580
\$\begingroup\$

Note to challenge writers as per meta consensus: This question was well-received when it was posted, but challenges like this, asking answerers to Do X without using Y are likely to be poorly received. Try using the sandbox to get feedback on if you want to post a similar challenge.


It's 2017 2018 2019 already, folks, go home.

So, now that it's 2014, it's time for a code question involving the number 2014.

Your task is to make a program that prints the number 2014, without using any of the characters 0123456789 in your code, and independently of any external variables such as the date or time or a random seed.

The shortest code (counting in bytes) to do so in any language in which numbers are valid tokens wins.


Leaderboard:

var QUESTION_ID=17005,OVERRIDE_USER=7110;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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  • 10
    \$\begingroup\$ Even though numbers are ignored in brainfuck, I thought I'd post one anyway. 32 Chars: ++++++[>++++++++<-]>++.--.+.+++. \$\endgroup\$ – Braden Best Apr 1 '15 at 21:37
  • 4
    \$\begingroup\$ Brainfuck isn't a valid language for this challenge. \$\endgroup\$ – Joe Z. Apr 1 '15 at 22:49
  • 1
    \$\begingroup\$ I know. That's why I posted it as a comment \$\endgroup\$ – Braden Best Apr 1 '15 at 22:51
  • 9
    \$\begingroup\$ I wonder if this question gets a small spike in popularity around New Year's. \$\endgroup\$ – Joe Z. Dec 26 '15 at 23:28
  • 3
    \$\begingroup\$ Waiting for "Come on folks, don't you realize it's 2016?" :) \$\endgroup\$ – padawan Jan 4 '16 at 23:35

274 Answers 274

1
\$\begingroup\$

Ruby 1.9, 10 bytes 

p 'ߞ'.ord
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  • \$\begingroup\$ nice! 7 chars if you are in the irb command prompt 'ߞ'.ord \$\endgroup\$ – Eduard Florinescu Jan 8 '14 at 16:54
  • \$\begingroup\$ @EduardFlorinescu Thanks, I know. \$\endgroup\$ – Timtech Jan 8 '14 at 21:38
  • \$\begingroup\$ These are 8 bytes in UTF8. \$\endgroup\$ – schmijos Jul 24 '17 at 14:42
1
\$\begingroup\$

vba (immediate window), 38 26 13

using regular ascii characters (no funny typing needed)

?&ha+&ha&&&he

26

?val("&hfbc")/-(true+true)

38

?year((cdbl(asc("ê"))*cdbl(asc("²"))))

find a date that can be represented as a number, and select the year from that (in this case, Jan, 13, 2014)

have to use cdbl, as it assumes signed int, and overflows

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  • \$\begingroup\$ How do you write that in the Immediate window? (I assume that is what you mean by “direct window”.) If I copy-paste it I get “?year((cdbl(asc("e^"))*cdbl(asc("^(2)"))))”. (Copy-pasting “ê” and “²” from charmap.exe results “?” both.) And of course, that way the calculation not gives 2014. \$\endgroup\$ – manatwork Jan 7 '14 at 16:08
  • \$\begingroup\$ I used ?chr(234),chr(178) to get the characters, or you can hold down the ALT key and type 234 (and 178) and release the ALT to get each character \$\endgroup\$ – SeanC Jan 7 '14 at 16:31
  • \$\begingroup\$ With Alt+234 I get “r”, with Alt+178 I get “¦”. Of course, it works with the chr() function. Anyway, nice trick to use year() this way. \$\endgroup\$ – manatwork Jan 7 '14 at 16:41
  • \$\begingroup\$ ok.. I was thinking back to DOS days - now it's 0234 and 0178, but I found another shorter way now \$\endgroup\$ – SeanC Jan 7 '14 at 16:44
  • \$\begingroup\$ Thanks, it works this way. Although here appears “ȩ” and “¸”, the calculation is correct. \$\endgroup\$ – manatwork Jan 7 '14 at 16:59
1
\$\begingroup\$

Clojure, no unicode tricks (49 characters/bytes)

Uses the fact that * called with no args evaluates to 1:

(let[b(inc(*))j(+(* b b b)b)](+(* b j j j)j b b))

Using the same trick and doing string concatenation instead of arithmetic, the lowest I could get was 51 chars:

(let[n(*)t(+ n n)z(+)f(+ t t)](print(str t z n f)))

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1
\$\begingroup\$

Python, 23

print ord("<DC3>")*ord("j")

<DC3> should be replaced with ASCII symbol 19 (device control 3).

