Loading Forever... Windows style

Question

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 code-golf, 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.

Answer 1

V, 17 16 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

Answer 2

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

Answer 3

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.

Answer 4

Python 3, 99 93 85 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!

Answer 5

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 :(

Answer 6

Mathematica, 67 77 Bytes

+10 Bytes as I forgot the square brackets.

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

Answer 7

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!

Answer 8

Javascript (ES6), 86 bytes

setInterval('with(console)clear(),log(`[${x=x[9]+x.slice(0,9)}]`)',100,x='...      .')

Answer 9

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>

Answer 10

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>

Answer 11

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)

Answer 12

PHP, 67 bytes

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

no comment

Answer 13

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();}}}}

Answer 14

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
         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 ['.',' '] = "...      ."

Answer 15

Pure bash, 68

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

Answer 16

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 [. ...].

Answer 17

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)

Answer 18

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:

Answer 19

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!

Answer 20

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.

Answer 21

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}

Answer 22

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

Answer 23

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

Answer 24

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.

Answer 25

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.

Answer 26

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
}

Answer 27

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.

Answer 28

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}

Answer 29

GNU sed (with exec extension), 64

Score includes +1 for -r flag.

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

Answer 30

c, 100

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

Answer 31

S.I.L.O.S, 282 bytes

canvas # #
def # 1000 ! Obj N new M move > 10
newObj 0 # #
pen 255 255 255
N! 0 > >0
M! 1 > >
N! 0 > >0
M! 2 560 >
x=8
lblx
x-1
N! 0 > >
if x x
M! 3 20 >
M! 4 20 >0
M! 5 550 >
M! 6 550 >0
lbla
x=4
lbly
A=6+x
B=60+50*(((x-1)+a)%>)
M! A B 50 
x-1
if x y
refresh
a+1
a%>
wait >0
GOTO a

TIO currently does not support graphical output, but the official interpreter does. I think this is the first time I have gotten a chance to test out the graphical output functionality of SILOS.

Answer 32

shortC, 85 72 71 66 65 64 bytes

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

Explanation:

f(i){                                                           function that takes integer
     U                                                          sleep
      dprintf(2,                                                print unbuffered
                "\r[%-10.10s]",                                 string with exactly 10 characters, wrapped with spaces
                               "....    ...."                   a string that, when printed with exactly 10 characters, will always have output that is in the loading screen
                                             +i%10              scroll through the string
                                                  )<<13);       sleep 10<<13 microseconds
                                                         f(i+9  recursion :D

Answer 33

Japt, 31 22 bytes

Saved 9 bytes thanks to @obakaron

Li@O¬OpT=T?Té :'.²²+6î

Try it online!

Answer 34

TI-Basic (TI-84 Plus CE), 58 bytes

"....      (6 spaces)
Ans+Ans
While 1
For(X,11,2,-1
ClrHome
Disp "["+sub(Ans,X,10)+"]
Wait .1
End
End

Answer 35

APL (Dyalog), 44 bytes

{∇r⊣⎕DL.1⊣⎕←⎕TC[20⍴2],1⌽'][',r←¯1⌽⍵}10↑4⍴'.'

Try it online with a slightly modified version: ⎕← replaced by ∆_ which outputs with timestamps.

4⍴'.' four periods

10↑ take the first ten characters of that (pads with spaces)

{…} apply the following function

 ¯1⌽ rotate one character from the tail to the head

 r← assign that to r

 '][', prepend the brackets

 1⌽ rotate one character from the head to the tail

 ⎕TC[…] index the list of Terminal Control characters with

  20⍴2 twenty repetitions of the number two (the second element is the newline)

 ⎕← output that

.1⊣ discard that in favour of the number 0.1

 ⎕DL Delay that many seconds (returns elapsed time)

 r⊣ discard that in favour of r (which is then returned by the function)

 ∇ recurse (tail calls do not work up a stack, and can thus be repeated forever)

Answer 36

Haskell, 132 bytes

import Control.Concurrent
main=mapM_(\x->putStrLn("\27[H\27[J["++x++"]")>>threadDelay(10^5))$iterate(\x->last x:init x)"....      "

First, we define a function r to rotate a string to the right. We begin with ".... " and construct an infinite list of its right-rotations, then for each x of these, print:

• An ANSI escape sequence to clear the screen,
• A left-bracket,
• x itself,
• A right-bracket,
• And a newline character so the terminal will flush.

