Count in octal: Difference between revisions
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The technique used is to convert the decimal number to binary, and separate the binary digits in groups of three, and then convert those binary groups (numbers) to decimal. |
The technique used is to convert the decimal number to binary, and separate the binary digits in groups of three, and then convert those binary groups (numbers) to decimal. |
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<lang rexx>/*REXX program counts in octal until the number exceeds |
<lang rexx>/*REXX program counts in octal until the number exceeds the number of program statements*/ |
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o= |
o= |
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do k=1 to length(_) by 3 |
do k=1 to length(_) by 3 |
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o=o'0'substr(_,k,3) |
o= o'0'substr(_, k, 3) |
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end /*k*/ |
end /*k*/ |
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say numIn 'base ten = ' right(#,w) numIn "octal = " right(b2x(o)+0,w+w) |
say numIn 'base ten = ' right(#,w) numIn "octal = " right( b2x(o) + 0, w + w) |
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if #>sourceline() then leave /*stop if #protons>pgm statements*/ |
if #>sourceline() then leave /*stop if # of protons > pgm statements*/ |
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end /*#*/ |
end /*#*/ |
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/*stick a fork in it, we're done.*/</lang> |
/*stick a fork in it, we're all done. */</lang> |
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{{out|output}} |
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<pre style="height: |
<pre style="height:38ex"> |
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number in base ten = 0 number in octal = 0 |
number in base ten = 0 number in octal = 0 |
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number in base ten = 1 number in octal = 1 |
number in base ten = 1 number in octal = 1 |
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number in base ten = 30 number in octal = 36 |
number in base ten = 30 number in octal = 36 |
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number in base ten = 31 number in octal = 37 |
number in base ten = 31 number in octal = 37 |
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number in base ten = 32 number in octal = 40 |
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number in base ten = 33 number in octal = 41 |
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</pre> |
</pre> |
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Revision as of 11:50, 26 March 2020
You are encouraged to solve this task according to the task description, using any language you may know.
- Task
Produce a sequential count in octal, starting at zero, and using an increment of a one for each consecutive number.
Each number should appear on a single line, and the program should count until terminated, or until the maximum value of the numeric type in use is reached.
- Related task
- Integer sequence is a similar task without the use of octal numbers.
0815
<lang 0815>}:l:> Start loop, enqueue Z (initially 0).
}:o: Treat the queue as a stack and <:8:= accumulate the octal digits /=>&~ of the current number. ^:o:
<:0:- Get a sentinel negative 1. &>@ Enqueue it between the digits and the current number. { Dequeue the first octal digit.
}:p: ~%={+ Rotate each octal digit into place and print it. ^:p:
<:a:~$ Output a newline. <:1:x{+ Dequeue the current number and increment it.
^:l:</lang>
360 Assembly
The program uses one ASSIST macro (XPRNT) to keep the code as short as possible. <lang 360asm>* Octal 04/07/2016 OCTAL CSECT
USING OCTAL,R13 base register B 72(R15) skip savearea DC 17F'0' savearea STM R14,R12,12(R13) prolog ST R13,4(R15) " ST R15,8(R13) " LR R13,R15 " LA R6,0 i=0
LOOPI LR R2,R6 x=i
LA R9,10 j=10 LA R4,PG+23 @pg
LOOP LR R3,R2 save x
SLL R2,29 shift left 32-3 SRL R2,29 shift right 32-3 CVD R2,DW convert octal(j) to pack decimal OI DW+7,X'0F' prepare unpack UNPK 0(1,R4),DW packed decimal to zoned printable LR R2,R3 restore x SRL R2,3 shift right 3 BCTR R4,0 @pg=@pg-1 BCT R9,LOOP j=j-1 CVD R2,DW binary to pack decimal OI DW+7,X'0F' prepare unpack UNPK 0(1,R4),DW packed decimal to zoned printable CVD R6,DW convert i to pack decimal MVC ZN12,EM12 load mask ED ZN12,DW+2 packed decimal (PL6) to char (CL12) MVC PG(12),ZN12 output i XPRNT PG,80 print buffer C R6,=F'2147483647' if i>2**31-1 (integer max) BE ELOOPI then exit loop on i LA R6,1(R6) i=i+1 B LOOPI loop on i
ELOOPI L R13,4(0,R13) epilog
LM R14,R12,12(R13) " XR R15,R15 " BR R14 exit LTORG
PG DC CL80' ' buffer DW DS 0D,PL8 15num ZN12 DS CL12 EM12 DC X'40',9X'20',X'2120' mask CL12 11num
YREGS END OCTAL</lang>
- Output:
0 00000000000 1 00000000001 2 00000000002 3 00000000003 4 00000000004 5 00000000005 6 00000000006 7 00000000007 8 00000000010 9 00000000011 10 00000000012 10 00000000012 11 00000000013 ... 2147483640 17777777770 2147483641 17777777771 2147483642 17777777772 2147483643 17777777773 2147483644 17777777774 2147483645 17777777775 2147483646 17777777776 2147483647 17777777777
Ada
<lang Ada>with Ada.Text_IO;
procedure Octal is
package IIO is new Ada.Text_IO.Integer_IO(Integer);
begin
for I in 0 .. Integer'Last loop IIO.Put(I, Base => 8); Ada.Text_IO.New_Line; end loop;
end Octal;</lang> First few lines of Output:
8#0# 8#1# 8#2# 8#3# 8#4# 8#5# 8#6# 8#7# 8#10# 8#11# 8#12# 8#13# 8#14# 8#15# 8#16# 8#17# 8#20#
Aime
<lang aime>integer o;
o = 0; do {
o_xinteger(8, o); o_byte('\n'); o += 1;
} while (0 < o);</lang>
ALGOL 68
<lang algol68>#!/usr/local/bin/a68g --script #
INT oct width = (bits width-1) OVER 3 + 1; main: (
FOR i TO 17 # max int # DO printf(($"8r"8r n(oct width)dl$, BIN i)) OD
)</lang> Output:
8r00000000001 8r00000000002 8r00000000003 8r00000000004 8r00000000005 8r00000000006 8r00000000007 8r00000000010 8r00000000011 8r00000000012 8r00000000013 8r00000000014 8r00000000015 8r00000000016 8r00000000017 8r00000000020 8r00000000021
ALGOL W
Algol W has built-in hexadecimal and decimal output, this implements octal output. <lang algolw>begin
string(12) r; string(8) octDigits; integer number; octDigits := "01234567"; number := -1; while number < MAXINTEGER do begin integer v, cPos; number := number + 1; v := number; % build a string of octal digits in r, representing number % % Algol W uses 32 bit integers, so r should be big enough % % the most significant digit is on the right % cPos := 0; while begin r( cPos // 1 ) := octDigits( v rem 8 // 1 ); v := v div 8; ( v > 0 ) end do begin cPos := cPos + 1 end while_v_gt_0; % show most significant digit on a newline % write( r( cPos // 1 ) ); % continue the line with the remaining digits (if any) % for c := cPos - 1 step -1 until 0 do writeon( r( c // 1 ) ) end while_r_lt_MAXINTEGER
end.</lang>
- Output:
0 1 2 3 4 5 6 7 10 11 12 ...
