89
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The Challenge

Write a complete program that writes twice as many bytes to standard output as the length of the program.

Rules

  • The program must write ASCII characters to the standard output.

  • The contents of the output doesn't matter.

  • The output, measured in bytes, must be exactly twice the length of the program, also measured in bytes, unless you fulfill the bonus.

  • Any trailing newline is included in the output's byte count.

Bonus

Your program can optionally take a number, n, as input. If so, the output must be exactly n * program length bytes. You can assume that n will always be a positive integer. If no input is provided, n must default to 2.

If you do this, you can subtract 25 bytes from your score.

Shortest program wins.

Restrictions

  • No standard loopholes.

  • The program must be at least 1 byte long.

  • No adding unnecessary whitespace to the source code to change its length. Similarly, comments don't count.

  • Unless you fulfill the bonus, the program must accept no input. If you do fulfill the bonus, the integer must be the only input.

Lowest score (program length in bytes - bonus) wins.

The shortest answer for each language wins for that language.

Leaderboards

Here is a Stack Snippet to generate both a regular leaderboard and an overview of winners by language.

To make sure that your answer shows up, please start your answer with a headline, using the following Markdown template:

# Language Name, N bytes

where N is the size of your submission. If you improve your score, you can keep old scores in the headline, by striking them through. For instance:

# Ruby, <s>104</s> <s>101</s> 96 bytes

If there you want to include multiple numbers in your header (e.g. because your score is the sum of two files or you want to list interpreter flag penalties separately), make sure that the actual score is the last number in the header:

# Perl, 43 + 2 (-p flag) = 45 bytes

You can also make the language name a link which will then show up in the leaderboard snippet:

