# Loading Forever… Windows style

Make a Windows style Loading bar by the following instructions.

(notice that this is different than Loading... Forever)

Your output should start by [.... ].

Every tick, you should wait 100 ms, then move each dots by one character right. if the dot is on the tenth character, move it to the first. Notice that you should clear the screen before outputting again. The output is ordered as the following:

[....      ]
[ ....     ]
[  ....    ]
[   ....   ]
[    ....  ]
[     .... ]
[      ....]
[.      ...]
[..      ..]
[...      .]


..Then it loops forever.

# Rules

• This is , so the shortest answer wins I doubt I would even accept a winning answer tho
• Please provide a gif file of the loading bar in action if possible.
• Can we output, say, twenty newlines before each output to 'clear' the screen? – Okx May 12 '17 at 11:11
• @Okx Yes, if your language has no other way of clearing the screen. – Matthew Roh May 12 '17 at 11:13
• How much error can the delay be?(e.g. +- 0.5 seconds) I'd suggest 250 milliseconds error.... – stevefestl May 12 '17 at 12:50
• Can I suggest not including the fixed time delay on future challenges? I find it's appeared on a lot of recent challenges, and each time I write the same ungolfable boilerplate to make the system wait n milleseconds. – xnor May 12 '17 at 20:23
• Is the use of \r allowed, instead of literally clearing the screen? – phyrfox May 12 '17 at 21:06

# V, 1716 15 bytes

i[´.¶ ]<esc>ògó$X|p  <esc> is 0x1b. And the hexdump: 00000000: 695b b42e b620 5d1b f267 f324 587c 70 i[... ]..g.$X|p


### Explanation

i                       " Insert
[                       " a [
´.                      " 4 .s
¶<space>                " 6 spaces
]<esc>                  " and a ]. Then return to normal mode
ò                       " Recursively do:
gó                     "  Sleep for 100 milliseconds
$" Go to the ] X " Delete the character to the left of the ] | " Go to the [ p " And paste the deleted character right after the [ " Implicit ending ò  • How to test in Vim? – Pavel May 12 '17 at 17:58 • @Phoenix i.... <esc>qq:sl 100m<CR>$X|P@qq@q should work (<esc> is obviously the escape key and <CR> is a linefeed) (there are 6 spaces after the 4 dots) – user41805 May 12 '17 at 18:00
• Glad to see the gó function being useful. Nice answer BTW :) – DJMcMayhem May 12 '17 at 20:19

## CSS/HTML, 202190 186 + 45 = 247235 231 bytes

pre{position:relative}x{position:absolute;display:inline-block;width:10ch;height:1em;overflow:hidden}x>x{width:14ch;left:-10ch;animation:1s steps(10,end)infinite l}@keyframes l{to{left:0
<pre>[<x><x>....      ....</x></x>          ]

Edit: Saved 12 14 bytes thanks to @Luke.

• Can't you save 6 bytes by renaming the animation to something like b? – Luke May 17 '17 at 15:38
• @Luke I can't believe I forgot to do that... – Neil May 17 '17 at 15:50
• You can save 2 more bytes by dropping the ch at the end; 0 doesn't need a unit. – Luke May 17 '17 at 17:00
• How about changing <x> to <span> (and in the CSS as well: x becomes span and x>x becomes span>*)? That saves the display:inline-block;, but costs only 15 bytes. So a total of 6B are saved. – Luke May 17 '17 at 19:58
• @Luke I don't care about the display but I do want to avoid repeating the position:absolute;. – Neil May 17 '17 at 20:19

# PowerShell, 67 66 Bytes

for($s='.'*4+' '*6;$s=-join($s[,9+0..8])){cls;"[$s]";sleep -m 100}


-1 by using shortened constructor thanks to Beatcracker

replaces the string with a copy of the string where the last char is put in front of the remaining chars, clears the screen, prints it, and then sleeps for 100 ms.

saved a lot of bytes by using the for loop constructor rather than wrap the logic inside the string.

