lang microsoftsmallbasic> For i = 0 To 100 Step 2
TextWindow.WriteLine(i)
EndFor </lang>
NS-HUBASIC
<lang NS-HUBASIC>10 FOR I=1 TO 10 STEP 2 20 PRINT I 30 NEXT</lang>
- Output:
1 3 5 7 9 11 13 15 17 19 21
Visual Basic
<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
<lang vbnet>Imports System.Console Module Program
Sub Main() For i = 2 To 8 Step 2 Write($"{i}, ") Next WriteLine("who do we appreciate?") End Sub
End Module</lang>
- Output:
2, 4, 6, 8, who do we appreciate?
<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
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 >
bc
<lang bc>for (i = 2; i <= 10; i += 2) {
i
}</lang>
Befunge
<lang befunge>1 >:.55+,v @_^#`9:+2<</lang>
C
This prints all odd digits: <lang c>int i; for(i = 1; i < 10; i += 2)
printf("%d\n", i);</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>
C++
This prints all odd digits: <lang cpp>for (int i = 1; i < 10; i += 2)
std::cout << i << std::endl;</lang>
Ceylon
<lang ceylon>shared void run() {
for(i in (2..8).by(2)) { process.write("``i`` "); } print("who do we appreciate?"); }</lang>
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
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>
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?
Using DO
<lang lisp> (do ((n 0 (incf n (+ (random 3) 2)))) ; Initialize to 0 and set random step-value 2, 3 or 4
((> n 20)) ; Break condition (print n)) ; On every loop print value
</lang>
- Output:
0 2 4 8 10 13 15 17 20
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
- 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;
{$APPTYPE CONSOLE}
var
i: Integer;
begin
i:=2; while i <= 8 do begin WriteLn(i); Inc(i, 2); end;
end.</lang>
- Output:
2 4 6 8
Dragon
<lang dragon>for(i = 2, i <= 8,i += 2){
show i + ", "
} showln "who do we appreciate?"</lang>
DWScript
<lang Delphi>var i : Integer;
for i := 2 to 8 step 2 do
PrintLn(i);</lang>
- Output:
2 4 6 8
E
There is no step in the standard numeric range object (a..b and a..!b) in E, which is typically used for numeric iteration. An ordinary while loop can of course be used:
<lang e>var i := 2 while (i <= 8) {
print(`$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>
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:
0 2 4 6 8 10
Elena
ELENA 4.x <lang elena>public program() {
for(int i := 2, i <= 8, i += 2 ) { 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>
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
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
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.
F#
<lang fsharp>for i in 2..2..8 do
printf "%d, " i
printfn "done"</lang>
- Output:
2, 4, 6, 8, done
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
<lang fortran>do i = 1,10,2
print *, i
end do</lang>
<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>
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>
Haxe
While Haxe's for-loop does not allow you to directly specify the step size, it is easy to create an iterator that allows you to do that.
<lang haxe>class Step {
var end:Int; var step:Int; var index:Int;
public inline function new(start:Int, end:Int, step:Int) { this.index = start; this.end = end; this.step = step; }
public inline function hasNext() return step > 0 ? end >= index : index >= end; public inline function next() return (index += step) - step;
}
class Main {
static function main() { for (i in new Step(2, 8, 2)) { Sys.print('$i '); } Sys.println('WHOM do we appreciate? GRAMMAR! GRAMMAR! GRAMMAR!'); }
}</lang>
- Output:
2 4 6 8 WHOM do we appreciate? GRAMMAR! GRAMMAR! GRAMMAR!