# [><>](http://esolangs.org/wiki/Fish), 121 bytes

var QUESTION_ID=59436,OVERRIDE_USER=41505;function answersUrl(e){return"https://api.stackexchange.com/2.2/questions/"+QUESTION_ID+"/answers?page="+e+"&pagesize=100&order=desc&sort=creation&site=codegolf&filter="+ANSWER_FILTER}function commentUrl(e,s){return"https://api.stackexchange.com/2.2/answers/"+s.join(";")+"/comments?page="+e+"&pagesize=100&order=desc&sort=creation&site=codegolf&filter="+COMMENT_FILTER}function getAnswers(){jQuery.ajax({url:answersUrl(answer_page++),method:"get",dataType:"jsonp",crossDomain:!0,success:function(e){answers.push.apply(answers,e.items),answers_hash=[],answer_ids=[],e.items.forEach(function(e){e.comments=[];var s=+e.share_link.match(/\-?\d+/);answer_ids.push(s),answers_hash[s]=e}),e.has_more||(more_answers=!1),comment_page=1,getComments()}})}function getComments(){jQuery.ajax({url:commentUrl(comment_page++,answer_ids),method:"get",dataType:"jsonp",crossDomain:!0,success:function(e){e.items.forEach(function(e){e.owner.user_id===OVERRIDE_USER&&answers_hash[e.post_id].comments.push(e)}),e.has_more?getComments():more_answers?getAnswers():process()}})}function getAuthorName(e){return e.owner.display_name}function process(){var e=[];answers.forEach(function(s){var r=s.body;s.comments.forEach(function(e){OVERRIDE_REG.test(e.body)&&(r="<h1>"+e.body.replace(OVERRIDE_REG,"")+"</h1>")});var a=r.match(SCORE_REG);a&&e.push({user:getAuthorName(s),size:+a[2],language:a[1],link:s.share_link})}),e.sort(function(e,s){var r=e.size,a=s.size;return r-a});var s={},r=1,a=null,n=1;e.forEach(function(e){e.size!=a&&(n=r),a=e.size,++r;var t=jQuery("#answer-template").html();t=t.replace("{{PLACE}}",n+".").replace("{{NAME}}",e.user).replace("{{LANGUAGE}}",e.language).replace("{{SIZE}}",e.size).replace("{{LINK}}",e.link),t=jQuery(t),jQuery("#answers").append(t);var o=e.language;/<a/.test(o)&&(o=jQuery(o).text()),s[o]=s[o]||{lang:e.language,user:e.user,size:e.size,link:e.link}});var t=[];for(var o in s)s.hasOwnProperty(o)&&t.push(s[o]);t.sort(function(e,s){return e.lang>s.lang?1:e.lang<s.lang?-1:0});for(var c=0;c<t.length;++c){var i=jQuery("#language-template").html(),o=t[c];i=i.replace("{{LANGUAGE}}",o.lang).replace("{{NAME}}",o.user).replace("{{SIZE}}",o.size).replace("{{LINK}}",o.link),i=jQuery(i),jQuery("#languages").append(i)}}var ANSWER_FILTER="!t)IWYnsLAZle2tQ3KqrVveCRJfxcRLe",COMMENT_FILTER="!)Q2B_A2kjfAiU78X(md6BoYk",answers=[],answers_hash,answer_ids,answer_page=1,more_answers=!0,comment_page;getAnswers();var SCORE_REG=/<h\d>\s*([^\n,]*[^\s,]),.*?(\-?\d+)(?=[^\n\d<>]*(?:<(?:s>[^\n<>]*<\/s>|[^\n<>]+>)[^\n\d<>]*)*<\/h\d>)/,OVERRIDE_REG=/^Override\s*header:\s*/i;
body{text-align:left!important}#answer-list,#language-list{padding:10px;width:290px;float:left}table thead{font-weight:700}table td{padding:5px}
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script> <link rel="stylesheet" type="text/css" href="//cdn.sstatic.net/codegolf/all.css?v=83c949450c8b"> <div id="answer-list"> <h2>Leaderboard</h2> <table class="answer-list"> <thead> <tr><td></td><td>Author</td><td>Language</td><td>Size</td></tr></thead> <tbody id="answers"> </tbody> </table> </div><div id="language-list"> <h2>Winners by Language</h2> <table class="language-list"> <thead> <tr><td>Language</td><td>User</td><td>Score</td></tr></thead> <tbody id="languages"> </tbody> </table> </div><table style="display: none"> <tbody id="answer-template"> <tr><td>{{PLACE}}</td><td>{{NAME}}</td><td>{{LANGUAGE}}</td><td>{{SIZE}}</td><td><a href="{{LINK}}">Link</a></td></tr></tbody> </table> <table style="display: none"> <tbody id="language-template"> <tr><td>{{LANGUAGE}}</td><td>{{NAME}}</td><td>{{SIZE}}</td><td><a href="{{LINK}}">Link</a></td></tr></tbody> </table>

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  • 1
    \$\begingroup\$ For the bonus, does the output have to be exactly n * program length bytes, or is that a minimum? \$\endgroup\$ – xnor Oct 2 '15 at 23:19
  • 2
    \$\begingroup\$ It has to be exact \$\endgroup\$ – Daniel M. Oct 2 '15 at 23:20
  • 3
    \$\begingroup\$ Looks like the code snippet has to be modified to handle negative scores. \$\endgroup\$ – El'endia Starman Oct 2 '15 at 23:49
  • 41
    \$\begingroup\$ A bonus of -25 is basically mandatory for some languages, since it lets them achieve a negative score. In the future, I'd suggest using a percent bonus, or just making the bonus the question if you really want answers to go for it. Or, just don't have a bonus. \$\endgroup\$ – xnor Oct 2 '15 at 23:58
  • 4
    \$\begingroup\$ For "no input is provided", do we assume the empty string is passed in? I can't see how one would deal with the user never typing in an input and the program just waiting. \$\endgroup\$ – xnor Oct 2 '15 at 23:59

267 Answers 267

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1
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Sass, 30 bytes

codegolf is absolutely awesome

if you try to compile this with sass the result is the following error

Invalid CSS after "...olutely awesome": expected "{", was "" 
/* 60 bytes */
| improve this answer | |
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  • \$\begingroup\$ I get invalid top-level expression on line 1 at column 1, but I also don't think this is a language as it's not turing-complete and it only compiles to CSS. \$\endgroup\$ – cat Dec 16 '15 at 13:04
  • \$\begingroup\$ @cat I'm pretty sure I could implement /// in Sass using just string manipulation and a while loop, and /// is turing complete. \$\endgroup\$ – Sparr Dec 17 '15 at 10:05
  • \$\begingroup\$ @Sparr fair enough \$\endgroup\$ – cat Dec 17 '15 at 11:57
1
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PlatyPar, 1 byte

#

# starts a number, but since no number is found after it, it is substituted with 59. Here's a "real" answer:

77^

Prints the result of 7^7, or 823543, which is of length 6.