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1
\$\begingroup\$

SAS, 34 characters/bytes

data a;x=put(' ',hex.);put x;run;

That puts it to the log, it's 6 longer if you need it to the output window. Note I'm not seeing the second character there; it is backwards-P, which is hex 14.

There should be a shorter solution with %sysfunc(putc(..., but I can't get that to work properly.

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1
\$\begingroup\$

JavaScript 45

alert(parseInt('bbc','twentyonefour'.length))
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1
\$\begingroup\$

Bash - 10 (or 8)

Well there have been a couple of answers that have been disqualified because they rely on the year. When golfing, one side goal is to see how close we can get to breaking the rules as currently written without breaking the letter of the rules (I include the clarifications by Joe Z in the 66 existing comments on the rules). The question very specifically states that I can not depend on 2014 being the current year. I instead rely on it being 8:14pm in my timezone.

date +%H%M 

When I ran it, it output 2014 exactly, thus it satisfies it No, it has to be 2014 exactly. comment. (Due to context people seem to misread it as ... 2014 always, but that was not what was written, even if that were perhaps what was intended.) This lets me beat the current Bash record, at least until this loophole is closed. This interpretation may seem too cheaty since all the existing popular answers assume that the rules really meant always. Indeed some of them exploit this and export something that isn't exactly 2014, but instead contains 2014. I am fine with that interpretation too since Bash can do:

cat /*/*

This is a mere 8 characters, which will concatenates a bunch of files including /dev/urandom/, and it generally takes my machine under a minute to find 2014 in /dev/urandom. Although my rule twisting golfing code of honour won't let me pick this solution since it violates the letter of Joe Z's clarification, the only objection Joe Z raised to the random approach in the 66 comments was that it was too long. At 8 characters this answer is actually shorter than my rules-lawyer answer.

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  • 1
    \$\begingroup\$ Normally entries that explore edge cases are popular. This is the only answer they obeys the letter of the rules that is down voted (all other down voted answers rely on year=2014). Anyone care to comment on what rubs them wrong about this answer? \$\endgroup\$ – gmatht Apr 8 '14 at 0:59
  • \$\begingroup\$ It follows the letter of the rules in a way that's almost universally considered boring and unoriginal. Additionally, it was posted almost three months after the original problem, meaning that it's way back on the 4th page where barely anybody will see it, let alone vote it up. \$\endgroup\$ – Joe Z. Apr 17 '14 at 0:45
  • \$\begingroup\$ For that matter, I've edited the question again to make my intentions for the problem even clearer. Is "independently of any external variables" good enough? Even then, this answer isn't the shortest in either category, so I still won't accept it. \$\endgroup\$ – Joe Z. Apr 17 '14 at 0:46
  • \$\begingroup\$ Thanks. For the record, when I commented above this answer had been voted down to -2, so the context was more "would someone like to comment on why they are voting it down?" rather than "why isn't it upvoted?". It also wasn't intended as a criticism of Joe Z's question specification. \$\endgroup\$ – gmatht May 4 '14 at 13:04
1
\$\begingroup\$

Python, 55 bytes (no math import and no char or unicode trickery!)

x=False;a=x**x;b=a+a;c=b+b;print c**c*(c+c)-b**(c+a)-b

Uses the fact that zero to the zeroth power is defined as one and False can be implicitly casted to 0. Hence a, b and c will contain 1, 2 and 4 respectively.

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  • \$\begingroup\$ Arne, True implicitly casts to 1, so you can start with a=True and save 8 characters. Also, please specify Python 2. \$\endgroup\$ – isaacg Jul 17 '14 at 5:52
1
\$\begingroup\$

Marbelous 14

CB
CE
CF
CD
~~

How it works

the first 4 lines are language literals, in hexadecimal. Their values are 203, 206, 207 and 205. They will fall down by one cell on each tick. If you perform an 8-bit binary not on those values (which is exactly what ~~ does) you get the following values: 52, 49, 48 and 50. These values happen to be the ascii values of 4, 1, 0 and 2 respectively. The literals then fall off the board which causes their corresponding ascii character to be printed to STDOUT.

