Lolololololololololololol

Question

Let us take a break from the brain-wrecking questions and answer some of the simpler ones

You have recently read something extremely funny, and want to express your laughter to the world! But how can you?

Task

You have to display the string: Lolololololololololololololololololololololololololololololololololololololololololololololololololololololololol...

...to STDOUT.

The string should be of infinite length, or will constantly be printed until the end of time.

It is just as simple!

But remember, this is code-golf, so the source code must be as short as possible!

Note: Some languages may throw errors since excecution can be timed out, or for other reasons. That is okay! It can be weird when you laugh forever!

Good luck!

By the way, standard site rules will allow functions which return the result (e.g. in an infinite generator) making the print to STDOUT part redundant unless you specify that this must be a full program (which I'd advise against, but is allowed). — Jonathan Allan, Commented Sep 12, 2020 at 13:21
I don't understand why this was closed. The other challenge is much simpler. — the default., Commented Sep 12, 2020 at 19:28
@thedefault. I don't understand your comment. The difference (requires an extra character to be outputted before the program, and outputs two characters instead of one) is fairly trivial. As you can see, if we remove the print L part and change ol to A, we get an SVL answer! — TwilightSparkle, Commented Sep 13, 2020 at 11:56
I have reopened this question. I believe the small difference presents interesting golfing opportunities in some languages and, furthermore, that simply porting a solution might well not be competitive. One only needs to look at a few of the answers to see some interesting differences. If people disagree then by all means vote to close again. — Jonathan Allan, Commented Sep 24, 2020 at 8:27

xnor · Accepted Answer · 2020-09-12 13:56:55Z

28

Python 3, 32 bytes

x='L'
while[print(end=x)]:x='ol'

Try it online!

In Python 3, the print function by default has end='\n' to put a newline after what you print. Rather than changing that to the empty string, we stick the value x that we want to be printed there, and don't provide any value to be printed.

We stick the printing in the while loop condition. Since print returns None by default but this is Falsey and won't continue the loop, we wrap it in a singleton list to make it Truthy.

I had tried to stick an infinite iterator into print like print(*iter(...),sep=''), but it looks like Python will consume the whole iterable first and never actually print.

edited Sep 12, 2020 at 13:56

answered Sep 12, 2020 at 13:42

xnor

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1

\$\begingroup\$ I wonder why I didn't think of this when I put the printing in the condition in my Jelly answer :/ \$\endgroup\$
– Jonathan Allan
Commented Sep 12, 2020 at 14:11
1

\$\begingroup\$ Ruby, 23 bytes - x='L';x='ol'while $><<x (Can't post because it's marked as duplicate. Nice trick!) \$\endgroup\$
– DrQuarius
Commented Sep 13, 2020 at 4:39

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ovs · Accepted Answer · 2020-09-12 13:49:31Z

24

Befunge-98 (PyFunge), 8 bytes

"loL",<,

Try it online!

How?

Lo:
Initially the IP is going east.
"loL" pushes l, o and L to the stack.
, prints L, < turns the IP west and , prints o.

lo forever:
The IP is now moving west.
"loL" pushes L, o and l to the stack.
The IP wraps around and ,<, prints l and o.
L is left on the stack, but since we usually assume infinite resources, this is fine.

edited Sep 12, 2020 at 13:49

answered Sep 12, 2020 at 13:40

ovs

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Bubbler · Accepted Answer · 2021-07-21 08:25:44Z

Hexagony, 5 bytes

L;o>l

Try it online!

Hexagony golfing language confirmed

For some reason I was looking at my own answer that prints "six" in 6 bytes and randomly thought "what if I remove @?", and exactly got this answer. 4 bytes is impossible because Lol; is already 4 bytes and it is impossible to alternate two chars and print both in a single loop without redirection.

Since there is no "halt" command in this program, the program flow looks like this: (It is recommended to read the docs on how > redirects the PC)

 A B
C > D
 E F

    [.....................................]  <= looping region
A B C > C B A D > A D B F > F C E A > D E F C > C ...
L ; o   o ; L l   L l ; .   . o . L   l . . o   o
^ ^     ^ ^         ^ ^

After the initial L;, o; and l; appear in the big loop in that order, therefore printing Lololol....