Then threadDelay waits 100000 microseconds before continuing onto the next element.

Since we are applying these operations to an infinite list, it will continue forever.

Answer 37

VBA (Excel), 109 bytes

Using Immediate Window and Cell A1 as output

a="...      .":Do:b=Right(a,1):a=b &Left(a,9):[A1]="["& a &"]":c=Timer+.1:Do While Timer<c:DoEvents:Loop:Loop

Answer 38

CJam, 39 bytes

'.4*S6*+{_a`'"-oDcoes{es1$m100<}g;1m>}h

Can't make a GIF right now...

How it works

'.4*S6*+       e# Push a string containing 4 dots and 6 spaces
{              e# Do:
 _             e#  Copy the string
 a`            e#  Wrap it in an array and take its string representation: ["....      "]
 '"-           e#  Remove quote characters from it: [....      ]
 o             e#  Print it
 Dco           e#  Print a carriage return (ASCII 13), moving the cursor back to the start
 es            e#  Push the current time in milliseconds
 {es1$m100<}g  e#  Loop until the current time is at least 100 more than that time
 ;             e#  Delete the old time
 1m>           e#  Rotate the string 1 space to the right
}h             e# Repeat while the TOS is truthy. The string being printed is non-empty, so
               e#   it is always truthy, and will repeat forever.

Answer 39

Pyth, 28 bytes

J+*4\.*6d#.d.1p"\r"pj=.>J1`Y

Answer 40

Perl 6, 61 bytes

$_='....      ';loop {"^M[$_]".print;s/(.*)(.)/$1$0/;sleep .1}

(That ^M is a single carriage return character.)

I really tried to base the output on a lazy infinite list, but couldn't nearly approach the terseness of this solution.

Answer 41

C, 74 73 bytes

i;f(){usleep(printf("\r[%-10.10s]","....      ...."+i%10)*7692);f(i+=9);}

Modified and improved version of https://codegolf.stackexchange.com/a/120321.

Displays correct output on MinGW.

Answer 42

C, 98 93 Bytes

main(){char b[]="x...      .";for(;;)*b=b[10],usleep(12<<printf("\r[%s]",memcpy(b+1,b,10)));}

View Video

Answer 43

Octave, 71 bytes

s="...       ."';do;printf('[%s]\r',s=circshift(s,1));pause(.1);until 0

For once my Octave answer is not MATLAB compatible. Specifically this solution uses

• Both " and ' for quotes, to transpose a string
  • MATLAB: ('asd')'
  • Octave: "asd"'
• Octave specific do ... until 0 structure, one char shorter than while 1 ... end
• Octave's printf,one char shorter than the fprintf in MATLAB

Answer 44

C, 91 bytes

main(i){for(char*s="....      ....    ";;i+=9)usleep(dprintf(2,"[%.10s]\r",s+i%10-1)<<13);}

If it doesn't have to begin at the left: (89 bytes)

main(i){for(char*s="....      ....    ";;i+=9)usleep(dprintf(2,"[%.10s]\r",s+i%10)<<13);}

Answer 45

R, 100 bytes

z="....      "
repeat{flush.console()
cat("\r[",z,"]",sep="")
z=sub("(.)(.+)(.)","\\3\\1\\2",z)
Sys.sleep(.1)}

Answer 46

F# (Mono), 122 114 118 110 bytes

let rec t s=printf"[%s]\r"s;System.Threading.Thread.Sleep(100);t((string s.[8])+s.Substring(0,8))
t"      ..."

Try it online!

Answer 47

JS (ES6), 179 bytes

x=`[....      ]
[ ....     ]
[  ....    ]
[   ....   ]
[    ....  ]
[     .... ]
[      ....]
[.      ...]
[..      ..]
[...      .]`.split`
`;c=0;setTimeout(y=>alert(x[++c]),100)

Answer 48

Powershell, 64 bytes

for(){cls;"[$('...      .'*2|% s*g(9-$i++%10)10)]";sleep -m 100}

Where $string|% s*g $From $Length is a shortcut to summon $string.SubString($From,$Length)

This solution has one small disadvantage:

• there will be an overflow exception after the Windows style Loading bar has worked for 988 million years.

Powershell, 65 bytes

for(){9..0|%{cls;"[$('...      .'*2|% s*g $_ 10)]";sleep -m 100}}

This Windows style Loading bar is an Immortal Eldar from Valinor.