ARM Assembly
<lang ARM Assembly> /* ARM assembly Raspberry PI */ /* program countoctal.s */
/************************************/ /* Constantes */ /************************************/ .equ STDOUT, 1 @ Linux output console .equ EXIT, 1 @ Linux syscall .equ WRITE, 4 @ Linux syscall
/*********************************/ /* Initialized data */ /*********************************/ .data sMessResult: .ascii "Count : " sMessValeur: .fill 11, 1, ' ' @ size => 11 szCarriageReturn: .asciz "\n"
/*********************************/
/* UnInitialized data */
/*********************************/
.bss
/*********************************/
/* code section */
/*********************************/
.text
.global main
main: @ entry of program
mov r4,#0 @ loop indice
1: @ begin loop
mov r0,r4 ldr r1,iAdrsMessValeur bl conversion8 @ call conversion octal ldr r0,iAdrsMessResult bl affichageMess @ display message add r4,#1 cmp r4,#64 ble 1b
100: @ standard end of the program
mov r0, #0 @ return code mov r7, #EXIT @ request to exit program svc #0 @ perform the system call
iAdrsMessValeur: .int sMessValeur iAdrszCarriageReturn: .int szCarriageReturn iAdrsMessResult: .int sMessResult
/******************************************************************/ /* 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 octal */ /******************************************************************/ /* r0 contains value and r1 address area */ /* r0 return size of result (no zero final in area) */ /* area size => 11 bytes */ .equ LGZONECAL, 10 conversion8:
push {r1-r4,lr} @ save registers mov r3,r1 mov r2,#LGZONECAL
1: @ start loop
mov r1,r0 lsr r0,#3 @ / by 8 sub r1,r0,lsl #3 @ compute remainder r1 - (r0 * 8) add r1,#48 @ digit strb r1,[r3,r2] @ store digit on area cmp r0,#0 @ stop if quotient = 0 subne r2,#1 @ else previous position bne 1b @ and loop @ and move digit from left of area mov r4,#0
2:
ldrb r1,[r3,r2] strb r1,[r3,r4] add r2,#1 add r4,#1 cmp r2,#LGZONECAL ble 2b @ and move spaces in end on area mov r0,r4 @ result length mov r1,#' ' @ space
3:
strb r1,[r3,r4] @ store space in area add r4,#1 @ next position cmp r4,#LGZONECAL ble 3b @ loop if r4 <= area size
100:
pop {r1-r4,lr} @ restaur registres bx lr @return
</lang>
AutoHotkey
<lang AHK>DllCall("AllocConsole") Octal(int){ While int out := Mod(int, 8) . out, int := int//8 return out } Loop { FileAppend, % Octal(A_Index) "`n", CONOUT$ Sleep 200 }</lang>
AWK
The awk extraction and reporting language uses the underlying C library to provide support for the printf command. This enables us to use that function to output the counter value as octal:
<lang awk>BEGIN {
for (l = 0; l <= 2147483647; l++) { printf("%o\n", l); }
}</lang>
BASIC
Some BASICs provide a built-in function to convert a number to octal, typically called OCT$
.
<lang qbasic>DIM n AS LONG FOR n = 0 TO &h7FFFFFFF
PRINT OCT$(n)
NEXT</lang>
However, many do not. For those BASICs, we need to write our own function.
<lang qbasic>WHILE ("" = INKEY$)
PRINT Octal$(n) n = n + 1
WEND END FUNCTION Octal$(what)
outp$ = "" w = what WHILE ABS(w) > 0 o = w AND 7 w = INT(w / 8) outp$ = STR$(o) + outp$ WEND Octal$ = outp$
END FUNCTION</lang>
See also: BBC BASIC, Liberty BASIC, PureBasic, Run BASIC
Applesoft BASIC
<lang ApplesoftBasic>10 N$ = "0"
100 O$ = N$ 110 PRINT O$ 120 N$ = "" 130 C = 1 140 FOR I = LEN(O$) TO 1 STEP -1 150 N = VAL(MID$(O$, I, 1)) + C 160 C = N >= 8 170 N$ = STR$(N - C * 8) + N$ 180 NEXT I 190 IF C THEN N$ = "1" + N$ 200 GOTO 100</lang>
Sinclair ZX81 BASIC
The octal number is stored and manipulated as a string, meaning that even with only 1k of RAM the program shouldn't stop until the number gets to a couple of hundred digits long. I have not left it running long enough to find out exactly when it does run out of memory. The SCROLL
statement is necessary: the ZX81 halts when the screen is full unless it is positively told to scroll instead.
<lang basic> 10 LET N$="0"
20 SCROLL 30 PRINT N$ 40 LET L=LEN N$ 50 LET N=VAL N$(L)+1 60 IF N=8 THEN GOTO 90 70 LET N$(L)=STR$ N 80 GOTO 20 90 LET N$(L)="0"
100 IF L=1 THEN GOTO 130 110 LET L=L-1 120 GOTO 50 130 LET N$="1"+N$ 140 GOTO 20</lang>
Batch File
<lang dos> @echo off
- {CTRL + C} to exit the batch file
- Send incrementing decimal values to the :to_Oct function
set loop=0
- loop1
call:to_Oct %loop% set /a loop+=1 goto loop1
- Convert the decimal values parsed [%1] to octal and output them on a new line
- to_Oct
set todivide=%1 set "fulloct="
- loop2
set tomod=%todivide% set /a appendmod=%tomod% %% 8 set fulloct=%appendmod%%fulloct% if %todivide% lss 8 (
echo %fulloct% exit /b
) set /a todivide/=8 goto loop2 </lang>
- Output:
0 1 2 3 4 5 6 7 10 ...
BBC BASIC
Terminate by pressing ESCape. <lang bbcbasic> N% = 0
REPEAT PRINT FN_tobase(N%, 8, 0) N% += 1 UNTIL FALSE END REM Convert N% to string in base B% with minimum M% digits: DEF FN_tobase(N%, B%, M%) LOCAL D%, A$ REPEAT D% = N% MOD B% N% DIV= B% IF D%<0 D% += B% : N% -= 1 A$ = CHR$(48 + D% - 7*(D%>9)) + A$ M% -= 1 UNTIL (N%=FALSE OR N%=TRUE) AND M%<=0 =A$
</lang>
bc
<lang bc>obase = 8 /* Output base is octal. */ for (num = 0; 1; num++) num /* Loop forever, printing counter. */</lang>
The loop never stops at a maximum value, because bc uses arbitrary-precision integers.