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1
\$\begingroup\$

C, 27 Bytes

main(){printf("%d",'\aÞ');}

Just a reminder that multi-character constants do exist :)

Alternatively, three bytes more:

main(){printf("%x",' i'-'U');}
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  • \$\begingroup\$ 1. int is not needed. 2. return doesn't print anything. \$\endgroup\$ – Dennis Sep 16 '14 at 15:07
  • \$\begingroup\$ 3. '^GÞ' (character codes 7 and 222) would require only one multi-character constant. \$\endgroup\$ – Dennis Sep 16 '14 at 15:16
  • \$\begingroup\$ My compiler does not agree :( \$\endgroup\$ – Gerwin Sep 16 '14 at 15:21
  • \$\begingroup\$ That's odd. 7 * 256 + 222 = 2014, so it should work. What do you get? \$\endgroup\$ – Dennis Sep 16 '14 at 15:23
  • \$\begingroup\$ 6178782, it seems like my compiler thinks of ^ as a separate character. \$\endgroup\$ – Gerwin Sep 16 '14 at 15:26
1
\$\begingroup\$

(Java, 553 bytes as .class, 112 bytes as it stands, 84 bytes after renaming the class to 'm' and removing whitespace.)

This probably isn't the kind of answer you're looking for, but there are a bunch of strings that share a hashcode of 2014.

public class make2014 {
    public static void main(String[] args){
        System.out.println("={".hashCode());
    }
}
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  • 1
    \$\begingroup\$ Consider adding the language name (Java, I assume) and byte count like the other answers do. \$\endgroup\$ – NinjaBearMonkey Sep 20 '14 at 16:31
  • \$\begingroup\$ Thanks, I forgot to do that. I didn't think my file size would matter much - the minimum file size of a .class is far greater than most of the attempts here. \$\endgroup\$ – John P Sep 20 '14 at 21:43
  • \$\begingroup\$ I got the .java down to 84 bytes without changing functionality. I see other people shaved off 9 bytes by printing special characters to produce 2014 instead of using hashcodes... I kind of thought it would be cheating to do it like that, since you're really printing 2014 as a number in a different format. Oh well. \$\endgroup\$ – John P Sep 20 '14 at 21:50
1
\$\begingroup\$

Xojo, 27 chars (all ASCII)

MsgBox Str(&hFBC/(&hC-&hA))
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1
\$\begingroup\$

Racket: 20 (19 chars)

(char->integer #\ߞ)

ߞ is a unicode character that has 2014 as it's code.

This abuses the fact that every top level form gets its evaluation printed to stdout. This is quite unique amongst LISPs which usualy only have this behaviour in the REPL and not when running programs.

Scheme: 29 bytes (28 chars)

(display(char->integer #\ߞ))
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  • \$\begingroup\$ So 29 bytes, then? \$\endgroup\$ – Joe Z. Jan 1 '14 at 18:47
  • 1
    \$\begingroup\$ @JoeZ. Yes. I was just about to update it :) \$\endgroup\$ – Sylwester Jan 1 '14 at 18:49
  • \$\begingroup\$ At least in Racket, you don’t need to call display — the REPL will print the result of char->integer automatically — which would save you another 9 chars. \$\endgroup\$ – Matthew Butterick Oct 19 '14 at 17:19
  • \$\begingroup\$ @MatthewButterick Unlike Scheme and CL #!racket actually display every top level result to stdout even when running a compiled program. It's quite annoying but I can abuse that. thanks :) \$\endgroup\$ – Sylwester Oct 19 '14 at 18:20
  • \$\begingroup\$ Still beat it, I’m afraid \$\endgroup\$ – Matthew Butterick Oct 19 '14 at 20:39
1
\$\begingroup\$

TinyMUSH, 16

We need more MUD language entries.

\encrypt($"#&,.)
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1
\$\begingroup\$

Lua - 30 bytes

b=#" "print(b..b-b..b/b..b+b)

# is the length operator, so b = 2.

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  • 1
    \$\begingroup\$ Would b/b work? \$\endgroup\$ – Lynn Nov 8 '14 at 16:20
  • \$\begingroup\$ @nooodl Good suggestion; indeed b/b can replace b-#" " and save 3 characters. \$\endgroup\$ – Seeker14491 Nov 10 '14 at 0:56
1
\$\begingroup\$

C++ 30

main(){cout<<('&')*(']'-'(');}
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  • \$\begingroup\$ Why do you need the brackets around '&'? Does main(){cout<<'&'*(']'-'(');} not work? \$\endgroup\$ – Joe Z. Mar 24 '15 at 7:39
1
\$\begingroup\$

x86 machine code, 19 bytes

B8 3A 0E 2C 08 CD 10 2C 02 CD 10 04 01 CD 10 04 04 CD 10

Assembly code equivalent:

mov ax, 0E3Ah; ah = 0Eh (bios teletype), al = 3Ah (ascii semicolon)
sub al, 08h; ascii 2
int 10h

sub al, 02h; ascii 0
int 10h

add al, 01h; ascii 1
int 10h

add al, 04h; ascii 5
int 10h

Yeah, I know: it logs 2015 rather than 2014.

