Chargement pour toujours… style Windows

36

Créez une barre de chargement de style Windows en suivant les instructions suivantes.

(remarquez que cela diffère de Loading ... Forever )

Votre sortie devrait commencer par [.... ].

Chaque tick, vous devriez attendre 100 ms, puis déplacez chaque point d'un caractère à droite. si le point est sur le dixième caractère, déplacez-le vers le premier. Notez que vous devez effacer l’écran avant de relancer la sortie. La sortie est ordonnée comme suit:

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

..Alors il boucle pour toujours.

Règles

  • C’est du , donc la réponse la plus courte gagne je doute d’accepter une réponse gagnante
  • S'il vous plaît fournir un fichier gif de la barre de chargement en action si possible.
Matthew Roh
la source
1
Pouvons-nous générer, disons, vingt nouvelles lignes avant chaque sortie pour «effacer» l'écran?
Okx
2
@Okx Oui, si votre langue n'a pas d'autre moyen d'effacer l'écran.
Matthew Roh
Combien d’erreurs le délai peut-il être? (Par exemple + - 0.5 secondes) Je suggérerais une erreur de 250 millisecondes ....
stevefestl
1
Puis-je suggérer de ne pas inclure le délai fixe pour les défis futurs? Je trouve que cela est apparu dans beaucoup de défis récents, et chaque fois que j’écris le même passe-partout non golfable pour faire attendre le système n milles secondes.
xnor
2
L'utilisation de l' \roption est-elle autorisée au lieu d'effacer littéralement l'écran?
phyrfox

Réponses:

19

V, 17 16 15 octets

i[´.¶ ]<esc>ògó$X|p

<esc>est 0x1b.

Et l'hexdump:

00000000: 695b b42e b620 5d1b f267 f324 587c 70    i[... ]..g.$X|p

Explication

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

gif

Kritixi Lithos
la source
Comment tester dans Vim?
Pavel
@Phoenix i.... <esc>qq:sl 100m<CR>$X|P@qq@qdevrait fonctionner ( <esc>est évidemment la touche d'échappement et <CR>un saut de ligne) (il y a 6 espaces après les 4 points)
Kritixi Lithos
3
Content de voir que la fonction est utile.
Bonne
19

CSS / HTML, 202 190 186 + 45 = 247 235 231 octets

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: Sauvegardé 12 14 octets grâce à @Luke.

Neil
la source
Ne pouvez-vous pas sauvegarder 6 octets en renommant l'animation en quelque chose comme b?
Luc
@Luke Je n'arrive pas à croire que j'ai oublié de le faire ...
Neil
Vous pouvez sauvegarder 2 autres octets en laissant tomber le chà la fin; 0n'a pas besoin d'une unité.
Luc
2
Que diriez - vous de changer <x>à <span>(et dans le CSS ainsi: xdevient spanet x>xdevient span>*)? Cela économise le display:inline-block;, mais ne coûte que 15 octets. Donc, un total de 6B sont sauvegardés.
Luc
1
@Luke Je me fiche de l 'affichage mais je veux éviter de répéter position:absolute;.
Neil
12

PowerShell, 67 à 66 octets

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

-1 en utilisant un constructeur raccourci grâce à Beatcracker

remplace la chaîne par une copie de la chaîne où le dernier caractère est placé devant les caractères restants, efface l'écran, l'imprime, puis se met en veille pendant 100 ms.

économisé de nombreux octets en utilisant le constructeur de la boucle for plutôt que d’envelopper la logique dans la chaîne.

enter image description here

Colsw
la source
1
+1 pour le fortour de boucle et me faire relire à propos de_Join .
beatcracker
1
PS Vous pouvez jouer au golf un octet de plus $s='.'*4+' '*6.
beatcracker
@beatcracker merci pour cela - mis à jour :)
colsw
Le script ne commence pas par [.... ]. Vous pouvez le réparer sans pénalité:for($s='.'*4+' '*6){cls;"[$s]";$s=-join($s[,9+0..8]);sleep -m 100}
mazzy
10

Python 3 , 99 93 85 83 + 2 ( drapeau -u ) octets

-12 octets grâce à ovs
-2 octets grâce à totalementhumain

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

Essayez-le en ligne!