Try it online!

| improve this answer | |
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1
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Z80 machine code, 16 bytes

3E 1F 0E 02 1E 2A F5 CD 05 00 F1 3D F2 02 01 C7

This was made on an Osborne Executive running CP/M version 3.0 using SID. Here is a disassembly (with comments following # characters for readability)

MVI A,1F     # Set register A equal to 0x1F (iterations-1)
MVI C,2      # Write is BDOS call 2 (CP/M's system calls)
MVI E,2a     # I'm writing '*' as output. Its ASCII value is 0x2a
PUSH PSW     # Push operates on register pairs. This pair contains A.
CALL 5       # Do the BDOS call (which destroys some registers)
POP PSW      # But we can get A back from the stack.
DCR A        # Decrement the loop counter
JP 102       # If last result is non-negative, go to address 0x102
RST 0        # Otherwise, exit

All programs are loaded at a fixed address of 0x100, so the jump to 0x102 is well-defined. After running this, CP/M wants to print a newline character. If we count that as output from the program, change the second byte from 1F to 1E.

| improve this answer | |
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1
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Fuzzy Octo Guacamole, 8 bytes

42*![o;]

Also could use:

8  ![o;]

Or

81*![o;]

But that is cooler.

Prints 8 "8"s and 8 newlines, including a trailing one.

Explanation:

4: Push 4 to the stack.

2: Push 2 to the stack.

*: Pop and multiply the top 2 items on the stack and push the result (8)

!: Set the loop counter to the top. Is now 8.

[: Start a loop that lasts 'loopcounter' (8) iterations.

o: Peek at the top of the stack and push it to the temp variable.

;: Print the temp var.

]: End the loop.

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1
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Pylongolf, 2 bytes

.;

. - Reset the stack
; - Debugally print both the stack and the variables.

The interpreter I use prints debugally by converting an 2 arrays into a string which has that string begin with [ and end with ] which prints:

[][]
| improve this answer | |
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1
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Lua, 35 - 25 = 10 bytes

Takes input from the command line, which is in the varargs.

print(("n"):rep(35*(...or 2)-1))
| improve this answer | |
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  • \$\begingroup\$ You can shave off 3 bytes with print(("n"):rep(35*(...or 2)-1)) \$\endgroup\$ – Katenkyo Mar 30 '16 at 9:55
1
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Labyrinth, 10 bytes - 25 = -15

Another Labyrinth collaboration with Sp3000.

?02
`\
~"@

This prints 10n linefeeds to STDOUT.

Try it online!

Explanation

The most interesting part is probably how the default value of 2 is handled.

If an input number is given, the following code is executed:

?   Read integer from STDIN.
0   Multiply by 10. This is because digits in Labyrinth work by multiplying the top of
    the stack by 10 and then adding themselves, such that multi-digit numbers can be
    written into the code more easily.
    The top of the stack is now positive (and contains the number of characters to be
    printed) so the instruction pointer (IP) turns right/south towards the \.

If no input number is given, ? pushes a 0 instead, and this happens:

?   Push 0.
0   Multiply by 10, which is still 0. Since the top of the stack is now 0, the IP
    keeps moving forward/east instead.
2   Multiply by 10, add 2, which sets the top of the stack to 2. The IP hits a
    dead end so it turns around.
0   Multiply by 10 to give 20. Now the top of the stack is positive and IP
    turns left/south towards the \. Again, the top of the stack is the number
    of characters to be printed.

