Count in octal
You are encouraged to solve this task according to the task description, using any language you may know.
The task is to 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.
- Integer sequence is a similar task without the use of octal numbers.
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#
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
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>
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.
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>
- include <iomanip>
using namespace std;
int main() {
unsigned i = 0;
do
{
cout << setbase(8) << i << endl;
++i;
} while(i != 0);
}</lang>
D
<lang d>import std.stdio;
void main() {
ubyte i;
do writefln("%o", i++);
while(i);
}</lang>
Delphi
<lang Delphi>program CountingInOctal;
{$APPTYPE CONSOLE}
uses SysUtils;
function DecToOct(aValue: Integer): string; var
lRemainder: Integer;
begin
Result := ; while aValue > 0 do begin lRemainder := aValue mod 8; Result := IntToStr(lRemainder) + Result; aValue := aValue div 8; end;
end;
var
i: Integer;
begin
for i := 1 to 20 do WriteLn(DecToOct(i));
end.</lang>
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
Forth
<lang forth>: octal
base @ 8 base ! 0 begin dup cr . 1+ dup 0= until drop base ! ;</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>
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 ...
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 [:,@:(":"0) 8&#.^:_1
(>:[disp)^:_ ]0</lang>
Java
<lang java>public class Count{
public static void main(String[] args){
for(int i = 0;i <= Integer.MAX_VALUE;i++){
System.out.println(Integer.toOctalString(i)); //optionally use "Integer.toString(i, 8)"
}
}
}</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>
Lua
<lang lua>for l=1,2147483647 do
print(string.format("%o",l))
end</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>
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>
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", $i++ while 1</lang>
Perl 6
<lang perl6>say .fmt: '%o' for 0 .. *;</lang>
PicoLisp
<lang PicoLisp>(for (N 0 T (inc N))
(prinl (oct N)) )</lang>
Python
<lang Python>import sys for n in xrange(sys.maxint):
print oct(n)</lang>
Retro
Integers in Retro are signed.
<lang Retro>octal 17777777777 [ putn cr ] iter</lang>
Seed7
<lang seed7>$ include "seed7_05.s7i";
const proc: main is func
local
var integer: i is 0;
begin
repeat
writeln(str(i, 8));
incr(i);
until FALSE;
end func;</lang>
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>
If the shell has $(( ... )) arithmetic, then we can increment a decimal counter and use printf(1) to print it in octal. Our loop stops when the counter overflows to negative.
<lang sh>num=0 while test 0 -le $num; do
printf '%o\n' $num num=$((num + 1))
done</lang>