hexiscript
<lang hexiscript>for let i 0; i <= 50; let i (i + 5)
println i
endfor</lang>
HicEst
<lang hicest>DO i = 1, 6, 1.25 ! from 1 to 6 step 1.25
WRITE() i
ENDDO</lang>
HolyC
This prints all odd digits: <lang holyc>U8 i; for (i = 1; i < 10; i += 2)
Print("%d\n", i);</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:
<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>
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>
langur
<lang langur>for .i = 1; .i < 10; .i += 2 {
writeln .i
}</lang>
- Output:
1 3 5 7 9
Lasso
<lang Lasso>loop(-to=100, -from=1, -by=2) => {^
loop_count '\r' // for formatting
^}</lang>
LIL
The inc command accepts a value to add to the variable, 1 if not specified. <lang tcl>for {set i 1} {$i < 15} {inc i 3} {print $i}</lang>
- Output:
# for {set i 1} {$i < 15} {inc i 3} {print $i} 1 4 7 10 13 #
Lingo
Lingo loops don't support a "step" parameter, so it has to be implemented manually: <lang lingo>step = 3 repeat with i = 0 to 10
put i i = i + (step-1)
end repeat</lang>
- Output:
-- 0 -- 3 -- 6 -- 9
Lisaac
<lang Lisaac>1.to 9 by 2 do { i : INTEGER;
i.print; '\n'.print;
};</lang>
LiveCode
<lang LiveCode>repeat with n = 0 to 10 step 2
put n after loopn if n is not 10 then put comma after loopn
end repeat put loopn</lang> Output<lang LiveCode>0,2,4,6,8,10</lang>
Logo
<lang logo>for [i 2 8 2] [type :i type "|, |] print [who do we appreciate?]</lang>
Lua
<lang lua> for i=2,9,2 do
print(i)
end </lang>
- Output:
2 4 6 8
M2000 Interpreter
A for loop
Str$(i) always return decimal separator as "." format$() use same as Locale number specify. So here we use "," from Locale 1036. Str$() place a space before number if it is positive. We can use str$(i, "") to trim lead space. Here we use a space in format$ before number (and for negative numbers)
For this task we use single float numbers, and we make the loop one time from lower to higher value, and one time form higher to lower value.
<lang M2000 Interpreter> Module LoopFor {
Locale 1036 Document doc$ \\ define i as a single def single i for i=1 to 21 step 5/3 Print i doc$=format$(" {0}", i) next i doc$={ } \\ make i as a single for i=21 to 1 step 5/3 Print i doc$=format$(" {0}", i) next i clipboard doc$ report doc$
} LoopFor </lang>
- Output:
1 2,66667 4,33333 6 7,66667 9,33333 11 12,66667 14,33333 16 17,66667 19,33333 21 21 19,33333 17,66667 16 14,33334 12,66667 11 9,33333 7,66667 6 4,33333 2,66667 1
Iterator step 2
<lang M2000 Interpreter> a=("A", "B", "C", "D", "E", "F", "Z") k=Each(a) While k {
Print Array$(k), k=Each(a, k^+2) ' set start again
} Print \\ a list of keys (unique keys allowed) Inventory b="A", "B", "C", "D", "E", "F", "Z" k=Each(b) While k {
Print Eval$(k), ' return keys as values, because no value exist yet for each key. k=Each(b, k^+2)
} Print </lang>
- Output:
A C E Z A C E Z
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>
MiniScript
<lang MiniScript>for i in range(1,20,4)
print i
end for</lang>
- Output:
1 5 9 13 17
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
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
Never
The increment step of the Never for expression can be simple or complex and need not be contiguous.
<lang fsharp>for (i = 0; i < 10; i += 3)</lang>
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
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>
- 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>
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>
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>
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) {
$i
}</lang>
Prolog
If you need a stepping iterator, write one: <lang prolog>for(Lo,Hi,Step,Lo) :- Step>0, Lo=<Hi. for(Lo,Hi,Step,Val) :- Step>0, plus(Lo,Step,V), V=<Hi, !, for(V,Hi,Step,Val).
example :-
for(0,10,2,Val), write(Val), write(' '), fail.
example.</lang>
?- example. 0 2 4 6 8 10 true.
Adding the following two rules lets you go backwards too: <lang prolog>for(Hi,Lo,Step,Hi) :- Step<0, Lo=<Hi. for(Hi,Lo,Step,Val) :- Step<0, plus(Hi,Step,V), Lo=<V, !, for(V,Lo,Step,Val).</lang>
Python
<lang python>for i in xrange(2, 9, 2):
print "%d," % i,
print "who do we appreciate?"</lang>
<lang python>for i in range(2, 9, 2):
print("%d, " % i, end="")
print("who do we appreciate?")</lang>
- Output:
2, 4, 6, 8, who do we appreciate?