Now all we need to do is print one character each while decrementing the top of the stack to zero. The cheapest character to be printed in Labyrinth is a linefeed, because \ prints one without affecting the stack at all. As an additional trick, we decrement via multiply by -1, bitwise NOT, to ensure that the top of the stack is negative in the top left corner of the loop (otherwise the IP would move towards the ? again).

The loop is then simply:

\   Print linefeed.
`   Multiply by -1.
~   Bitwise NOT.
"   No-op. This cell acts a junction. While the top of the stack is positive
    the IP will turn left/north, otherwise it will move forward/east.

When the IP leaves the loop it hits the @ which terminates the program.


For completeness, here are also two 7-byte versions without the bonus:

7:(
@`!

prints

-7-6-5-4-3-2-1

Try it online!

And

>11!:
@

prints

11001011111011

Try it online!

The former is a very simply (but compact) loop which prints -n while decrementing n from 7 down to 0.

The latter is a simple modification of Sp3000's solution to this challenge (which makes the execution a little bit crazier though).

| improve this answer | |
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1
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Hexagony, 6 bytes

o!!!!@

Prints

111111111111

Try it online!

Explanation

Probably one of the simplest Hexagony programs I've written. The unfolded code is

 o !
! ! !
 @ .

and is simply executed in reading order. o sets the current memory edge to 111 (any letter from d to z would do). Then the ! print that four times and @ terminates the program.

I might try for the bonus later, but I have some doubts that it will fit in side-length 4 (and side-length 5 might end up costing more than the bonus gives).

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  • \$\begingroup\$ How about this? TIO \$\endgroup\$ – Adyrem May 28 '18 at 12:21
  • \$\begingroup\$ @Adyrem That only works if measured in characters (whereas the challenge requires counting bytes). And by the time you get to 5-digit code points (in order to fix that), UTF-8 requires 3 bytes for the character. \$\endgroup\$ – Martin Ender May 28 '18 at 13:09
1
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APL, 3 bytes

1e5

Print 100000...

| improve this answer | |
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1
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dc, 7 bytes

2oFddnf

2o         Set the output radix to 2: write to stdout using the binary digits [01]
  F        Push 0xF on the stack, equivalent to 1111b
   dd      Duplicate the top-of-stack, then duplicate the top-of-stack: 1111b, 1111b, 1111b
     n     Pop the topmost item from the stack (1111b) and write it (using binary, per
               the above) to stdout. Do not follow with a newline.
      f    Dump the contents of the stack (1111b, 1111b), following each item with a
               newline.

Visible characters comprise 12 bytes; add two (2) newlines for a total of 14.

Edit: Since I'm the only dc answer with a natural number for a score, why not post the following?

dc, 8 bytes

cccccccP

Clears the stack seven times, then attempts to pop the top (non-existent) value and print it as text (i.e., a number with output-radix UCHAR_MAX+1). Since the stack is empty, this results in a fifteen-byte error message followed by a single newline. (Works for GNU dc 1.2)

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  • \$\begingroup\$ Hello, and welcome to PPCG! This is a cool answer. Can you explain what exactly o, F, et cetera do? \$\endgroup\$ – NoOneIsHere Jun 21 '16 at 23:27
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PD, 204 bytes

#N canvas 1 7 1 1 1;
#X obj 1 6 loadbang;
#X msg 1 1 \; pd quit;
#X obj 2 5 print;
#X obj 2 2 metro 10;
#X obj 1 7 del 340;
#X connect 0 0 3 0;
#X connect 0 0 4 0;
#X connect 3 0 2 0;
#X connect 4 0 1 0;

run with pd -nogui patchname.pd 2>&1. The program will print the String print: bang (12 bytes including the newline) every 10ms. Then the program terminates after 340ms which will result in the string being printed 34 times (34 * 12 = 408 Bytes output).

