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Challenges

Looping counter

+9
−0

Looping counter

Create an infinite loop that outputs lines of asterisks, with each line containing one more asterisk. Instead of the asterisk, any printable, non-whitespace character can be used. However all lines need to use the same character.

The beginning of the output looks like this:

*
**
***
****
*****
******
*******
********
*********
**********
***********
************

An ungolfed Python implementation:

from itertools import count

for i in count(1):
    print('*'*i)
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2 comment threads

Minimum maximum (1 comment)
Output line 0. Input for starting line (1 comment)

19 answers

+5
−0

Lua, 31 bytes

_=''::_::_=_..'*'print(_)goto _

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You can shorten your loop by using goto. As a bonus, this makes your code look even more cryptic (: ... (2 comments)
+3
−0

Ruby, 18 bytes

loop{$><<$/=?*+$/}

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+3
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Vyxal, 5 4 bytes

¤{‹…

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This is so stupid and I love it

Explained

¤{‹…
¤     # Push an empty string
 {    # Forever:
  ‹   #   Append a "-" to the top of the stack
   …  #   And print it without popping
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+2
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Haskell, 35 bytes

main=mapM putStrLn$iterate('*':)"*"

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55 -> 39, with orthoplex's idea.

39 -> 35 from orthoplex.

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Neat solution, but that import is quite expensive. Maybe try something like ```haskell x="*":map('*... (3 comments)
+2
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brainfuck, 31 bytes

++++++++++[[>]>-[<+>---]<[.<]>]

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+2
−0

MATL, 16 bytes

1`1yh1$lZctDwQt]

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The TIO output fetch will likely fail, so make sure to click the run button to stop the execution.

1`1yh1$lZctDwQt]
1`             ] - Push 1 onto stack, enter do-while
  1              - Push 1 onto stack
   y             - duplicate n-th level of stack, default is 2
    h            - horzcat, 1 10h -> [1;10]
     1$l         - ones, specifying arguments with 1$
        Zc       - strjoin
          t      - duplicate top of stack
           D     - display
            w    - swap
             Q   - increment by 1
              t  - duplicate

Matlab can loop infinitely using k=1:inf, but MATL cannot perform [1:Y] where Y is shorthand for inf.

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+1
−0

JavaScript (Node.js), 31 bytes

for(i='';;console.log(i+='*'));

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+1
−0

Bash, 17 bytes

echo x$1;./$0 x$1

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Thanks to celtschk‭ for spotting my mistake :)

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Wrong output (1 comment)
+1
−0

JavaScript, 25 bytes

Could be 21 but calling the function like that feels like cheating.

(f=s=>f(s+=8,print(s)))``

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22 bytes

Didn't want to post this as my main solution as it's pretty much the same as Moshi's.

for(s=``;;)print(s+=8)

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+1
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C (gcc), 47 bytes

i,n;f(){for(i=n+++2;i--;)putchar(i?42:13);f();}

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+1
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J, 13 char

'*',^:(<_)'*'

How it works: What's in parenthesis indicates to the verb ^: that the verb to the left of ^: has to be performed to the object on the right of the ^: an infinite number of times and that it has to list each intermediate result. '_' is infinity, and '<' boxes it to force listing the intermediate results.

   '*',^:(<4)'*'
*   
**  
*** 
****
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+1
−0

Embed ESCR, about 32 characters depending on how you count

loop
  append s "*"
  show s
  endloop

The indentation is not required, but shown for clarity. Declaring the variable S is also not included, since similar stuff doesn't seem to be included in other examples.

S is a string variable. In ESCR, storage for a string can grow dynamically at run time as needed. The string initialially starts empty, with one "*" appended to it each iteration.

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+1
−0

Python 2, 26 bytes

_=1
while 1:_*=10;print~-_

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3.x compatibility (1 comment)
+1
−0

Python 3, 28 bytes

_=""
while 1:_+="*";print(_)

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+1
−0

Japt, 6 bytes

ßOpP±Q

Test it

ßOpP±Q
ß          :Recursive call
 Op        :Output with trailing newline
   P       :Empty string, initially
    ±      :Append
     Q     :Quotation mark
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+1
−0

Fig, 5 bytes

The actual score is $5\log_{256}(96)\approx$ 4.116 bytes. The leaderboard only likes ints in the header.

(J,Q0

-1 char thanks to Seggan

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Fractional byte lang created by Seggan. Is this even allowed here?

(J,Q0
  ,Q    : Print and return last return value, initialized as 0
 J  0   : Add 0 to end of result
(       : Loop forever
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The leaderboard thinks this is 116 bytes lol. You might want to change it to 4 or 5 (2 comments)
+1
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><>, 10 bytes

1:naoa*1+!

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+1
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dc, 13 bytes

[r1+pA*rdx]dx

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+1
−0

C, 30 + 1-4 byte

f(m){f(printf("%0*i\n",m,0));}

Has this limitations:

  • counter stops working at INT_MAX or at stack overflow.
  • Needs a input value to indicate on which line we start. 0 starts on the line with one character.
  • It uses 0 and not * as output character

Because of the additional input number, i added the + 1 in the title. If you call the program, you need to write f(0), this is why the +4 in the title.

Fully working program, 36 byte:

main(m){main(printf("%0*i\n",m,0));}
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