Befunge
This is almost identical to the Binary digits sample, except for the change of base and the source coming from a loop rather than a single input. <lang befunge>:0\55+\:8%68>*#<+#8\#68#%/#8:_$>:#,_$1+:0`!#@_</lang>
Bracmat
Stops when the user presses Ctrl-C or when the stack overflows. The solution is not elegant, and so is octal counting. <lang bracmat>
( oct = . !arg:<8 & (!arg:~<0|ERROR) | str$(oct$(div$(!arg.8)) mod$(!arg.8)) )
& -1:?n & whl'(1+!n:?n&out$(!n oct$!n)); </lang>
Brainf***
<lang bf>+[ Start with n=1 to kick off the loop [>>++++++++<< Set up {n 0 8} for divmod magic [->+>- Then [>+>>]> do [+[-<+>]>+>>] the <<<<<<] magic >>>+ Increment n % 8 so that 0s don't break things >] Move into n / 8 and divmod that unless it's 0 -< Set up sentinel ‑1 then move into the first octal digit [++++++++ ++++++++ ++++++++ Add 47 to get it to ASCII
++++++++ ++++++++ +++++++. and print it
[<]<] Get to a 0; the cell to the left is the next octal digit >>[<+>-] Tape is {0 n}; make it {n 0} >[>+] Get to the ‑1 <[[-]<] Zero the tape for the next iteration ++++++++++. Print a newline [-]<+] Zero it then increment n and go again</lang>
C
<lang C>#include <stdio.h>
int main() {
unsigned int i = 0; do { printf("%o\n", i++); } while(i); return 0;
}</lang>
C#
<lang csharp>using System;
class Program {
static void Main() { var number = 0; do { Console.WriteLine(Convert.ToString(number, 8)); } while (++number > 0); }
}</lang>
C++
This prevents an infinite loop by counting until the counter overflows and produces a 0 again. This could also be done with a for or while loop, but you'd have to print 0 (or the last number) outside the loop.
<lang cpp>#include <iostream>
int main() {
unsigned i = 0; do { std::cout << std::oct << i << std::endl; ++i; } while(i != 0); return 0;
}</lang>
Clojure
<lang clojure>(doseq [i (range)] (println (format "%o" i)))</lang>
COBOL
<lang cobol> >>SOURCE FREE IDENTIFICATION DIVISION. PROGRAM-ID. count-in-octal.
ENVIRONMENT DIVISION. CONFIGURATION SECTION. REPOSITORY.
FUNCTION dec-to-oct .
DATA DIVISION. WORKING-STORAGE SECTION. 01 i PIC 9(18).
PROCEDURE DIVISION.
PERFORM VARYING i FROM 1 BY 1 UNTIL i = 0 DISPLAY FUNCTION dec-to-oct(i) END-PERFORM .
END PROGRAM count-in-octal.
IDENTIFICATION DIVISION.
FUNCTION-ID. dec-to-oct.
DATA DIVISION. LOCAL-STORAGE SECTION. 01 rem PIC 9.
01 dec PIC 9(18).
LINKAGE SECTION. 01 dec-arg PIC 9(18).
01 oct PIC 9(18).
PROCEDURE DIVISION USING dec-arg RETURNING oct.
MOVE dec-arg TO dec *> Copy is made to avoid modifying reference arg. PERFORM WITH TEST AFTER UNTIL dec = 0 MOVE FUNCTION REM(dec, 8) TO rem STRING rem, oct DELIMITED BY SPACES INTO oct DIVIDE 8 INTO dec END-PERFORM .
END FUNCTION dec-to-oct.</lang>
CoffeeScript
<lang coffeescript> n = 0
while true
console.log n.toString(8) n += 1
</lang>
Common Lisp
<lang lisp>(loop for i from 0 do (format t "~o~%" i))</lang>
Component Pascal
BlackBox Component Builder <lang oberon2> MODULE CountOctal; IMPORT StdLog,Strings;
PROCEDURE Do*; VAR i: INTEGER; resp: ARRAY 32 OF CHAR; BEGIN FOR i := 0 TO 1000 DO Strings.IntToStringForm(i,8,12,' ',TRUE,resp); StdLog.String(resp);StdLog.Ln END END Do; END CountOctal.
</lang>
Execute: ^Q CountOctal.Do
Output:
0%8 1%8 2%8 3%8 4%8 5%8 6%8 7%8 10%8 11%8 12%8 13%8 14%8 15%8 16%8 17%8 20%8 21%8 22%8
Crystal
<lang ruby># version 0.21.1
- using unsigned 8 bit integer, range 0 to 255
(0_u8..255_u8).each { |i| puts i.to_s(8) }</lang>
- Output:
0 1 2 3 4 5 6 7 10 11 12 ... 374 375 376 377
D
<lang d>void main() {
import std.stdio;
ubyte i; do writefln("%o", i++); while(i);
}</lang>
Dc
Named Macro
A simple infinite loop and octal output will do. <lang Dc>8o0[p1+lpx]dspx</lang>
Anonymous Macro
Needs r
(swap TOS and NOS):
<lang Dc>8 o 0 [ r p 1 + r dx ] dx</lang>
Pushing/poping TOS to a named stack can be used instead of swaping:
<lang Dc>8 o 0 [ S@ p 1 + L@ dx ] dx</lang>
DCL
<lang DCL>$ i = 0 $ loop: $ write sys$output f$fao( "!OL", i ) $ i = i + 1 $ goto loop</lang>
- Output:
00000000000 00000000001 00000000002 ... 17777777777 20000000000 20000000001 ... 37777777777 00000000000 00000000001 ...
Delphi
<lang Delphi>program CountingInOctal;
{$APPTYPE CONSOLE}
uses SysUtils;
function DecToOct(aValue: Integer): string; var
lRemainder: Integer;
begin
Result := ; repeat lRemainder := aValue mod 8; Result := IntToStr(lRemainder) + Result; aValue := aValue div 8; until aValue = 0;
end;
var
i: Integer;
begin
for i := 0 to 20 do WriteLn(DecToOct(i));
end.</lang>
Elixir
<lang elixir>Stream.iterate(0,&(&1+1)) |> Enum.each(&IO.puts Integer.to_string(&1,8))</lang> or <lang elixir>Stream.unfold(0, fn n ->
IO.puts Integer.to_string(n,8) {n,n+1}
end) |> Stream.run</lang> or <lang elixir>f = fn ff,i -> :io.fwrite "~.8b~n", [i]; ff.(ff, i+1) end f.(f, 0)</lang>
Emacs Lisp
Displays in the message area interactively, or to standard output under -batch
.
<lang lisp>(dotimes (i most-positive-fixnum) ;; starting from 0
(message "%o" i))</lang>
Erlang
The fun is copied from Integer sequence#Erlang. I changed the display format. <lang Erlang> F = fun(FF, I) -> io:fwrite("~.8B~n", [I]), FF(FF, I + 1) end. </lang> Use like this:
F( F, 0 ).
Euphoria
<lang euphoria>integer i i = 0 while 1 do
printf(1,"%o\n",i) i += 1
end while</lang>
Output:
... 6326 6327 6330 6331 6332 6333 6334 6335 6336 6337
F#
<lang fsharp>let rec countInOctal num : unit =
printfn "%o" num countInOctal (num + 1)
countInOctal 1</lang>
Factor
<lang factor>USING: kernel math prettyprint ; 0 [ dup .o 1 + t ] loop</lang>
Forth
Using INTS from Integer sequence#Forth <lang forth>: octal ( -- ) 8 base ! ; \ where unavailable
octal ints</lang>
Fortran
<lang fortran>program Octal
implicit none integer, parameter :: i64 = selected_int_kind(18) integer(i64) :: n = 0
! Will stop when n overflows from ! 9223372036854775807 to -92233720368547758078 (1000000000000000000000 octal)
do while(n >= 0) write(*, "(o0)") n n = n + 1 end do
end program</lang>
FreeBASIC
<lang freebasic>' FB 1.05.0 Win64
Dim ub As UByte = 0 ' only has a range of 0 to 255 Do
Print Oct(ub, 3) ub += 1
Loop Until ub = 0 ' wraps around to 0 when reaches 256 Print Print "Press any key to quit" Sleep</lang>
Futhark
Futhark cannot print. Instead we produce an array of integers that look like octal numbers when printed in decimal.