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1
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Python, 24-25= 0 -1 bytes

print'a'*int(input())*24

It takes input, converts it to an integer, multiplies it by 24 (the length of my code) and multiplies the character a by it

Thanks to @EamonOlive for reducing 1 byte

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  • 1
    \$\begingroup\$ You can eliminate the space between the print and the '. You would also have to change the 25 to a 24. \$\endgroup\$ – Wheat Wizard Aug 4 '16 at 3:45
  • \$\begingroup\$ @EamonOlive I didn't know you could remove the space! Thanks for the tip \$\endgroup\$ – vikarjramun Aug 4 '16 at 14:13
  • \$\begingroup\$ It looks as if you may have forgotten to make the change, the space is still there. \$\endgroup\$ – Wheat Wizard Aug 4 '16 at 15:03
  • \$\begingroup\$ @EamonOlive I did everything I intended to except remove the space - thanks for pointing it out! :) \$\endgroup\$ – vikarjramun Aug 4 '16 at 15:11
  • \$\begingroup\$ Is there a default value? \$\endgroup\$ – Titus Jan 27 '17 at 6:13
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Cubix, 10 bytes

Cubix is a 2D esolang with a twist: the source code is wrapped around the outside of a cube.

>..(NU@?O/

Test it online! This maps to the following cube:

    > .
    . (
N U @ ? O / . .
. . . . . . . .
    . .
    . .

The output is

10998877665544332211

Don't even ask how it works... though if you'd like to watch it in action, run it here.

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1
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PHP, 32 30 bytes -25 = 5

<?=str_pad(_,30*$argv[1]?:60);

prints an underscore, fills up with spaces

fancier, but longer (38 bytes):

<?=date(str_pad(r,3*$argv[1]?:6,MYr));

gives ("rMY" repeated N times) as argument to date(), which returns an RFC 2822 formatted date (e.g. Thu, 26 Jan 2017 23:32:31 -0800, length=31) followed by 3 letters of the month name and the 4 digit year - repeated N times. <?= prints the result.

| improve this answer | |
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1
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Ruby, 24 bytes (with bonus)

->m{m.to_i.times {49.times {print "a"}}}[gets||2]
| improve this answer | |
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  • \$\begingroup\$ Did you try that with input? Assignments have a lower precedence than the ternary operator, so you probably should add parentheses. \$\endgroup\$ – Titus Jan 27 '17 at 4:40
  • \$\begingroup\$ Hmm ... can you use gets||2? \$\endgroup\$ – Titus Jan 27 '17 at 8:33
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Pushy, 2 bytes

H#

Try it online! - this prints 100 followed by a trailing newline, 4 bytes of output.

H  \ Push 100 to the stack
 # \ Print with a trailing newline
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1
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Haskell, 25-25 = 0 bytes

f n=putStr$[1..25*n]>>"*"

prints n*25 Asterisks

| improve this answer | |
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  • \$\begingroup\$ Is this a "complete program"? It looks like just a function to me. \$\endgroup\$ – dfeuer Apr 6 '19 at 21:28
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PHP, 45 bytes

<?php print number_format(pow(10,66))."a"; ?>

output:

1,000,000,000,000,000,132,394,543,446,603,018,655,781,305,157,705,474,440,625,207,115,776a

<?php              //PHP start tag
print              //Prints what's ahead
number_format(arg) //format what's inside the parentheses as a number, with separators. (Used as escape function for having scientific notation result 
pow(base,exp)      //Gets the value of the `base` raised to the power of `exp`
.                  //Concatenation operator
"a"                //String of "a"
?>                 //End tag for PHP
| improve this answer | |
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1
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Forth (gforth), 36 - 25 = 11 bytes

: x depth 0= if 2 then 36 * spaces ;

How it works:

Checks if no value is on the stack (depth 0=). If that is the case, push 2. Then just print n * 36 spaces.

Forth (gforth), 2 bytes

.S

Only works if the stack is empty. Prints <0> (<, 0, > and space)

| improve this answer | |
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1
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Braingolf, 1 byte

Prints 0 and a newline.

l

Try it online!

Explanation

l

l      push length of the stack to the stack
       implicit output
       implicit newline printed at end of program
| improve this answer | |
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1
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><>, 15 + 2 (-v flag) - 25 = -8 bytes

2{f*:?!;0n1-30.

Explanation:

2{ puts 2 at the bottom of the stack. If no input was provided, this means it is at the top. Otherwise, the provided input will be at the top.

f* multiplies the input by 15 (the length of the program)

We then go into a loop: :?!;0n1-30.