• +1 for the for loop trick and making me re-read about_Join. – beatcracker May 13 '17 at 10:36
• P.S. You can golf one more byte using $s='.'*4+' '*6. – beatcracker May 14 '17 at 17:10 • @beatcracker thanks for that - updated :) – colsw May 14 '17 at 17:29 • The script does not start by [.... ]. You can fix it without penalty: for($s='.'*4+' '*6){cls;"[$s]";$s=-join($s[,9+0..8]);sleep -m 100} – mazzy Nov 6 '18 at 12:14 # Python 3, 999385 83+2 (-u flag) bytes -12 bytes thanks to ovs -2 bytes thanks to totallyhuman import time s=4*'.'+6*' ' while 1:print(end='\r[%s]'%s);time.sleep(.1);s=s[9]+s[:9]  Try it online! • Why do you have flush=True? It works without for me – L3viathan May 12 '17 at 13:48 • @L3viathan because my (ubuntu) terminal wasn't flushing. This flushing behaviour is OS dependent =/ – Rod May 12 '17 at 13:51 • Save some bytes with print(end='\r[%s]'%s,flush=1) – ovs May 12 '17 at 13:57 • You can remove flush entirely by using the -u command line flag. Related SO question – ovs May 12 '17 at 14:06 • You can also save some bytes with s[9]+s[:9]. – totallyhuman May 13 '17 at 17:08 # Windows batch , 201 181 bytes Turns out using the old-school method actually saves bytes! @for %%p in (".... " " .... " " .... " " .... " " .... " " .... " " ...." ". ..." ".. .." "... .")do @echo [%%~p]&timeout 0 >nul&cls @%0  Note: get-screenrecorder.level - low grade get-gpu.level - horrible if get-screenrecorder.level == low grade || get-gpu.level == horrible { say("GIF may not be accurate"); }  Please note that my GIF recorder skipped a few frames, making the loading bar jumps :( • Rather than calculating the number of dots, if you just kept a variable with the dots and spaces and performed string manipulation on it you could probably get this down to 100 bytes. – Neil May 12 '17 at 15:54 • I would try work on this, thanks for your tips :)! – stevefestl May 12 '17 at 22:56 • timeout/t 0 >nul instead of ping 1.1 -n 1 -w 100>nul will be within the 100ms +/- 250ms timing requirement (should be around 25 - 100ms normally) so can save a few bytes there (ss64.com/nt/timeout.html) – Liam Daly May 15 '17 at 6:35 • Also removing the @echo off and replacing the do with do @(echo %%~p&timeout/t 0 >nul&cls) will also work and should save 11 characters (200 bytes on my computer) – Liam Daly May 15 '17 at 6:51 # Javascript (ES6), 86 bytes setInterval('with(console)clear(),log([${x=x[9]+x.slice(0,9)}])',100,x='...      .')

## Mathematica, 67 77 Bytes

+10 Bytes as I forgot the square brackets.

Animate["["<>"....      "~StringRotateRight~n<>"]",{n,1,10,1},RefreshRate->10]

• Really, Mathematica has a built-in Animate? :| – Mr. Xcoder May 13 '17 at 10:08
• Yup, it will animate just about anything over a given variable. :) – Ian Miller May 13 '17 at 13:06
• This doesn't seem to include the rectangular brackets that most other answers do. – Mark S. May 13 '17 at 14:26
• Oh rats, didn't look closely enough. Ok fixed. – Ian Miller May 14 '17 at 0:18

# C (gcc), 126125124123122121119118117114 115 bytes

This one uses a bitmask to keep track of where the dots are.

I had to add another byte as I was only outputting 5 spaces before.

m=30;f(i){usleep(3<<15);system("clear");for(i=1;i<1920;i*=2)putchar(i^1?i&m?46:32:91);m+=m&512?m+1:m;f(puts("]"));}


Try it online!

• WHY is your command prompt font Comic Sans MS?!?!?! – MD XF May 12 '17 at 23:51
• Why has no one asked why it's pretending to be a Windows command prompt? – Oskar Skog Jun 10 at 17:37

# JavaScript (ES6) + HTML, 10485 83 bytes

f=(s="....      ")=>setTimeout(f,100,s[9]+s.slice(0,9),o.value=[${s}]) <input id=o  • Saved 2 bytes thanks to Johan's suggestion that I use an input instead of a pre. ## Try It Requires a closing > on the input tag in order to function in a Snippet. (f=(s=".... ")=>setTimeout(f,100,s[9]+s.slice(0,9),o.value=[${s}]))()
<input id=o>