<lang Futhark> fun octal(x: int): int =
loop ((out,mult,x) = (0,1,x)) = while x > 0 do let digit = x % 8 let out = out + digit * mult in (out, mult * 10, x / 8) in out
fun main(n: int): [n]int =
map octal (iota n)
</lang>
FutureBasic
<lang futurebasic> include "ConsoleWindow defstr word
dim as short i
for i = &o000000 to &o000031 // 0 to 25 in decimal
print oct$(i); " in octal ="; i
next </lang>
Output:
000000 in octal = 0 000001 in octal = 1 000002 in octal = 2 000003 in octal = 3 000004 in octal = 4 000005 in octal = 5 000006 in octal = 6 000007 in octal = 7 000010 in octal = 8 000011 in octal = 9 000012 in octal = 10 000013 in octal = 11 000014 in octal = 12 000015 in octal = 13 000016 in octal = 14 000017 in octal = 15 000020 in octal = 16 000021 in octal = 17 000022 in octal = 18 000023 in octal = 19 000024 in octal = 20 000025 in octal = 21 000026 in octal = 22 000027 in octal = 23 000030 in octal = 24 000031 in octal = 25
Go
<lang go>package main
import (
"fmt" "math"
)
func main() {
for i := int8(0); ; i++ { fmt.Printf("%o\n", i) if i == math.MaxInt8 { break } }
}</lang> Output:
0 1 2 3 4 5 6 7 10 11 12 ... 175 176 177
Note that to use a different integer type, code must be changed in two places. Go has no way to query a type for its maximum value. Example: <lang go>func main() {
for i := uint16(0); ; i++ { // type specified here fmt.Printf("%o\n", i) if i == math.MaxUint16 { // maximum value for type specified here break } }
}</lang> Output:
... 177775 177776 177777
Note also that if floating point types are used for the counter, loss of precision will prevent the program from from ever reaching the maximum value. If you stretch interpretation of the task wording "maximum value" to mean "maximum value of contiguous integers" then the following will work: <lang go>import "fmt"
func main() {
for i := 0.; ; { fmt.Printf("%o\n", int64(i)) /* uncomment to produce example output if i == 3 { i = float64(1<<53 - 4) // skip to near the end fmt.Println("...") } */ next := i + 1 if next == i { break } i = next }
}</lang> Output, with skip uncommented:
0 1 2 3 ... 377777777777777775 377777777777777776 377777777777777777 400000000000000000
Big integers have no maximum value, but the Go runtime will panic when memory allocation fails. The deferred recover here allows the program to terminate silently should the program run until this happens. <lang go>import (
"big" "fmt"
)
func main() {
defer func() { recover() }() one := big.NewInt(1) for i := big.NewInt(0); ; i.Add(i, one) { fmt.Printf("%o\n", i) }
}</lang> Output:
0 1 2 3 4 5 6 7 10 11 12 13 14 ...
Groovy
Size-limited solution: <lang groovy>println 'decimal octal' for (def i = 0; i <= Integer.MAX_VALUE; i++) {
printf ('%7d %#5o\n', i, i)
}</lang>
Unbounded solution: <lang groovy>println 'decimal octal' for (def i = 0g; true; i += 1g) {
printf ('%7d %#5o\n', i, i)
}</lang>
Output:
decimal octal 0 00 1 01 2 02 3 03 4 04 5 05 6 06 7 07 8 010 9 011 10 012 11 013 12 014 13 015 14 016 15 017 16 020 17 021 ...
Haskell
<lang haskell>import Numeric
main = mapM_ (putStrLn . flip showOct "") [1..]</lang>
Icon and Unicon
<lang unicon>link convert # To get exbase10 method
procedure main()
limit := 8r37777777777 every write(exbase10(seq(0)\limit, 8))
end</lang>
J
Solution: <lang J> disp=.([smoutput) ' '(-.~":)8&#.inv
(1+disp)^:_]0x</lang>
The full result is not displayable, by design. This could be considered a bug, but is an essential part of this task. Here's how it starts:
<lang j> (1+disp)^:_]0x 0 1 2 3 4 5 6 7 10 11 ...</lang>
The important part of this code is 8&#.inv which converts numbers from internal representation to a sequence of base 8 digits. (We then convert this sequence to characters and remove the delimiting spaces - this gives us the octal values we want to display.)
So then we define disp as a word which displays its argument in octal and returns its argument as its result (unchanged).
Finally, the ^:_
clause tells J to repeat this function forever, with (1+disp)
adding 1 to the result each time it is displayed (or at least tha clause tells J to keep repeating that operation until it gives the same value back twice in a row - which won't happen - or to stop when the machine stops - like if the power is turned off - or if J is shut down - or...).
We use arbitrary precision numbers, not because there's any likelihood that fixed width numbers would ever overflow, but just to emphasize that this thing is going to have to be shut down by some mechanism outside the program.
Java
<lang java>public class Count{
public static void main(String[] args){ for(int i = 0;i >= 0;i++){ System.out.println(Integer.toOctalString(i)); //optionally use "Integer.toString(i, 8)" } }
}</lang>
JavaScript
<lang javascript>for (var n = 0; n < 1e14; n++) { // arbitrary limit that's not too big
document.writeln(n.toString(8)); // not sure what's the best way to output it in JavaScript
}</lang>
Julia
<lang Julia> for i in one(Int64):typemax(Int64)
print(oct(i), " ") sleep(0.1)
end
</lang>
I slowed the loop down with a sleep
to make it possible to see the result without being swamped.
- Output:
1 2 3 4 5 6 7 10 11 12 13 14 15 16 17 20 21 22 23 24 25 26 27 30 31 32 33 34 35 36 ^C
Kotlin
<lang scala>// version 1.1
// counts up to 177 octal i.e. 127 decimal fun main(args: Array<String>) {
(0..Byte.MAX_VALUE).forEach { println("%03o".format(it)) }
}</lang>
- Output:
First ten lines:
000 001 002 003 004 005 006 007 010 011
LabVIEW
LabVIEW contains a Number to Octal String function. The following image shows the front panel and block diagram.
Lang5
<lang lang5>'%4o '__number_format set 0 do dup 1 compress . "\n" . 1 + loop</lang>
langur
We have to use an arbitrary limit for this.
We use the :8x interpolation modifier to create a string in base 8 (may use base 2 to 36).