640KB · Accepted Answer · 2020-09-25 14:11:36Z

18

x86-16 machine code, IBM PC DOS, 13 10 bytes

00000000: b04c cd29 0c20 3403 ebf8                 .L.). 4...

Listing:

B0 4C       MOV  AL, 'L'        ; start off with capital L
        PRINT:
CD 29       INT  29H            ; write to console
0C 20       OR   AL, 20H        ; lowercase it
34 03       XOR  AL, 3          ; swap between 'l' (0x6c) and 'o' (0x6f)
EB F8       JMP  PRINT          ; loop forever

Try it online!

A standalone PC DOS executable COM program. Output to console.

-3 bytes thx to @nununoisy's very clever use of XOR to swap between l and o.

Runtime:

Forever and ever...

edited Sep 25, 2020 at 14:11

answered Sep 12, 2020 at 14:29

640KB

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\$\begingroup\$ 10 bytes: B0 4C mov al, 'L' PRINT: CD 29 int 29h 0C 20 or al,0x20 34 03 xor al,3 EB F8 jmp PRINT \$\endgroup\$
– nununoisy
Commented Sep 23, 2020 at 21:13
2

\$\begingroup\$ @nununoisy oh that's very clever! Updated with credit given! \$\endgroup\$
– 640KB
Commented Sep 23, 2020 at 21:32

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DLosc · Accepted Answer · 2020-09-25 00:25:31Z

12

Bash + coreutils, 23 bytes

(echo L&yes)|tr "
y" ol

Try it online!

Explanation

We can generate infinite output using yes: without arguments, it outputs an infinite stream of y separated by newlines. echo L & yes outputs an L first, so our output stream looks like

L
y
y
y
y

To turn this into the output we want, we just need to change newline to o and y to l. tr "\ny" ol does this transliteration, and we can save a further byte by using an actual newline in place of \n.

answered Sep 25, 2020 at 0:25

DLosc

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2

\$\begingroup\$ Although '\ny' is longer, you can use \\ny for the same byte count. \$\endgroup\$
– Neil
Commented Oct 18, 2020 at 18:08

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RezNesX · Accepted Answer · 2020-09-12 17:03:57Z

11

brainfuck, 41 40 bytes

-1 byte thanks @ovs

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

Try it online!

edited Sep 12, 2020 at 17:03

answered Sep 12, 2020 at 14:20

RezNesX

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1

\$\begingroup\$ You don't the initial >, at least for the interpreter you linked, since it uses a wrapping tape. \$\endgroup\$
– ovs
Commented Sep 12, 2020 at 16:56
2

\$\begingroup\$ Thanks i didn't know that. \$\endgroup\$
– RezNesX
Commented Sep 12, 2020 at 17:00
4

\$\begingroup\$ I found a shorter version using KSab's bfbrute: Try it online! \$\endgroup\$
– ovs
Commented Sep 12, 2020 at 18:36
\$\begingroup\$ @ovs Do you have any tips on how to use bfbrute. \$\endgroup\$
– RezNesX
Commented Sep 12, 2020 at 18:56
2

\$\begingroup\$ @JoKing the bfbrute result I commented is 30 bytes long. \$\endgroup\$
– ovs
Commented Sep 24, 2020 at 15:54

| Show 2 more comments

Bubbler · Accepted Answer · 2020-09-27 23:02:14Z

Labyrinth, 7 bytes

762
8.3

Try it online!

Adds 623 % 256 == 111 to the previous answers below.

76   Push 76 and turn right (three-way junction)
.    Pop and print % 256 as char (L)
     Now the top is 0, so it should go straight, but instead it reflects to North
623  Turn right at 6 and push 623, going around the corners
.    Pop and print % 256 as char (o); go straight (three-way junction)
876  Push 876
     Loop forever, printing "lo"

10 bytes

76
8.1
 11

Try it online!