• Shouldn't there be 10 characters between the []s? – Neil May 12 '17 at 13:46
• You're right, @Neil; there are 6 spaces - if I'm going to count things by eye, the least I could do is wear my glasses! – Shaggy May 12 '17 at 13:48
• Can't you use an <input> instead of <pre>and then value instead of innerText? – Johan Karlsson May 12 '17 at 14:15
• Good call, @JohanKarlsson; that saves 2 bytes. – Shaggy May 12 '17 at 14:18
• Hey! This is the same byte count: s='.... ';setInterval(f=>{o.value='[${s=s[9]+s.slice(0,9)}]'},100);<input id=o, maybe someone can improve it (replace quotation mark with ) – Thomas Wagenaar May 12 '17 at 18:05 # Noodel, 161514 13 bytes [ CỤ‘Ṁ~ððÐ]ʠḷẸḍt ]ʠ[Ð.×4¤×6⁺ḷẸḍt ]ʠ⁶¤⁴.ȧ[ėÐḷẸḍt Try it:) ## How it works ]ʠ⁶¤⁴.ȧ[ėÐḷẸḍt ]ʠ⁶¤⁴.ȧ[ėÐ # Set up for the animation. ] # Pushes the literal string "]" onto the stack. ʠ # Move the top of the stack down by one such that the "]" will remain on top. ⁶¤ # Pushes the string "¤" six times onto the stack where "¤" represents a space. ⁴. # Pushes the string "." four times onto the stack. ȧ # Take everything on the stack and create an array. [ # Pushes on the string literal "[". ė # Take what is on the top of the stack and place it at the bottom (moves the "[" to the bottom). Ð # Pushes the stack to the screen which in Noodel means by reference. ḷẸḍt # The main animation loop. ḷ # Loop endlessly the following code. Ẹ # Take the last character of the array and move it to the front. ḍt # Delay for a tenth of a second. # Implicit end of loop.  ## Update [Ð]ıʠ⁶¤⁴.ḷėḍt  Try it:) Don’t know why this took me a while to think of. Anyways, this places it at 13 bytes. [Ð]ıʠ⁶¤⁴.ḷėḍt [Ð]ıʠ⁶¤⁴. # Sets up the animation. [ # Push on the character "[" Ð # Push the stack as an array (which is by reference) to the screen. ] # Push on the character "]" ı # Jump into a new stack placing the "[" on top. ʠ # Move the top of the stack down one. ⁶¤ # Push on six spaces. ⁴. # Push on four dots. ḷėḍt # The main loop that does the animation. ḷ # Loop the following code endlessly. ė # Take the top of the stack and put it at the bottom. ḍt # Delay for a tenth of a second.  <div id="noodel" code="[Ð]ıʠ⁶¤⁴.ḷėḍt" input="" cols="12" rows="2"></div> <script src="https://tkellehe.github.io/noodel/noodel-latest.js"></script> <script src="https://tkellehe.github.io/noodel/ppcg.min.js"></script> • Never heard of Noodel before, but it seems to be the right tool for the right job! +1 – Kevin Cruijssen May 12 '17 at 13:22 • @KevinCruijssen, ETHProductions has a good list with languages for code golfing:) – tkellehe May 12 '17 at 13:31 • Just when I thought I outgolfed you, I notice you have already golfed your solution twice – user41805 May 12 '17 at 15:12 • @KritixiLithos, I was scared you were going to beat me so I spent forever trying to get to 14 bytes. But, now you are close again!! Dang!! – tkellehe May 12 '17 at 15:16 • @nitro2k01 Noodel uses its own code-page with 256 characters, which are all saved as a single byte in their own encoding. Similar as some other golfing languages do, like Jelly or 05AB1E. If you would save these characters as default UTF-8 encoding, they will indeed be 2 or 3 bytes instead, but in their own encoding they are 1 byte each. – Kevin Cruijssen May 15 '17 at 9:07 # PHP, 67 bytes for($s="...      .";$s=substr($s.$s,9,10);usleep(1e5))echo"\r[$s]";


no comment

# C#, 162 157 bytes

()=>{for(string o="[....      ]";;){o=o.Insert(1,o[10]+"").Remove(11,1);System.Console.Write(o);System.Threading.Thread.Sleep(100);System.Console.Clear();}};


or as whole program for 177 bytes

namespace System{class P{static void Main(){for(string o="[....      ]";;){o=o.Insert(1,o[10]+"").Remove(11,1);Console.Write(o);Threading.Thread.Sleep(100);Console.Clear();}}}}