:?!; ends the program if the counter is 0.

Otherwise, 0n prints 0, 1- decrements the counter, and then 30. goes back to the start of the loop.

| improve this answer | |
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1
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05AB1E, -23 bytes (non-competing)

Try it online!

| improve this answer | |
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  • \$\begingroup\$ This doesn't default to 2 for no input \$\endgroup\$ – Jo King Aug 23 '18 at 2:59
  • \$\begingroup\$ @JoKing it used to, 05AB1E basically got a massive overhaul. It for sure used to default to 10. \$\endgroup\$ – Magic Octopus Urn Aug 23 '18 at 4:32
  • \$\begingroup\$ Doesn't work for 1. I don't see how this works, so may you explain your program? \$\endgroup\$ – user85052 Dec 22 '19 at 14:15
1
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TI-Basic, 2 bytes

::

(Ab)uses the fact that programs output Done when nothing is evaluated on the last line.

| improve this answer | |
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  • \$\begingroup\$ This doesn't work. You need two colons after the last expression to display Done. \$\endgroup\$ – lirtosiast Jul 11 '17 at 1:33
  • \$\begingroup\$ @lirtosiast Fixed \$\endgroup\$ – Timtech Jul 11 '17 at 2:19
1
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Java 8, 16 15 bytes

Golfed a byte by using a for loop instead of a while loop. (back to where I was before :P)

u->{for(int x=0;++x<40*u;)out.print(1);}

Using a static import to reduce the code by 7 bytes, this prints out 1 40u times, where u is the integer taken from input. Since the program code is 40 bytes long and I incorporated the bonus, that leads me with 40 - 25 = 15 bytes.

Try it online!

| improve this answer | |
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  • \$\begingroup\$ On TIO, this outputs in scientific notation, which means that the output for u.c(1) is ~6 bytes. Is that also how it works for you locally? If so, I'm not sure this is working as intended. \$\endgroup\$ – Stephen Aug 14 '17 at 15:52
  • \$\begingroup\$ I, for some reason, thought the output simply had to represent double the length of the source code. Thanks for pointing it out; fixed accordingly. \$\endgroup\$ – NotBaal Aug 15 '17 at 15:22
1
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Ly, 17 - 25 = -8 bytes

"9>n[<&s&ol>1-]<;

Try it online!

A simple quine variant. Outputs 9>n[<&s&ol>1-]<; input times. (there's a tab at the end)

3 bytes, no bonus

"&|

Outputs:

38 124

Try it online!

| improve this answer | |
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1
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Pyke, 1 byte

T

Try it here!

Outputs 10

| improve this answer | |
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1
\$\begingroup\$

Pyke, 5 bytes -25 = -20

2|}hV

Try it here!

| improve this answer | |
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1
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Pyke, 3 bytes -25 = -22

2|S

Try it here!

2|  -  input or 2
2|S - range(1, ^)
| improve this answer | |
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  • \$\begingroup\$ Not sure if this is valid, input of 0 acts as if n=2, but I guess there isn't really a way to avoid that \$\endgroup\$ – ASCII-only Apr 20 '18 at 4:28
  • \$\begingroup\$ It says input will be a positive integer though it doesn't specify if 0 is valid or not \$\endgroup\$ – Blue Apr 20 '18 at 7:33
  • \$\begingroup\$ Oh, fair enough \$\endgroup\$ – ASCII-only Apr 20 '18 at 7:33
1
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VBA, 2 1 -3 20 - 25 = -5 Bytes

?Spc([Max(A1,2)*20])

Which outputs [A1] (analagous to n) else 1 times 22 (length of the code) spaces

Previous Version 1 Byte

?

Output

 
| improve this answer | |
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1
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Windows Batch, 148 144 73 30 bytes

@echo %OS%%OS%%OS%%OS%%OS%%OS%

The %OS% system variable should be Windows_NT on most Windows NT systems.

6 of this %OS%(i.e. Windows_NT) is exactly 60 character, which is codeLength * 2.

| improve this answer | |
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