Same idea, but using 876 % 256 == 108. Turns out that going from an uppercase to lowercase is just a matter of prepending a 8 because 800 % 256 == 32. At the center junction, the top is always 0 right after pop and print, so the IP goes straight (first from north to south, and second from east to west). All the numbers are corners which turn the IP 90 degrees, so the overall path is infinity-shaped 76.111.876.111. .... This form is one byte shorter than the naive square loop:

11 bytes

76.
8 1
.11

Try it online!

14 bytes

Try it online!

How it works

First, the flow: the execution starts at the first valid command, which is 7 on top. The only junction is 6, and since it causes the top of stack to be positive, the IP always turns right. It turns 180 degrees if it hits a dead end. So the sequence of commands executed in order is:

76.111.13676.111.13676.111. ...(runs indefinitely)

The stack has implicit zeros, and each of 0-9 adds that digit to the end of the top of the stack (in other words, n changes the top number x to 10x + n). So the program is supposed to print the characters with charcode 76, 111, 13676, 111, 13676, ...

But 13676 is not l! Actually, Labyrinth's character output is done modulo 256.

How did I find such a number? With the path designed like this

 7
.6?
1 ?
11.

The problem is to find a number ??676 that is same as l (108) modulo 256. Note that, the equation ??xxx == yyy modulo 256 (x and y are givens and ?s are unknown) is solvable if xxx == yyy modulo 8, and if so, it always has a solution within two or fewer digits, in particular 0 <= ?? < 32. Since 676 % 8 == 108 % 8 == 4, this is solvable, and the solution here is 13.

flawr · Accepted Answer · 2020-09-24 08:41:57Z

9

Haskell, 13 bytes

'L':cycle"ol"

Try it online!

answered Sep 24, 2020 at 8:41

flawr

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0xff · Accepted Answer · 2021-08-19 16:05:27Z

9

Java (JDK), 51 48 47 bytes

a->{for(var s="L";;s="ol")System.out.print(s);}

Try it online!

thanks to user and Jo King helping me save 4 and 1 byte(s) respectively!

edited Aug 19, 2021 at 16:05

answered Aug 19, 2021 at 13:27

0xff

8296 silver badges12 bronze badges

\$\begingroup\$ Does this count? \$\endgroup\$
– enzo
Commented Nov 29, 2023 at 16:05

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Sam · Accepted Answer · 2020-09-24 11:55:16Z

8

05AB1E, 8 6 bytes

ovs' far superior 6-byter:

„Lo[?l

Try it online!

Explanation:

„Lo     Push 2-char string onto stack ('lo')
   [    Begin infinite loop
    ?   Output with no newline
     l  Push lowercase of top of stack ('Lo' -> 'lo')
        (Implicitly close infinite loop)

05AB1E, 8 bytes

My pitiful 8-byter:

'L?„ol[?

Try it online!

~~It might still be golfable, perhaps if there's a way to compress 'Lol' even further.~~

Explanation:

'L        Push 'L' onto stack
  ?       Print without newline ('L')
   „ol    Push 2-char string onto stack ('ol')
      [   Loop Forever
       ?  Print without newline ('ol')
          (Implicitly close infinite loop)

edited Sep 24, 2020 at 11:55

answered Sep 24, 2020 at 9:37

Sam

3713 silver badges7 bronze badges

\$\begingroup\$ Ah of course, How could I have missed that. Thank you! \$\endgroup\$
– Sam
Commented Sep 24, 2020 at 11:48
\$\begingroup\$ Generally, low length strings are too much of a hassle to compress without sacrificing golfiness. \$\endgroup\$
– Razetime
Commented Sep 25, 2020 at 2:44
2

\$\begingroup\$ My idea was „olÞ'Lš, good job! \$\endgroup\$
– Makonede
Commented May 10, 2021 at 17:39

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jonatjano · Accepted Answer · 2020-09-25 07:09:02Z