• +1 Something to golf: for(string o="[.... ]";;) can be golfed to var o="[.... ]";for(;;). Or you can us a port of my Java 7 answer to golf the total some more: ()=>{var o=".... "for(;;){o=(o+o).Substring(9,10);System.Console.Write("["+o+"]\n");System.Threading.Thread.Sleep(100);System.Console.Clear();}}; – Kevin Cruijssen May 12 '17 at 13:21
• Would string interpolation trim anymore off? Something like $"[{o}]\n" – Marie May 12 '17 at 13:54 • If you replace System.Console.Write(o) with System.Console.Write(o+"\r") you can remove the System.Console.Clear(); – grabthefish May 13 '17 at 20:03 # Pure bash, 68 s=${1:-....      }
printf "[$s]\r" sleep .1 exec$0 "${s: -1}${s%?}"


# MATL, 24 bytes

1&Xx'['897B@YS46*93hhDT


Try it at MATL Online! Or see a gif from the offline compiler:

### Explanation

        % Do...while
1&Xx   %   Pause for 0.1 s and clear screen
'['    %   Push this character
897B   %   Push [1 1 1 0 0 0 0 0 0 1]
@      %   Push current iteration index, 1-based
YS     %   Circularly shift the array by that amount
46*    %   Multiply by 46 (code point of '.')
93     %   Push 93 (code point of ']')
hh     %   Concatenate horizontally twice. Numbers are interpreted as chars
%   with the corresponding code points
D      %   Display
T      %   Push true. Used as loop condition. Results in an infinite loop
% End (implicit)

• Your link crashes, meaning that I cannot kill it. – Leaky Nun May 12 '17 at 11:29
• @LeakyNun What do you mean exactly that it crashes? It works for me, and I can kill it. Sometimes there are timeout issues. If it doesn't start, try refreshing the page – Luis Mendo May 12 '17 at 11:45

# Jelly, 28 27 bytes

ṙ©-j@⁾[]ṭ”ÆȮœS.1®ß
897ṃ⁾. Ç


### How?

ṙ©-j@⁾[]ṭ”ÆȮœS.1®ß - Link 1, the infinite loop: list of characters s
ṙ                  - rotate s left by:
-                -   -1 (e.g. "...      ." -> "....      ")
©                 -   copy to the register and yield
⁾[]           - literal ['[',']']
j@              - join with reversed @rguments
”Æ        - literal '\r'
ṭ          - tack (append the display text to the '\r')
Ȯ       - print with no newline ending
.1   - literal 0.1
œS     - sleep for 0.1 seconds then yield the printed text (unused)
®  - recall the value from the register
ß - call this link (1) again with the same arity

897ṃ⁾. Ç - Main link: no arguments
897      - literal 897
⁾.   - literal ['.',' ']
ṃ     - base decompression: convert 897 to base ['.',' '] = "...      ."


# C (gcc), 2021981961899699888679777574 73 bytes

Saved 7 8 bytes thanks to Digital Trauma.

f(i){usleep(dprintf(2,"\r[%-10.10s]","....      ...."+i%10)<<13);f(i+9);}


Or, if your system's stdout doesn't need to be flushed after every write without a newline:

# C (gcc), 70 bytes

f(i){usleep(printf("\r[%-10.10s]","....      ...."+i%10)<<13);f(i+9);}


### How it works

• usleep( sleeps for the next return value in microseconds.
• dprintf(2, prints to file descriptor 2, or stderr. This is necessary because while stdout is line-buffered (meaning output will not show until it prints a newline), stderr is character-buffered (all output is shown immediately).
• "\r prints a carriage return (clears the current line).
• [%-10.10s]" is the printf format specifier for a string with exact length 10 (no matter what string provided the output will always be a string with length 10), padded with spaces to the right if necessary. This will be enclosed with brackets.
• ".... ...." is the loading bar.
• +i%10 offsets the loading bar by the current index modulo 10. For example, if i == 3, i % 10 is equal to 3. Offsetting the loading bar by 3 makes it equal to ". ....".
• When the offset-ed string is passed to the printf format specifier, it limits to a length of 10 if necessary and adds spaces to the end if necessary. Therefore, the loading bar will always be between [.... ] and [. ...].
• i;f(){for(;;i=++i%10)usleep(7500*dprintf(2,"\r[%-10.10s]",".... ...."-i+10));} should work. – Christoph May 15 '17 at 11:08
• Great golfing! Save 1 more byte with f(i){usleep(dprintf(2,"\r[%-10.10s]",".... ...."+i%10)<<13);f(i+9);} – Digital Trauma May 15 '17 at 19:05
• @DigitalTrauma The spaces in your code didn't render properly. However, I see what you meant, and thanks for the help! – MD XF May 15 '17 at 19:17

# Java 7, 139 124 bytes

String s="....      ";void c()throws Exception{System.out.print("["+s+"]\r");s=(s+s).substring(9,19);Thread.sleep(100);c();}

• Mentioning of \r thanks to @Phoenix.