<lang langur>val .limit = 70000
for .i = 0; .i <= .limit; .i += 1 {
writeln $"10x\.i; == 8x\.i:8x;"
}</lang>
- Output:
10x0 == 8x0 10x1 == 8x1 10x2 == 8x2 10x3 == 8x3 10x4 == 8x4 10x5 == 8x5 10x6 == 8x6 10x7 == 8x7 10x8 == 8x10 10x9 == 8x11 10x10 == 8x12 10x11 == 8x13 10x12 == 8x14 10x13 == 8x15 10x14 == 8x16 10x15 == 8x17 10x16 == 8x20 10x17 == 8x21 10x18 == 8x22 10x19 == 8x23 10x20 == 8x24 10x21 == 8x25 10x22 == 8x26 10x23 == 8x27 10x24 == 8x30 10x25 == 8x31 10x26 == 8x32 10x27 == 8x33 10x28 == 8x34 10x29 == 8x35 10x30 == 8x36 10x31 == 8x37 10x32 == 8x40 10x33 == 8x41 10x34 == 8x42 10x35 == 8x43 10x36 == 8x44 10x37 == 8x45 10x38 == 8x46 10x39 == 8x47 10x40 == 8x50 10x41 == 8x51 10x42 == 8x52 10x43 == 8x53 10x44 == 8x54 10x45 == 8x55 10x46 == 8x56 10x47 == 8x57 10x48 == 8x60 10x49 == 8x61 10x50 == 8x62 10x51 == 8x63 10x52 == 8x64 ... 10x69982 == 8x210536 10x69983 == 8x210537 10x69984 == 8x210540 10x69985 == 8x210541 10x69986 == 8x210542 10x69987 == 8x210543 10x69988 == 8x210544 10x69989 == 8x210545 10x69990 == 8x210546 10x69991 == 8x210547 10x69992 == 8x210550 10x69993 == 8x210551 10x69994 == 8x210552 10x69995 == 8x210553 10x69996 == 8x210554 10x69997 == 8x210555 10x69998 == 8x210556 10x69999 == 8x210557 10x70000 == 8x210560
LFE
<lang lisp>(: lists foreach
(lambda (x) (: io format '"~p~n" (list (: erlang integer_to_list x 8)))) (: lists seq 0 2000))
</lang>
Liberty BASIC
Terminate these ( essentially, practically) infinite loops by hitting <CTRL<BRK> <lang lb>
'the method used here uses the base-conversion from RC Non-decimal radices/Convert 'to terminate hit <CTRL<BRK>
global alphanum$ alphanum$ ="01234567"
i =0
while 1 print toBase$( 8, i) i =i +1 wend
end
function toBase$( base, number) ' Convert decimal variable to number string. maxIntegerBitSize =len( str$( number)) toBase$ ="" for i =10 to 1 step -1 remainder =number mod base toBase$ =mid$( alphanum$, remainder +1, 1) +toBase$ number =int( number /base) if number <1 then exit for next i toBase$ =right$( " " +toBase$, 10) end function
</lang> As suggested in LOGO, it is easy to work on a string representation too. <lang lb>
op$ = "00000000000000000000"
L =len( op$)
while 1
started =0
for i =1 to L m$ =mid$( op$, i, 1) if started =0 and m$ ="0" then print " "; else print m$;: started =1 next i print
for i =L to 1 step -1 p$ =mid$( op$, i, 1) if p$ =" " then v =0 else v =val( p$) incDigit = v +carry if i =L then incDigit =incDigit +1 if incDigit >=8 then replDigit =incDigit -8 carry =1 else replDigit =incDigit carry =0 end if op$ =left$( op$, i -1) +chr$( 48 +replDigit) +right$( op$, L -i) next i
wend
end </lang> Or use a recursive listing of permutations with the exception that the first digit is not 0 (unless listing single-digit numbers). For each digit-place, list numbers with 0-7 in the next digit-place.
<lang lb> i = 0
while 1
call CountOctal 0, i, i > 0 i = i + 1
wend
sub CountOctal value, depth, startValue
value = value * 10 for i = startValue to 7 if depth > 0 then call CountOctal value + i, depth - 1, 0 else print value + i end if next i
end sub </lang>
Logo
No built-in octal-formatting, so it's probably more efficient to just manually increment a string than to increment a number and then convert the whole thing to octal every time we print. This also lets us keep counting as long as we have room for the string.
<lang logo>to increment_octal :n
ifelse [empty? :n] [ output 1 ] [ local "last make "last last :n local "butlast make "butlast butlast :n make "last sum :last 1 ifelse [:last < 8] [ output word :butlast :last ] [ output word (increment_octal :butlast) 0 ] ]
end
make "oct 0 while ["true] [
print :oct make "oct increment_octal :oct
]</lang>
LOLCODE
LOLCODE has no conception of octal numbers, but we can use string concatenation (SMOOSH) and basic arithmetic to accomplish the task. <lang LOLCODE>HAI 1.3
HOW IZ I octal YR num
I HAS A digit, I HAS A oct ITZ "" IM IN YR octalizer digit R MOD OF num AN 8 oct R SMOOSH digit oct MKAY num R QUOSHUNT OF num AN 8 NOT num, O RLY? YA RLY, FOUND YR oct OIC IM OUTTA YR octalizer
IF U SAY SO
IM IN YR printer UPPIN YR num
VISIBLE I IZ octal YR num MKAY
IM OUTTA YR printer
KTHXBYE</lang>
Lua
<lang lua>for l=1,2147483647 do
print(string.format("%o",l))
end</lang>
M4
<lang M4>define(`forever',
`ifelse($#,0,``$0, `pushdef(`$1',$2)$4`'popdef(`$1')$0(`$1',eval($2+$3),$3,`$4')')')dnl
forever(`y',0,1, `eval(y,8) ')</lang>
Maple
<lang Maple> octcount := proc (n)
seq(printf("%a \n", convert(i, octal)), i = 1 .. n); end proc;
</lang>
Mathematica
<lang Mathematica>x=0; While[True,Print[BaseForm[x,8];x++]</lang>
MATLAB / Octave
<lang Matlab> n = 0;
while (1) dec2base(n,8) n = n+1; end; </lang>
Or use printf: <lang Matlab> n = 0;
while (1) printf('%o\n',n); n = n+1; end; </lang>
If a predefined sequence should be displayed, one can use <lang Matlab> seq = 1:100;
dec2base(seq,8)</lang>
or <lang Matlab> printf('%o\n',seq);</lang>
Mercury
<lang>
- - module count_in_octal.
- - interface.
- - import_module io.
- - pred main(io::di, io::uo) is det.
- - implementation.
- - import_module int, list, string.
main(!IO) :-
count_in_octal(0, !IO).
- - pred count_in_octal(int::in, io::di, io::uo) is det.
count_in_octal(N, !IO) :-
io.format("%o\n", [i(N)], !IO), count_in_octal(N + 1, !IO).