8

Marbelous, 14 31 bytes

6C
6F@0
@0/\4C

Marbelous is a language based on marble machines

@n (n from 0 to Z) is a portal which teleport the marble to another portal with the same value
00-FF initiate a marble with this value
/\ create a duplicate passing marble to it's left and right
.. is a noop
marbles going out of the machine from the bottom are implicitly outputed

interpretor

-17 bytes thanks to DLosc

edited Sep 25, 2020 at 7:09

answered Sep 24, 2020 at 11:48

jonatjano

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\$\begingroup\$ I think this three-line version should work: 6C, 6F@0, @0/\4C (14 bytes). I'm not 100% sure because I tried running Marbelous by putting the source code into Try It Online's Python 2 interpreter, where it seemed oddly nondeterministic--sometimes it just output L, sometimes Loo, sometimes Loooooo infinitely, and sometimes Lololololol infinitely. But since your original code did the same things, I'm reasonably confident the golfed version works correctly. \$\endgroup\$
– DLosc
Commented Sep 25, 2020 at 0:05
\$\begingroup\$ @DLosc you're right it works, thank :D \$\endgroup\$
– jonatjano
Commented Sep 25, 2020 at 7:05

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jdt · Accepted Answer · 2021-08-05 00:32:20Z

8

C (gcc), 27 bytes

f(i){f(!printf("Lol"+!i));}

Try it online!

edited Aug 5, 2021 at 0:32

answered Aug 4, 2021 at 13:16

jdt

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Jonathan Allan · Accepted Answer · 2020-09-12 13:41:55Z

7

Python 3, 37 34 bytes

-3 thanks to xnor - noting that print may have no unnamed argument!

x='L'
while x:x=print(end=x)or'ol'

Try it online!

edited Sep 12, 2020 at 13:41

answered Sep 12, 2020 at 13:28

Jonathan Allan

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2

\$\begingroup\$ Nice assignment trick! You can do print(end=x). \$\endgroup\$
– xnor
Commented Sep 12, 2020 at 13:39

Add a comment |

JosiahRyanW · Accepted Answer · 2020-10-28 15:23:44Z

7

Poetic, 147 bytes

haha l-o-l funny!i saw a thingy,it was soo funny
o?what
i am crying
o?what
i am dying,literal CHOKING
dying r-n?goddamn
just see
ohhhhhh heh,lol ig

Try it online!

In the form of a text conversation between two people collectively laughing at something they found online.

answered Oct 28, 2020 at 15:23

JosiahRyanW

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NoOorZ24 · Accepted Answer · 2021-05-11 07:43:24Z

7

PHP, 24 bytes

L<?php for(;;)echo'ol';

edited May 11, 2021 at 7:43

answered Sep 28, 2020 at 5:59

NoOorZ24

2511 silver badge4 bronze badges

1

\$\begingroup\$ -1 byte using a for loop. \$\endgroup\$
– Razetime
Commented Sep 28, 2020 at 6:51
\$\begingroup\$ -2 by using ol instead of 'ol' and a for loop. \$\endgroup\$
– Makonede
Commented May 10, 2021 at 17:45

Add a comment |

whqwert · Accepted Answer · 2021-12-22 18:54:46Z

7

convey, 10 bytes

['L'}'lo'}

Try it online!

Visualization (separated)

answered Dec 22, 2021 at 18:54

whqwert

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\$\begingroup\$ Welcome to Code Golf, and nice first answer! \$\endgroup\$
– emanresu A
Commented Dec 22, 2021 at 18:59
\$\begingroup\$ Seems to be an interesting bug where having two string on the same line make both of them generators (e.g. 'lo'>'L'}) \$\endgroup\$
– Jo King ♦
Commented Jan 26, 2022 at 0:36

Add a comment |

Jonathan Allan · Accepted Answer · 2020-09-12 15:23:15Z

6

Jelly, 7 bytes

”L⁾olȮ¿

A full program which prints an L then repeatedly prints ol.

Try it online!

How?

”L⁾olȮ¿ - Main Link: no arguments
”L      - set the left argument to 'L'
      ¿ - while...
     Ȯ  - ...condition: print & yield the left argument
  ⁾ol   - ...do: set the left argument to "ol"

edited Sep 12, 2020 at 15:23

answered Sep 12, 2020 at 13:10

Jonathan Allan

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a stone arachnid · Accepted Answer · 2021-05-16 04:28:24Z

6

Hexagony, 11 6 bytes

-5 thanks to @Bubbler

L;o;l;

Try it online!