R
<lang R>for(a in seq(2,8,2)) {
cat(a, ", ")
} cat("who do we appreciate?\n")</lang>
Here the loop may be done implicitly by first concatenating the string and then printing:
<lang R>cat(paste(c(seq(2, 8, by=2), "who do we appreciate?\n"), collapse=", "))</lang>
Racket
<lang racket>
- lang racket
(for ([i (in-range 2 9 2)])
(printf "~a, " i))
(printf "who do we appreciate?~n") </lang>
Raku
(formerly Perl 6)
With at least two values on the left-hand side, the sequence operator (...
) can infer an arithmetic series. (With at least three values, it can infer a geometric sequence, too.)
<lang perl6>for 2, 4 ... 8 {
print "$_, ";
}
say 'whom do we appreciate?';</lang>
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.
version 3
<lang rexx>Do v=1 by 3/2 While v**2<30
Say v End
Say '('v'**2) is greater than 30 (30.25)'</lang>
- Output:
1 2.5 4.0 (5.5**2) is greater than 30 (30.25)
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:
2 4 6 8 10
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:
0 0.50 1 1.50 2 2.50 3 3.50 4 4.50 5 5.50 6 6.50 7 7.50 8 8.50 9 9.50 10
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?
Rust
For Rust 1.28 and later: <lang rust>fn main() {
for i in (2..=8).step_by(2) { print!("{}", i); } println!("who do we appreciate?!");
} </lang>
An alternative which also works in earlier versions of Rust: <lang rust>fn main() {
let mut i = 2; while i <= 8 { print!("{}, ", i); i += 2; } 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>
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:
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
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>
Spin
<lang spin>con
_clkmode = xtal1 + pll16x _clkfreq = 80_000_000
obj
ser : "FullDuplexSerial.spin"
pub main | n
ser.start(31, 30, 0, 115200)
repeat n from 0 to 19 step 3 ser.dec(n) ser.tx(32)
waitcnt(_clkfreq + cnt) ser.stop cogstop(0)</lang>
- Output:
0 3 6 9 12 15 18
SPL
<lang spl>> n, 1..10,2
#.output(n)
<</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>
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.
Bourne Shell
<lang bash>x=2 while test $x -le 8; do echo $x x=`expr $x + 2` || exit $? done</lang>
<lang bash>for x in `jot - 2 8 2`; do echo $x; done</lang>
Korn Shell
<lang bash>x=2 while $x -le 8; do echo $x ((x=x+2)) done</lang>
<lang bash>x=2 while ((x<=8)); do echo $x ((x+=2)) done</lang>
Bourne Again Shell
<lang bash>for (( x=2; $x<=8; x=$x+2 )); do
printf "%d, " $x
done</lang>
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
<lang csh>foreach x (`jot - 2 8 2`) echo $x end</lang>
Ursa
<lang ursa>decl int i for (set i 2) (< i 9) (set i (int (+ i 2))) out i ", " console end for out "who do we appreciate?" endl console</lang>
Vala
<lang vala>for (int i = 1; i < 10; i += 2)
stdout.printf("%d\n", i);</lang>
VAX Assembly
<lang VAX Assembly> 0000 0000 1 .entry main,0
50 D4 0002 2 clrf r0 ;init to 0.0 0004 3 loop: 01 0004 4 nop ;do nothing FFF9 50 0A 3E 4F 0005 5 acbf #112.0, #1.25, r0, loop ;limit, step 000B 6 04 000B 7 ret 000C 8 .end main</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
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>
Wren
There is currently no direct way to incorporate a step into a for loop but we can simulate it by declaring a second variable at the start of the loop which maps the loop variable to the value we want or we can simply using a while loop instead. <lang ecmascript>// Print odd numbers under 20. for (i in 1..10) {
var j = 2*i - 1 System.write("%(j) ")
}
System.print("\n")
// Do the same using a 'while' loop. var k = 1 while (k < 20) {
System.write("%(k) ") k = k + 2
} System.print() </lang>
- Output:
1 3 5 7 9 11 13 15 17 19 1 3 5 7 9 11 13 15 17 19
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>
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