Barre
la source
Pourquoi avez - vous flush=True? Cela fonctionne sans pour moi
L3viathan
3
@ L3viathan parce que mon terminal (ubuntu) ne vidait pas. Ce comportement de rinçage dépend du système d'exploitation = /
Rod
1
Économisez quelques octets avecprint(end='\r[%s]'%s,flush=1)
ovs
2
Vous pouvez supprimer entièrement le vidage en utilisant l' -uindicateur de ligne de commande. SO question
connexe
1
Vous pouvez également enregistrer des octets avec s[9]+s[:9].
totalement humain
10

Lot Windows, 201 181 octets

Il s'avère que l'utilisation de la méthode old-school économise réellement des octets!

@for %%p in ("....      " " ....     " "  ....    " "   ....   " "    ....  " "     .... " "      ...." ".      ..." "..      .." "...      .")do @echo [%%~p]&timeout 0 >nul&cls
@%0

Remarque:

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");
}

GIF!

Veuillez noter que mon enregistreur GIF a sauté quelques images, faisant sauter la barre de chargement :(

stevefestl
la source
1
Plutôt que de calculer le nombre de points, si vous gardiez simplement une variable avec les points et les espaces et que vous effectuiez une manipulation de chaîne dessus, vous pourriez probablement la réduire à 100 octets.
Neil
Je voudrais essayer de travailler sur cela, merci pour vos conseils :)!
stevefestl
timeout / t 0> nul au lieu de ping 1.1 -n 1 -w 100> nul sera dans les 100 ms +/- 250 ms (devrait être environ 25 - 100 ms normalement) afin de pouvoir économiser quelques octets ( ss64.com/ nt / timeout.html )
Liam Daly
1
Le fait de supprimer @echo offet de remplacer le do @(echo %%~p&timeout/t 0 >nul&cls)travail par fonctionne également et devrait enregistrer 11 caractères (200 octets sur mon ordinateur)
Liam Daly
8

Mathematica, 67 77 octets

+10 octets car j'ai oublié les crochets.

Animate["["<>"....      "~StringRotateRight~n<>"]",{n,1,10,1},RefreshRate->10]
Ian Miller
la source
1
Really, Mathematica has a built-in Animate? :|
Mr. Xcoder
Yup, it will animate just about anything over a given variable. :)
Ian Miller
This doesn't seem to include the rectangular brackets that most other answers do.
Mark S.
Oh rats, didn't look closely enough. Ok fixed.
Ian Miller
8

C (gcc), 126 125 124 123 122 121 119 118 117 114 115 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!

enter image description here

cleblanc
la source
48
WHY is your command prompt font Comic Sans MS?!?!?!
MD XF
6

JavaScript (ES6) + HTML, 104 85 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>

Shaggy
la source
1
Shouldn't there be 10 characters between the []s?
Neil
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
1
Can't you use an <input> instead of <pre>and then value instead of innerText?
Johan Karlsson
Good call, @JohanKarlsson; that saves 2 bytes.
Shaggy
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 W
5

Noodel, 16 15 14 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>

tkellehe
la source
2
Never heard of Noodel before, but it seems to be the right tool for the right job! +1
Kevin Cruijssen
1
@KevinCruijssen, ETHProductions has a good list with languages for code golfing:)
tkellehe
6
Just when I thought I outgolfed you, I notice you have already golfed your solution twice
Kritixi Lithos
@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
1
@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
4

PHP, 67 bytes

for($s="...      .";$s=substr($s.$s,9,10);usleep(1e5))echo"\r[$s]";

no comment

Titus
la source
4

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();}}}}
LiefdeWen
la source
+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
Would string interpolation trim anymore off? Something like $"[{o}]\n"
Marie
1
If you replace System.Console.Write(o) with System.Console.Write(o+"\r") you can remove the System.Console.Clear();
grabthefish
4

Pure bash, 68

s=${1:-....      }
printf "[$s]\r"
sleep .1
exec $0 "${s: -1}${s%?}"
Digital Trauma
la source
4