</lang>
min
min has no support for octal or base conversion (it is a minimalistic language, after all) so we need to do that ourselves. <lang min>(
(dup 0 ==) (pop () 0 shorten) (((8 mod) (8 div)) cleave) 'cons linrec reverse 'print! foreach newline
) :octal
0 (dup octal succ) 9.223e18 int times ; close to max int value</lang>
МК-61/52
<lang>ИП0 П1 1 0 / [x] П1 Вx {x} 1 0 * 7 - x=0 21 ИП1 x#0 28 БП 02 ИП0 1 + П0 С/П БП 00 ИП0 lg [x] 1 + 10^x П0 С/П БП 00</lang>
Modula-2
<lang modula2>MODULE octal;
IMPORT InOut;
VAR num : CARDINAL;
BEGIN
num := 0; REPEAT InOut.WriteOct (num, 12); InOut.WriteLn; INC (num) UNTIL num = 0
END octal.</lang>
Nanoquery
Even though all integers are arbitrary precision, the maximum value that can be represented as octal using the format function is 2^64 - 1. Once this value is reached, the program terminates. <lang nanoquery>i = 0 while i < 2^64
println format("%o", i) i += 1
end</lang>
NetRexx
<lang NetRexx>/* NetRexx */ options replace format comments java crossref symbols binary
import java.math.BigInteger
-- allow an option to change the output radix. parse arg radix . if radix.length() == 0 then radix = 8 k_ = BigInteger k_ = BigInteger.ZERO
loop forever
say k_.toString(int radix) k_ = k_.add(BigInteger.ONE) end
</lang>
NewLISP
<lang NewLISP>; file: ocount.lsp
- url
- http://rosettacode.org/wiki/Count_in_octal
- author
- oofoe 2012-01-29
- Although NewLISP itself uses a 64-bit integer representation, the
- format function relies on underlying C library's printf function,
- which can only handle a 32-bit octal number on this implementation.
(for (i 0 (pow 2 32)) (println (format "%o" i)))
(exit)</lang>
Sample output:
0 1 2 3 4 5 6 7 10 11 12 ...
Nim
<lang nim>import strutils for i in 0 ..< int.high:
echo toOct(i, 16)</lang>
Oberon-2
<lang oberon2> MODULE CountInOctal; IMPORT
NPCT:Tools, Out := NPCT:Console;
VAR
i: INTEGER;
BEGIN
FOR i := 0 TO MAX(INTEGER) DO; Out.String(Tools.IntToOct(i));Out.Ln END
END CountInOctal. </lang>
- Output:
00000000000 00000000001 00000000002 00000000003 00000000004 00000000005 00000000006 00000000007 00000000010 00000000011 00000000012 00000000013 00000000014 00000000015 00000000016 00000000017 00000000020 00000000021 ... 00000077757 00000077760 00000077761 00000077762 00000077763 00000077764 00000077765 00000077766 00000077767 00000077770 00000077771 00000077772 00000077773 00000077774 00000077775 00000077776 00000077777
OCaml
<lang ocaml>let () =
for i = 0 to max_int do Printf.printf "%o\n" i done</lang>
- Output:
0 1 2 3 4 5 6 7 10 11 12 ... 7777777775 7777777776 7777777777
PARI/GP
Both versions will count essentially forever; the universe will succumb to proton decay long before the counter rolls over even in the 32-bit version.
Manual: <lang parigp>oct(n)=n=binary(n);if(#n%3,n=concat([[0,0],[0]][#n%3],n));forstep(i=1,#n,3,print1(4*n[i]+2*n[i+1]+n[i+2]));print; n=0;while(1,oct(n);n++)</lang>
Automatic:
<lang parigp>n=0;while(1,printf("%o\n",n);n++)</lang>
Pascal
See Delphi or
old string incrementer for Turbo Pascal transformed, same as in http://rosettacode.org/wiki/Count_in_octal#Logo, about 100x times faster than Dephi-Version, with the abilty to used preformated strings leading zeroes. Added a Bit fiddling Version IntToOctString, nearly as fast. <lang pascal>program StrAdd; {$Mode Delphi} {$Optimization ON} uses
sysutils;//IntToStr
const
maxCntOct = (SizeOf(NativeUint)*8+(3-1)) DIV 3;
procedure IntToOctString(i: NativeUint;var res:Ansistring); var
p : array[0..maxCntOct] of byte; c,cnt: LongInt;
begin
cnt := maxCntOct; repeat c := i AND 7; p[cnt] := (c+Ord('0')); dec(cnt); i := i shr 3; until (i = 0); i := cnt+1; cnt := maxCntOct-cnt; //most time consuming with Ansistring //call fpc_ansistr_unique setlength(res,cnt); move(p[i],res[1],cnt);
end;
procedure IncStr(var s:String;base:NativeInt); var
le,c,dg:nativeInt;
begin
le := length(s); IF le = 0 then Begin s := '1'; EXIT; end;
repeat dg := ord(s[le])-ord('0') +1; c := ord(dg>=base); dg := dg-(base AND (-c)); s[le] := chr(dg+ord('0')); dec(le); until (c = 0) or (le<=0);
if (c = 1) then begin le := length(s); setlength(s,le+1); move(s[1],s[2],le); s[1] := '1'; end;
end;
const
MAX = 8*8*8*8*8*8*8*8*8;//8^9
var
sOct, s : AnsiString; i : nativeInt; T1,T0: TDateTime;
Begin
sOct := ; For i := 1 to 16 do Begin IncStr(sOct,8); writeln(i:10,sOct:10); end; writeln;
For i := 1 to 16 do Begin IntToOctString(i,s); writeln(i:10,s:10); end;
sOct := ; T0 := time; For i := 1 to MAX do IncStr(sOct,8); T0 := (time-T0)*86400; writeln(sOct);
T1 := time; For i := 1 to MAX do IntToOctString(i,s); T1 := (time-T1)*86400; writeln(s); writeln; writeln(MAX); writeln('IncStr ',T0:8:3); writeln('IntToOctString ',T1:8:3);
end. </lang>
- Output:
1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 10 9 11 10 12 11 13 12 14 13 15 14 16 15 17 16 20 1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 10 9 11 10 12 11 13 12 14 13 15 14 16 15 17 16 20 1000000000 1000000000 134217728 IncStr 0.944 secs IntToOctString 2.218 secs
Perl
Since task says "system register", I take it to mean "no larger than machine native integer limit": <lang perl>use POSIX; printf "%o\n", $_ for (0 .. POSIX::UINT_MAX);</lang> Otherwise: <lang perl>use bigint; my $i = 0; printf "%o\n", $i++ while 1</lang> The above count in binary or decimal and convert to octal. This actually counts in octal. It will run forever or until the universe ends, whichever comes first. <lang perl>#!/usr/bin/perl
$_ = 0; s/([^7])?(7*)$/ $1 + 1 . $2 =~ tr!7!0!r /e while print "$_\n";</lang>
Phix
<lang Phix>integer i = 0 constant ESC = #1B while not find(get_key(),{ESC,'q','Q'}) do
printf(1,"%o\n",i) i += 1
end while </lang>
PHP
<lang php><?php for ($n = 0; is_int($n); $n++) {
echo decoct($n), "\n";
} ?></lang>
PicoLisp
<lang PicoLisp>(for (N 0 T (inc N))
(prinl (oct N)) )</lang>
Pike
<lang Pike> int i=1; while(true)
write("0%o\n", i++);
</lang>
- Output:
01 02 ...