  L ;   Executed once: Set memory to 'L' (76) and print
 o ; l  Executed repeatedly: Set memory to 'o' (111), print, set memory to 'l' (108)
  ; .   Executed repeatedly: Print

After the first row is executed once, the second and third rows are repeatedly executed, infinitely printing ol.

edited May 16, 2021 at 4:28

answered May 12, 2021 at 17:01

a stone arachnid

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\$\begingroup\$ Simply L;o;l; works. Not sure if it can be reduced to 5 bytes, but it might be possible. \$\endgroup\$
– Bubbler
Commented May 16, 2021 at 3:46

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emanresu A · Accepted Answer · 2021-08-18 20:26:37Z

6

Cascade, 22 14 bytes

@o|
.l@
L"^
/"

Try it online!

-8 thanks to Jo King.

edited Aug 18, 2021 at 20:26

answered Aug 16, 2021 at 10:27

emanresu A

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\$\begingroup\$ 18 bytes, taking advantage of multiple starting points \$\endgroup\$
– Jo King ♦
Commented Aug 18, 2021 at 15:59
\$\begingroup\$ Wait, what am i doing, we should be using " to print instead: 14 bytes \$\endgroup\$
– Jo King ♦
Commented Aug 18, 2021 at 16:38

Add a comment |

Febriyanto Nugroho · Accepted Answer · 2020-09-26 03:14:54Z

x86-16 machine code MS-DOS - 23 bytes

This answer inspired by 640KB answer.

 000000: B4 02 B7 4C 8A D7 CD 21  B2 6F CD 21 8A D7 80 F2   ...L...!.o.!....
 000010: 20 CD 21 EB F3 CD 20                                .!...

Listing:

      6 0100  B4 02              MOV AH, 02H
      7 0102  B7 4C              MOV BH, 'L'
      8 0104  8A D7              MOV DL, BH
      9 0106  CD 21              INT 21H
     10
     11 0108                 PRINT:
     12 0108  B2 6F              MOV DL, 'o'
     13 010A  CD 21              INT 21H
     14 010C  8A D7              MOV DL, BH
     15 010E  80 F2 20           XOR DL, 20H
     16 0111  CD 21              INT 21H
     17 0113  EB F3              JMP PRINT
     18
     19 0115  CD 20              INT 20H

Output:

The INT 20H at the end seems unnecessary. Also, MOV DL, BH (with or without the XOR DL, 20H) takes far too many bytes. Just move the constant into DH directly. Also move the PRINT: back one instruction to save yourself an INT 21H. — Neil, Commented Oct 18, 2020 at 18:12

daniero · Accepted Answer · 2020-09-27 22:33:53Z

5

braingasm, 12 bytes

76.28524+[.]

Prints the byte streams 76 once, then 28524 forever

answered Sep 27, 2020 at 22:33

daniero

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Pedro Maimere · Accepted Answer · 2021-05-16 00:32:15Z

5

AWK, two answers of 33 bytes

BEGIN{for(;;)printf s=s?"ol":"L"}

and

BEGIN{for(printf"L";;)printf"ol"}

Try it online! (uncomment a line to try it)

First answer: a simple ternary condition upon the assignment of the variable s. On the first run, it is evaluated as false. As it is not assigned yet, it is parsed as a null string, which is false.

Second answer: just a for loop.

answered May 16, 2021 at 0:32

Pedro Maimere

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Aaroneous Miller · Accepted Answer · 2021-07-26 14:05:54Z

5

Flipbit, ³⁵ ³⁰ 28 bytes

Thanks to Bubbler for -5 by shortening the loop

Thanks to ovs for -2 by being big smort

^>>>^>^>>.<<<<<^>>>[>^>^.<<]

Try it online!