The carriage return \r resets the 'cursor' back to the begin of the line, which can then be overwritten. Unfortunately, online compilers nor the Eclipse IDE doesn't support this, so I've added a gif at the end of this answer to show it from Windows Command Prompt.

Try it here. (Slightly modified so you won't have to wait for the time-out before viewing the result. Also, the TIO doesn't support carriage returns, so every line is printed without overwriting the previous line.)

Explanation:

String s="....      ";            // Starting String "....      " on class level
void c()                          // Method without parameter nor return-type
throws Exception{                // throws-clause/try-catch is mandatory for Thread.sleep
System.out.print("["+s+"]\r");  //  Print the String between square-brackets,
//  and reset the 'cursor' to the start of the line
s=(s+s).substring(9,19);        //  Set s to the next String in line
Thread.sleep(100);              //  Wait 100 ms
c();                            //  Recursive call to same method
}                                 // End of method


Output gif:

• You can clear the line by replacing println with print and outputting a carriage return. Might not work in your IDE's terminal, but it would work in any other sane one. – Pavel May 12 '17 at 18:00
• @Phoenix By carriage return you mean \r\n? How does System.out.print(someString+"\r\n); clear the console.. It's the same as using System.out.println(someString);.. It simply goes to the next line, but doesn't remove any previous line printed.. :S – Kevin Cruijssen May 12 '17 at 19:31
• No, I mean \r, without \n. That resets the "cursor" to the beginning of the line so printing anything will overwrite that line. – Pavel May 12 '17 at 22:21
• @Phoenix Ah of course. Thanks. Modified my answer and added a gif to show the result. Too bad online compilers nor the Eclipse IDE aren't supporting this.. >.> – Kevin Cruijssen May 13 '17 at 10:36

# Python 2, 81 78 bytes

-1 byte (noticing I missed use of %s when Rod submitted an almost identical Python 3 version at the same time!)
-2 bytes (using totallyhuman's idea - replace s[-1]+s[:-1] with s[9]+s[:9])

import time
s='.'*4+' '*6
while s:print'\r[%s]'%s,;s=s[9]+s[:9];time.sleep(.1)


• How it's flushing the output? this is the reason why I'm using python3 on my answer (it would take more bytes to flush on python2) – Rod May 12 '17 at 12:29
• @Rod the \r overwrites the line and the , makes it print a tuple rather than a string - I saw it a while back somewhere and have used it before too. – Jonathan Allan May 12 '17 at 12:33
• Yes, this is what I was doing, but the output wasn't being printed in real time (had to use sys.stdout.flush()) – Rod May 12 '17 at 12:36
• Found the culprit : my ubuntu terminal :c – Rod May 12 '17 at 12:39

# Go, 150145132129 124 bytes

-5 bytes thanks to sudee.

I feel like I don't see enough Go here... But my answer is topping C so... pls halp golf?

package main
import(."fmt"
."time")
func main(){s:="....      ";for{Print("\r["+s+"]");Sleep(Duration(1e8));s=(s+s)[9:19];}}


Try it online!

• Not familiar with Go, but I would assume you can convert 100000000 to 10^8 to save 5 bytes. – Grant Miller May 13 '17 at 5:33
• @goatmeal I tried that but it's apparently bitwise negation. I also tried 10**8 which also gives an error. – totallyhuman May 13 '17 at 12:50
• You can use scientific notation: 1e8. – sudee May 14 '17 at 16:20
• @sudee Aha, that would be the way to use large numbers. Thanks! – totallyhuman May 14 '17 at 16:24
• @MDXF I should've phrased that differently, I meant my answer is being out-golfed by C. – totallyhuman May 15 '17 at 16:39

# VBA 32-bit, 159 157 143 141 134 Bytes

VBA does not have a built in function that allows for waiting for time periods less than one second so we must declare a function from kernel32.dll

32 Bit Declare Statement (41 Bytes)

Declare Sub Sleep Lib"kernel32"(ByVal M&)


64 Bit Declare Statement (49 Bytes)

Declare PtrSafe Sub Sleep Lib"kernel32"(ByVal M&)


Additionally, we must include a DoEvents flag to avoid the infinite loop from making Excel appear as non-responsive. The final function is then a subroutine which takes no input and outputs to the VBE immediate window.