PL/I
Version 1: <lang pli>/* Do the actual counting in octal. */ count: procedure options (main);
declare v(5) fixed(1) static initial ((5)0); declare (i, k) fixed;
do k = 1 to 999; call inc; put skip edit ( (v(i) do i = 1 to 5) ) (f(1)); end;
inc: proc;
declare (carry, i) fixed binary;
carry = 1; do i = 5 to 1 by -1; v(i) = v(i) + carry; if v(i) > 7 then do; v(i) = v(i) - 8; if i = 1 then stop; carry = 1; end; else carry = 0; end;
end inc;
end count;</lang> Version 2: <lang pli>count: procedure options (main); /* 12 Jan. 2014 */
declare (i, j) fixed binary;
do i = 0 upthru 2147483647; do j = 30 to 0 by -3; put edit (iand(isrl(i, j), 7) ) (f(1)); end; put skip; end;
end count;</lang>
- Output:
(End of) Output of version 1 00000001173 00000001174 00000001175 00000001176 00000001177 00000001200 00000001201 00000001202 00000001203 00000001204 00000001205 00000001206 00000001207 00000001210 00000001211 00000001212 00000001213 00000001214 00000001215 00000001216
PowerShell
<lang PowerShell>[int64]$i = 0 While ( $True )
{ [Convert]::ToString( ++$i, 8 ) }</lang>
Prolog
Rather than just printing out a list of octal numbers, this code will generate a sequence. octal/1 can also be used to tell if a number is a valid octal number or not. octalize will keep producing and printing octal number, there is no limit.
<lang Prolog>o(O) :- member(O, [0,1,2,3,4,5,6,7]).
octal([O]) :- o(O). octal([A|B]) :- octal(O), o(T), append(O, [T], [A|B]), dif(A, 0).
octalize :- forall( octal(X), (maplist(write, X), nl) ).</lang>
PureBasic
<lang PureBasic>Procedure.s octal(n.q)
Static Dim digits(20) Protected i, j, result.s For i = 0 To 20 digits(i) = n % 8 n / 8 If n < 1 For j = i To 0 Step -1 result + Str(digits(j)) Next Break EndIf Next ProcedureReturn result
EndProcedure
Define n.q If OpenConsole()
While n >= 0 PrintN(octal(n)) n + 1 Wend Print(#CRLF$ + #CRLF$ + "Press ENTER to exit"): Input() CloseConsole()
EndIf </lang> Sample output:
0 1 2 3 4 5 6 7 10 11 12 ... 777777777777777777767 777777777777777777770 777777777777777777771 777777777777777777772 777777777777777777773 777777777777777777774 777777777777777777775 777777777777777777776 777777777777777777777
Python
<lang Python>import sys for n in xrange(sys.maxint):
print oct(n)</lang>
Racket
<lang racket>
- lang racket
(for ([i (in-naturals)])
(displayln (number->string i 8)))
</lang> (Racket has bignums, so this loop will never end.)
Raku
(formerly Perl 6) <lang perl6>say .base(8) for ^Inf;</lang>
- Output:
0
Here we arbitrarily show as many lines of output as there are lines in the program. :-)
REXX
If this REXX program wouldn't be stopped, it would count forever.
The technique used is to convert the decimal number to binary, and separate the binary digits in groups of three, and then convert those binary groups (numbers) to decimal. <lang rexx>/*REXX program counts in octal until the number exceeds the number of program statements*/
/*┌────────────────────────────────────────────────────────────────────┐ │ Count all the protons (and electrons!) in the universe. │ │ │ │ According to Sir Arthur Eddington in 1938 at his Tamer Lecture at │ │ Trinity College (Cambridge), he postulated that there are exactly │ │ │ │ 136 ∙ 2^256 │ │ │ │ protons in the universe, and the same number of electrons, which │ │ is equal to around 1.57477e+79. │ │ │ │ [Although, a modern estimate is around 10^80.] │ └────────────────────────────────────────────────────────────────────┘*/
numeric digits 100000 /*handle almost any sized big numbers. */ numIn= right('number in', 20) /*used for indentation of the output. */ w= length( sourceline() ) /*used for formatting width of numbers.*/
do #=0 to 136 * (2**256) /*Sir Eddington, here we come ! */ != x2b( d2x(#) ) _= right(!, 3 * (length(_) % 3 + 1), 0) o= do k=1 to length(_) by 3 o= o'0'substr(_, k, 3) end /*k*/
say numIn 'base ten = ' right(#,w) numIn "octal = " right( b2x(o) + 0, w + w) if #>sourceline() then leave /*stop if # of protons > pgm statements*/ end /*#*/ /*stick a fork in it, we're all done. */</lang>
- output:
number in base ten = 0 number in octal = 0 number in base ten = 1 number in octal = 1 number in base ten = 2 number in octal = 2 number in base ten = 3 number in octal = 3 number in base ten = 4 number in octal = 4 number in base ten = 5 number in octal = 5 number in base ten = 6 number in octal = 6 number in base ten = 7 number in octal = 7 number in base ten = 8 number in octal = 10 number in base ten = 9 number in octal = 11 number in base ten = 10 number in octal = 12 number in base ten = 11 number in octal = 13 number in base ten = 12 number in octal = 14 number in base ten = 13 number in octal = 15 number in base ten = 14 number in octal = 16 number in base ten = 15 number in octal = 17 number in base ten = 16 number in octal = 20 number in base ten = 17 number in octal = 21 number in base ten = 18 number in octal = 22 number in base ten = 19 number in octal = 23 number in base ten = 20 number in octal = 24 number in base ten = 21 number in octal = 25 number in base ten = 22 number in octal = 26 number in base ten = 23 number in octal = 27 number in base ten = 24 number in octal = 30 number in base ten = 25 number in octal = 31 number in base ten = 26 number in octal = 32 number in base ten = 27 number in octal = 33 number in base ten = 28 number in octal = 34 number in base ten = 29 number in octal = 35 number in base ten = 30 number in octal = 36 number in base ten = 31 number in octal = 37 number in base ten = 32 number in octal = 40 number in base ten = 33 number in octal = 41
Ring
<lang ring> size = 30 for n = 1 to size
see octal(n) + nl
next
func octal m
output = "" w = m while fabs(w) > 0 oct = w & 7 w = floor(w / 8) output = string(oct) + output end return output
</lang>
Ruby
From the documentation: "A Fixnum holds Integer values that can be represented in a native machine word (minus 1 bit). If any operation on a Fixnum exceeds this range, the value is automatically converted to a Bignum."