Prints L, gets the tape set up for l, then makes use of the fact that o and l differ by only their two least significant bits to create a short loop to print both characters repeatedly.

edited Jul 26, 2021 at 14:05

answered Jul 24, 2021 at 4:22

Aaroneous Miller

9,6901 gold badge24 silver badges39 bronze badges

\$\begingroup\$ 30 bytes \$\endgroup\$
– Bubbler
Commented Jul 24, 2021 at 14:03
\$\begingroup\$ @Bubbler Thanks! I wondered if the loop could be shortened, but I couldn’t figure out how to do it. \$\endgroup\$
– Aaroneous Miller
Commented Jul 24, 2021 at 15:55
\$\begingroup\$ 28 bytes by setting up l instead of o before the loop. \$\endgroup\$
– ovs
Commented Jul 24, 2021 at 16:13

Add a comment |

DLosc · Accepted Answer · 2021-08-10 00:34:28Z

5

Forte (Forter), 25 bytes

5PRINT"ol";:LET5=1
1PUT76

Try it online!

Explanation

An undefined behavior exploit courtesy of Jo King:

1 PUT 76

Print an L character.

5 PRINT "ol"; : LET 5 = 1

Print ol without a final newline, and set 5 to 1.

Apparently, this overwrites the former line 1 with the former line 5, and thus (lacking an END statement) the PRINT "ol"; line is executed repeatedly forever. But I don't really understand how this works--especially since it only works if you write line 5 before line 1, not vice versa...

Here's a 44-byte version that doesn't use any undefined behavior:

1PUT76
5PRINT"ol";:LET6=6+0
6LET0=2:LET5=5+2

Try it online!

Explanation

1 PUT 76

Print an L character.

5 PRINT "ol"; : LET 6 = 6 + 0

Print ol without a final newline, and set 6 to 6.

6 LET 0 = 2 : LET 5 = 5 + 2

Set 0 to 2, and set 5 to 7.

5 7 PRINT "ol"; : LET 6 = 6 + 0 2

Print ol again, and set 6 to 8.

6 8 LET 0 2 = 2 : LET 5 7 = 5 7 + 2

Set 2 to 2, and set 7 to 9.

5 7 9 PRINT "ol"; : LET 6 8 = 6 8 + 0 2

Print ol again, and set 8 to 10.

And so on forever.

edited Aug 10, 2021 at 0:34

answered Aug 7, 2021 at 16:34

DLosc

39.2k5 gold badges83 silver badges141 bronze badges

\$\begingroup\$ Or take advantage of undefined behaviour: Try it online! \$\endgroup\$
– Jo King ♦
Commented Aug 7, 2021 at 22:22
1

\$\begingroup\$ @JoKing Haha! I was just going by the Esolangs article, which sort of implies that "undefined behavior" == "error", so I didn't try it. (Reassigning the current line number is undefined behavior too.) \$\endgroup\$
– DLosc
Commented Aug 10, 2021 at 0:24
\$\begingroup\$ @JoKing Upon looking at it more closely, I find your second one really weird--do you know why it only works when line 5 is written before line 1? \$\endgroup\$
– DLosc
Commented Aug 10, 2021 at 0:36
\$\begingroup\$ it looks like the ruby interpreter has this line: @lines.min_by { |line| posclamp(evaluate(line.number) - num) }, which will take the first line in the program that matches the criteria \$\endgroup\$
– Jo King ♦
Commented Aug 10, 2021 at 1:44

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Aaroneous Miller · Accepted Answer · 2021-09-15 15:37:48Z

Vim `+startinsert`, ¹⁴ 11 bytes

ol<C-v><C-r>"<C-o>ddL<C-r>"

Try it online!

In TIO, when the recursive register expansion (RRE) times out, nothing is outputted since it didn't finish expanding. In Vim, you may not see anything right away since the console might freeze during the RRE, but if you Ctrl+C, you can stop the expansion and see that it does work.

Explanation:

                              # '+startinsert' flag - start vim in insert mode; TIO equivalent is 'i' in the header
ol<C-v><C-r>"                 # Insert 'ol<C-r>"'
              <C-o>           # Run normal mode command, then return to insert mode:
                   dd         #   Delete 'ol<C-r>"' into unnamed register
                     L        # Insert 'L'
                      <C-r>"  # Insert unnamed register, which recursively expands to 'ol' + unnamed register

M Virts · Accepted Answer · 2021-12-18 15:20:00Z

5

Neither beats Johan du Toit's answer, but I enjoy that they are full programs, plus I like the way the longjmp answer works.

