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Challenges

Looping counter

+8
−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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18 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
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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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+3
−0

Ruby, 18 bytes

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

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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)
+1
−0

Python 3, 28 bytes

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

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+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
−0

C (gcc), 47 bytes

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

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+1
−0

Python 2, 26 bytes

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

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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

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

><>, 10 bytes

1:naoa*1+!

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+1
−0

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

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

brainfuck, 31 bytes

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

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+1
−0

dc, 13 bytes

[r1+pA*rdx]dx

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+1
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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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+0
−0

Fig, 4.116 bytes

$5\log_{256}(96)\approx$ 4.116 bytes.

(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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