<lang ruby>n = 0 loop do
puts "%o" % n n += 1
end
- or
for n in 0..Float::INFINITY
puts n.to_s(8)
end
- or
0.upto(1/0.0) do |n|
printf "%o\n", n
end
- version 2.1 later
0.step do |n|
puts format("%o", n)
end</lang>
Run BASIC
<lang runbasic>input "Begin number:";b input " End number:";e
for i = b to e
print i;" ";toBase$(8,i)
next i end
function toBase$(base,base10) for i = 10 to 1 step -1
toBase$ = str$(base10 mod base) +toBase$ base10 = int(base10 / base) if base10 < 1 then exit for
next i end function</lang>
Rust
<lang rust>fn main() {
for i in 0..std::usize::MAX { println!("{:o}", i); }
}</lang>
Salmon
Salmon has built-in unlimited-precision integer arithmetic, so these examples will all continue printing octal values indefinitely, limited only by the amount of memory available (it requires O(log(n)) bits to store an integer n, so if your computer has 1 GB of memory, it will count to a number with on the order of octal digits).
<lang Salmon>iterate (i; [0...+oo])
printf("%o%\n", i);;</lang>
or
<lang Salmon>for (i; 0; true)
printf("%o%\n", i);;</lang>
or
<lang Salmon>variable i := 0; while (true)
{ printf("%o%\n", i); ++i; };</lang>
Scala
<lang scala>Stream from 0 foreach (i => println(i.toOctalString))</lang>
Scheme
<lang scheme>(do ((i 0 (+ i 1))) (#f) (display (number->string i 8)) (newline))</lang>
Scratch
Seed7
This example uses the radix operator to write a number in octal.
<lang seed7>$ include "seed7_05.s7i";
const proc: main is func
local var integer: i is 0; begin repeat writeln(i radix 8); incr(i); until FALSE; end func;</lang>
Sidef
<lang ruby>var i = 0; loop { say i++.as_oct }</lang>
Simula
<lang simula> BEGIN
PROCEDURE OUTOCT(N); INTEGER N; BEGIN PROCEDURE OCT(N); INTEGER N; BEGIN IF N > 0 THEN BEGIN OCT(N//8); OUTCHAR(CHAR(RANK('0')+MOD(N,8))); END; END OCT; IF N < 0 THEN BEGIN OUTCHAR('-'); OUTOCT(-N); END ELSE IF N = 0 THEN OUTCHAR('0') ELSE OCT(N); END OUTOCT;
INTEGER I; WHILE I < MAXINT DO BEGIN OUTINT(I,0); OUTTEXT(" => "); OUTOCT(I); OUTIMAGE; I := I+1; END;
END. </lang>
Sparkling
<lang sparkling>for (var i = 0; true; i++) {
printf("%o\n", i);
}</lang>
Standard ML
<lang sml>local
fun count n = (print (Int.fmt StringCvt.OCT n ^ "\n"); count (n+1))
in
val _ = count 0
end</lang>
Swift
<lang swift>import Foundation
func octalSuccessor(value: String) -> String {
if value.isEmpty { return "1" } else { let i = value.startIndex, j = value.endIndex.predecessor() switch (value[j]) { case "0": return value[i..<j] + "1" case "1": return value[i..<j] + "2" case "2": return value[i..<j] + "3" case "3": return value[i..<j] + "4" case "4": return value[i..<j] + "5" case "5": return value[i..<j] + "6" case "6": return value[i..<j] + "7" case "7": return octalSuccessor(value[i..<j]) + "0" default: NSException(name:"InvalidDigit", reason: "InvalidOctalDigit", userInfo: nil).raise(); return "" } }
}
var n = "0" while strtoul(n, nil, 8) < UInt.max {
println(n) n = octalSuccessor(n)
}</lang>
- Output:
The first few lines. anyway:
0 1 2 3 4 5 6 7 10 11 12 13 14 15 16 17 20 21 22 23
Tcl
<lang tcl>package require Tcl 8.5; # arbitrary precision integers; we can count until we run out of memory! while 1 {
puts [format "%llo" [incr counter]]
}</lang>
UNIX Shell
We use the bc calculator to increment our octal counter:
<lang sh>#!/bin/sh num=0 while true; do
echo $num num=`echo "obase=8;ibase=8;$num+1"|bc`
done</lang>
Using printf
Increment a decimal counter and use printf(1)
to print it in octal. Our loop stops when the counter overflows to negative.
<lang sh>#!/bin/sh num=0 while test 0 -le $num; do
printf '%o\n' $num num=`expr $num + 1`
done</lang>
Various recent shells have a bultin $(( ... ))
for arithmetic rather than running expr
, in which case
<lang sh>num=0 while test 0 -le $num; do
printf '%o\n' $num num=$((num + 1))
done</lang>
VBA
With i defined as an Integer, the loop will count to 77777 (32767 decimal). Error handling added to terminate nicely on integer overflow.
<lang VBA> Sub CountOctal() Dim i As Integer i = 0 On Error GoTo OctEnd Do
Debug.Print Oct(i) i = i + 1
Loop OctEnd: Debug.Print "Integer overflow - count terminated" End Sub </lang>
VBScript
<lang vb> For i = 0 To 20 WScript.StdOut.WriteLine Oct(i) Next </lang>
Vim Script
<lang vim>let counter = 0 while counter >= 0
echon printf("%o\n", counter) let counter += 1
endwhile</lang>
Whitespace
This program prints octal numbers until the internal representation of the current integer overflows to -1; it will never do so on some interpreters.
<lang Whitespace>
</lang>
It was generated from the following pseudo-Assembly.
<lang asm>push 0
- Increment indefinitely.
0:
push -1 ; Sentinel value so the printer knows when to stop. copy 1 call 1 push 10 ochr push 1 add jump 0
- Get the octal digits on the stack in reverse order.
1:
dup push 8 mod swap push 8 div push 0 copy 1 sub jn 1 pop
- Print them.
2:
dup jn 3 ; Stop at the sentinel. onum jump 2
3:
pop ret</lang>
XPL0
XPL0 doesn't have built-in routines to handle octal; instead it uses hex. <lang XPL0>include c:\cxpl\codes; \intrinsic code declarations
proc OctOut(N); \Output N in octal int N; int R; [R:= N&7; N:= N>>3; if N then OctOut(N); ChOut(0, R+^0); ];
int I; [I:= 0; repeat OctOut(I); CrLf(0);
I:= I+1;
until KeyHit or I=0; ]</lang>
Example output:
0 1 2 3 4 5 6 7 10 11 12 13 14 15 16 17 20 21
zig
<lang zig>const std = @import("std"); const fmt = std.fmt; const warn = std.debug.warn;
pub fn main() void {
var i: u8 = 0; var buf: [3]u8 = undefined;
while (i < 255) : (i += 1) { _ = fmt.formatIntBuf(buf[0..], i, 8, false, 0); // buffer, value, base, uppercase, width warn("{}\n", buf); }
}</lang>
zkl
<lang zkl>foreach n in ([0..]){println("%.8B".fmt(n))}</lang>
- Output:
0 1 2 3 4 5 6 7 10 11 12
ZX Spectrum Basic
<lang zxbasic>10 PRINT "DEC. OCT." 20 FOR i=0 TO 20 30 LET o$="": LET n=i 40 LET o$=STR$ FN m(n,8)+o$ 50 LET n=INT (n/8) 60 IF n>0 THEN GO TO 40 70 PRINT i;TAB 3;" = ";o$ 80 NEXT i 90 STOP 100 DEF FN m(a,b)=a-INT (a/b)*b</lang>
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