C (gcc), 39 bytes

As straight forward as it gets, uses for initializer to print L, then it's ols from there on out.

main(){for(printf("L");printf("ol"););}

Try it online!

C (gcc), 50 bytes

Definitely more fun, use an int array as jmp_buf, then add the return of setjmp to the string pointer so that the first iteration adds 0, then pass 1 to longjmp so the rest add 1 and skip the L.

Thanks to jdt for -4 bytes!

j[9];main(){printf("Lol"+setjmp(j));longjmp(j,1);}

Try it online!

edited Dec 18, 2021 at 15:20

answered Aug 11, 2021 at 3:44

M Virts

1,0754 silver badges9 bronze badges

1

\$\begingroup\$ +1 for the longjmp. This also works for a full program main(i){main(!printf("Lol"+!i));} \$\endgroup\$
– jdt
Commented Aug 11, 2021 at 14:14
1

\$\begingroup\$ 50 bytes \$\endgroup\$
– jdt
Commented Sep 1, 2021 at 14:29

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emanresu A · Accepted Answer · 2022-03-17 07:45:31Z

5

Pyramid Scheme, 200 bytes

  ^     ^
 / \   / \
/out\ /do \
-----^-----^
    / \   ^-
   /chr\ / \
  ^-----/out\
 / \   ^-----^
/76 \ / \   ^-
-----/chr\ / \
    ^-----/chr\
   / \   ^-----
  /111\ / \
  -----/108\
       -----

Try it online!

This seems to be the shortest way that I can find. This language isn't easy to golf in.

answered Mar 17, 2022 at 7:45

emanresu A

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Mukundan314 · Accepted Answer · 2020-09-12 13:24:47Z

4

Pyth, 8 bytes

p\L#p"ol

Try it online!

Explanation

p\L#p"ol
p\L        : print "L"
   #       : while True:
    p"ol   :     print "ol"

answered Sep 12, 2020 at 13:24

Mukundan314

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lyxal · Accepted Answer · 2020-09-26 04:53:05Z

4

Keg, 5 bytes

L,{ǪȽ

Try it online!

Finally! A reasonable use for the push'n'print commands!

Explained

Print the letter "L" (L,)
While true: ({)
---- Print the letter "o" (Ǫ)
---- Print the letter "l" (Ƚ)

answered Sep 26, 2020 at 4:53

lyxal

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3

\$\begingroup\$ There's a push and print command for every letter?! \$\endgroup\$
– Razetime
Commented Sep 26, 2020 at 5:33
5

\$\begingroup\$ @Razetime yep.... It's a result of Bad Choices™ \$\endgroup\$
– lyxal
Commented Sep 26, 2020 at 6:13

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Esolanging Fruit · Accepted Answer · 2020-09-26 05:16:37Z

Flobnar, 18 bytes

og,!<
\l@>\<
2:L!_

Try it online! (requires the -i flag)

Explanation

Flobnar is a 2D language where expressions are laid out geometrically. For example, for a program that computes the number 10, you might write:

5
+  @
5

Here, @ indicates the entry point for the program, and has the effect of evaluating the term to its west; + evaluates the terms to the north and south and returns their sum; etc.

The basic idea for this program is to embed the characters 'o', 'l', and 'L' in the source code at coordinates (0, 0), (1, 1), and (2, 2) respectively. Execution proceeds roughly like this:

def step(n):
    step(!n if (!putchar(get(n, n))) else "impossible since putchar() returns 0")

step(2)

Here's what the program looks like ungolfed:

o    >>>>v
 l   ^ \ < \ @
  L    v   2
     :!_
  :
  g , !<
  :

See the specification for more detailed information about what each term does.

Note that this requires -i, so I don't think it works for the reference implementation. — Wezl, Commented Oct 5, 2020 at 19:41
@Wezl I don't think the reference implementation even supports character IO. — Esolanging Fruit, Commented Oct 5, 2020 at 20:05

Lolololololololololololol

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Explanation

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