Immediate Window function, 93 Bytes

Anonymous VBE immediate window function that takes no input and outputs to the range A1 on the ActiveSheet

s="...      ....      .":Do:DoEvents:Sleep 100:[A1]="["&Mid(s,10-i,10)&"]":i=(i+1)Mod 10:Loop


### Old Version, 109 Bytes

Immediate window function that takes no input and outputs to the VBE immediate window.

s="...      ....      .":i=0:Do:DoEvents:Sleep 100:Debug.?"["&Mid(s,10-i,10)&"]":i=(i+1) Mod 10:Loop


Ungolfted and formatted

Declare PtrSafe Sub Sleep Lib "kernel32" (ByVal M&)
Sub a()
Dim i As Integer, s As String
s = "...      ....      ."
i = 0
Do
Debug.Print [REPT(CHAR(10),99]; "["; Mid(s, 10 - i, 10); "]"
DoEvents
Sleep 100
i = (i + 1) Mod 10
Loop
End Sub


-2 Bytes for removing whitespace

-30 Bytes for counting correctly

-14 Bytes for converting to immediate window function

Output

The gif below uses the full subroutine version because I was too lazy to rerecord this with the immediate window function.

• What's that a at the top of the output? – MD XF May 15 '17 at 15:44
• @MDXF that is the call to run subroutine a as is listed above; this is functionally equivalant to the more verbose call a(). – Taylor Scott May 16 '17 at 3:27
• Ah, my bad. Just looking for bad submissions. Your's isn't, then, so +1 – MD XF May 16 '17 at 3:28

# 05AB1E, 23 bytes

'.4×ð6×J[D…[ÿ],Á¶т×,т.W


Try it online!

Explanation

'.4×ð6×J                  # push the string "....      "
[                 # forever do:
D                # duplicate
…[ÿ],           # interpolate the copy between brackets and print
Á          # rotate the remaining copy right
¶т×,      # print 100 newlines
т.W   # wait 100ms


# Batch, 99 98 bytes

Saved 1 byte thanks to SteveFest!

(I could remove \r from the code, but in the spirit of batch golfing, I won't.)

@SET s=....
:g
@CLS
@ECHO [%s%]
@SET s=%s:~-1%%s:~,-1%
@ping 0 -n 1 -w 100>nul
@GOTO g


There are four spaces after the first line.

The main logic is modifying the string. %s:~-1% is the last character of %s% and %s:~0,-1% is all but the last character of %s%. Thus, we are moving the last character to the front of the string, which rotates the string.

• Aw... I've been looking for this... – stevefestl May 12 '17 at 23:05
• Golf 1 byte: the 0 in the variable substring can be removed – stevefestl May 12 '17 at 23:12
• You use cmder. Nice job. – MD XF May 13 '17 at 3:21
• @SteveFest Huh, TIL. Thanks! – Conor O'Brien May 13 '17 at 3:45
• @MDXF It's the only reason I'm still sane :P – Conor O'Brien May 13 '17 at 3:46

s=?.*4+' '*6;loop{$><<"[%s]\r"%s=s[-1]+s.chop;sleep 0.1}  Heavily influenced by other answers here. Saved a byte thanks to @manatwork. Also apparently I have trouble counting characters -- I use ST3 and apparently it will include newlines in the count of characters in the line if you're not attentive. • How does it work? Does this assume that the input is stored in s? – Rɪᴋᴇʀ May 12 '17 at 22:47 • @Riker He defines s at the beginning of the program as 4 .s and a few spaces – Conor O'Brien May 12 '17 at 22:50 • s[0..8]s.chop – manatwork May 15 '17 at 8:53 # Perl, 69 bytes -3 bytes thanks to @Dom Hastings. $_="....".$"x6;{print"\ec[$_]
";select$a,$a,!s/(.*)(.)/$2$1/,.1;redo}


That select undef,undef,undef,.1 is the shortest way to sleep less than 1 second in Perl, and it takes a lot of bytes...