MATL, 24 bytes

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

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

enter image description here

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)
Luis Mendo
la source
Your link crashes, meaning that I cannot kill it.
Leaky Nun
1
@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
4

Jelly, 28 27 bytes

ṙ©-j@⁾[]ṭ”ÆȮœS.1®ß
897ṃ⁾. Ç

Example run

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
         Ӯ        - 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 ['.',' '] = "...      ."
Jonathan Allan
la source
4

C (gcc), 202 198 196 189 96 99 88 86 79 77 75 74 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 [. ...].
MD XF
la source
i;f(){for(;;i=++i%10)usleep(7500*dprintf(2,"\r[%-10.10s]",".... ...."-i+10));} should work.
Christoph
1
Great golfing! Save 1 more byte with f(i){usleep(dprintf(2,"\r[%-10.10s]",".... ...."+i%10)<<13);f(i+9);}
Digital Trauma
@DigitalTrauma The spaces in your code didn't render properly. However, I see what you meant, and thanks for the help!
MD XF
3

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:

enter image description here

Kevin Cruijssen
la source
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
@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
4
No, I mean \r, without \n. That resets the "cursor" to the beginning of the line so printing anything will overwrite that line.
Pavel
@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
3

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)

Example run

Jonathan Allan
la source
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
@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
1
Yes, this is what I was doing, but the output wasn't being printed in real time (had to use sys.stdout.flush())
Rod
1
Found the culprit : my ubuntu terminal :c
Rod
3

Go, 150 145 132 129 124 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!

totallyhuman
la source
1
Not familiar with Go, but I would assume you can convert 100000000 to 10^8 to save 5 bytes.
Grant Miller
@goatmeal I tried that but it's apparently bitwise negation. I also tried 10**8 which also gives an error.
totallyhuman
3
You can use scientific notation: 1e8.
sudee
1
@sudee Aha, that would be the way to use large numbers. Thanks!
totallyhuman
2
@MDXF I should've phrased that differently, I meant my answer is being out-golfed by C.
totallyhuman
3

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.

VBA loading Gif

Taylor Scott
la source
What's that a at the top of the output?
MD XF
@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
Ah, my bad. Just looking for bad submissions. Your's isn't, then, so +1
MD XF
2

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
Emigna
la source
2

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

Recorded with LICEcap

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.

Conor O'Brien
la source
Aw... I've been looking for this...
stevefestl
1
Golf 1 byte: the 0 in the variable substring can be removed
stevefestl
You use cmder. Nice job.
MD XF
1
@SteveFest Huh, TIL. Thanks!
Conor O'Brien
1
@MDXF It's the only reason I'm still sane :P
Conor O'Brien
2

Ruby, 57 56 bytes

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.

canhascodez
la source
How does it work? Does this assume that the input is stored in s?
Rɪᴋᴇʀ
@Riker He defines s at the beginning of the program as 4 .s and a few spaces
Conor O'Brien
s[0..8]s.chop
manatwork
2

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}
Dada
la source
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
@DomHastings Nice, thanks! You still know how to golf :D
Dada
2

Bash, 93 90 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

marcosm
la source
I intended to post a quite similar solution, but is pointless now. But may give some inspiration to improve your: pastebin.com/Ld6rryNX
manatwork
much better! i'm not stealing from you, i knew i had to work out this one...
marcosm
edited, printf padding can't help in shortening s. wraping the string as @DigitalTrauma looks better.
marcosm
1

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
}
staticmethod
la source
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
1
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
1

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.


la source
1

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}
Value Ink
la source
1

GNU sed (with exec extension), 64

Score includes +1 for -r flag.

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

c, 100

char *s="....      ....     ";main(i){for(i=0;;i=(i+9)%10)dprintf(2,"[%.10s]\r",s+i),usleep(3<<15);}
Digital Trauma
la source
Why print to stderr using dprintf and not just use printf?
MD XF
@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
Good point, but actually, you wouldn't have to #include <stdio.h> to flush STDOUT. fflush(0) flushes all buffers.
MD XF
1
Save three bytes by renaming main to f, that counts.
MD XF