Slightly longer (79 bytes), there is:

@F=((".")x4,($")x6);{print"\ec[",@F,"]\n";@F=@F[9,0..8];select$a,$a,$a,.1;redo}

• Evening, managed to get this down a little more 69 (or 68 with a literal ESC): gist.github.com/dom111/e3ff41c8bc835b81cbf55a9827d69992 I feel like the tried to use !print but you need parens so it ends up the same length :/ – Dom Hastings May 15 '17 at 18:30
• @DomHastings Nice, thanks! You still know how to golf :D – Dada May 16 '17 at 7:44

# Bash, 9390 96 bytes

s="...      ....      ."
for((;;)){ for i in {0..9};do printf "\r[\${s:10-i:10}]";sleep .1;done;}


view here

couldn't get nested { } in for syntax

• I intended to post a quite similar solution, but is pointless now. But may give some inspiration to improve your: pastebin.com/Ld6rryNX – manatwork May 12 '17 at 14:06
• much better! i'm not stealing from you, i knew i had to work out this one... – marcosm May 12 '17 at 14:19
• edited, printf padding can't help in shortening s. wraping the string as @DigitalTrauma looks better. – marcosm May 15 '17 at 13:27

# Groovy, 72 bytes

s="*"*4+" "*6
for(;;){print("["+s+"]"+"\n"*20);s=s[9]+s[0..8];sleep 100}


Explaination

s="*"*4+" "*6 //creates the string "****      "
for(;;){ //classic infinite loop
print("["+s+"]"+"\n"*20) //prints the string with [ at the beginning and ] at the end. After that some newlines
s=s[9]+s[0..8] //appends the final char of the string to beginning, creating a cycling illusion
sleep 100 //100 ms delay
}

• Didn't know a proper way to clear the console in Groovy/Java. If someone has a way of doing it, please tell me – staticmethod May 12 '17 at 18:48
• You can use \r to return the cursor to the start of the line. It appears that at least several answers are doing this. From there, you could delete the *20, saving 3 bytes. – phyrfox May 12 '17 at 21:09

# Haskell (Windows), 159 bytes

import System.Process
import Control.Concurrent
main=mapM id[do system"cls";putStrLn('[':["....      "!!mod(i-n)10|i<-[0..9]]++"]");threadDelay(10^5)|n<-[0..]]


Explanation

mapM id             sequentially perform each IO action in the following list
[                   start a list comprehension where each element is...
do                  an IO operation where
system "cls";       we clear the screen by calling the windows builtin "cls"
putStrLn(           then display the string...
'[':                with '[' appended to
[                   a list comprehension where each character is...
"....      "!!       the character in literal string "....      " at the index
mod(i-n)10          (i - n) % 10
|i<-[0..9]]         where i goes from 0 to 9
++"]"             and that ends with ']'
);
threadDelay(10^5)   then sleep for 100,000 microseconds (100 ms)
|n<-[0..]]          where n starts at 0 and increments without bound


Haskell's purity made generating the cycling dot pattern somewhat complex. I ended up creating a nested list comprehension that generated an infinite list of strings in the order they should be output, then went back added the appropriate IO operations.

# Ruby, 61 bytes

If the spec were for the dots to scroll left instead of right, it would save 1 byte because rotate! with no arguments shifts the array once to the left.

s=[?.]*4+[' ']*6
loop{print ?[,*s,"]\r";s.rotate!9;sleep 0.1}


# GNU sed (with exec extension), 64

Score includes +1 for -r flag.

s/^/[....      ]/
:
esleep .1
s/[^. ]*(.+)(.)].*/\c[c[\2\1]/p
b


# c, 100

char *s="....      ....     ";main(i){for(i=0;;i=(i+9)%10)dprintf(2,"[%.10s]\r",s+i),usleep(3<<15);}

• Why print to stderr using dprintf and not just use printf? – MD XF May 12 '17 at 23:58
• @MDXF Because by default stderr is character buffered, whereas stdout is line buffered. Since I don't want to print any \n, then with printf() I'd have to explicitly fflush(stdout) as well as #include <stdio.h> – Digital Trauma May 13 '17 at 0:27
• Good point, but actually, you wouldn't have to #include <stdio.h> to flush STDOUT. fflush(0) flushes all buffers. – MD XF May 13 '17 at 0:28
• Save three bytes by renaming main to f`, that counts. – MD XF May 13 '17 at 3:19