But seeing that this challenge is old and now the year is 2015, it seemed more appropriate to use the current year (it's my excuse for not "going home" :) )

Note: This was tested using DOSBOX

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  • 1
    \$\begingroup\$ You have numbers in your source. \$\endgroup\$ – Elliot A. Jan 31 '16 at 11:06
  • \$\begingroup\$ @ElliotA. Read the challenge: "without using any of the characters 0123456789"; numbers = characters representing numbers. \$\endgroup\$ – SirPython Jan 31 '16 at 21:49
1
\$\begingroup\$

JavaScript (19)

Obvious cheating, but these expression ran in REPL print strings "2014" and "2015":

''+'ߞ'.charCodeAt() // 2014
''+'ߟ'.charCodeAt() //2015

TIL: .charCodeAt implicitly converts it's first argument to 0.

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1
\$\begingroup\$

Vitsy, 5 4 Bytes, 3 Characters

When in Rome...

'Nߞ

Get the character with the value 2014 and then print it as a number. Simple.

More Interesting Version (12 10 9 Bytes):

"ca-^b-N-

My language supports hexadecimal, too. ;)

"         Capture the entire source as string by looping around the source.
 ca-      Push 2 to the stack
    ^     45^2
     b-   -11
       N  Output as a number.
        - Only here for character value 45.
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  • \$\begingroup\$ Am I right that this language is newer than the question? Besides, I believe this question is already 'closed'(winner chosen, and even an edit at the start discouraging more replies) \$\endgroup\$ – Sanchises Nov 1 '15 at 11:42
  • \$\begingroup\$ @sanchises You're absolutely right - I'll pull my edit request. I'd still like to add it to the list, though, even if marked as a new language.. :D \$\endgroup\$ – Addison Crump Nov 1 '15 at 11:46
  • \$\begingroup\$ Just put it under 'invalid' I guess, there's a section for that at the end. \$\endgroup\$ – Sanchises Nov 1 '15 at 11:49
  • \$\begingroup\$ Changed the edit suggestion. :P Forgot about the 'newer than question' thing, but I should've considered that. Thanks, @sanchises \$\endgroup\$ – Addison Crump Nov 1 '15 at 11:50
1
\$\begingroup\$

Perl 5, 8 28 bytes

say 38*53

Seems to do it.

Oh, without cheating ?

$z=ord("!")*ord("=");say++$z
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  • \$\begingroup\$ D'oh. I thought it was without any of those numbers. \$\endgroup\$ – Dale Johnson Dec 2 '15 at 4:24
1
\$\begingroup\$

T-SQL 27 bytes

PRINT ASCII('')*ASCII('j') 

Note that the character that isn't rendered here is the DC3 (CHAR(19)) in the first set of quote marks. It's unicode U+009F which, it would appear, doesn't copy and paste here too well but I can assure you it works in SQL Management Studio.

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1
\$\begingroup\$

Retina, 24 bytes (newer than challenge)

Note the trailing space on lines 2 and 3. Language is newer than the challenge.


xx  x xxxx 
+`(x)* 
$#+

Try it online

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1
\$\begingroup\$

Quetzalcoatl, 11 4 5 bytes

Noncompeting because this language is from 2016.

::ord('ߞ')

The box should be replaced by Unicode character 2014.

Edit

This is for an old version of Quetzalcoatl. New version:

'ߞ'O
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  • \$\begingroup\$ This language is new though, this year. \$\endgroup\$ – NoOneIsHere Mar 2 '16 at 22:53
  • \$\begingroup\$ There is no ASCII character 2014... it has to be Unicode to go that high. \$\endgroup\$ – mbomb007 Mar 4 '16 at 19:40
  • \$\begingroup\$ If Quetzalcoatl is newer than the challenge, you're answer is non-competing and should say so in its body. Also, I'm not aware of any encoding in which ߞ would be a single byte. \$\endgroup\$ – Dennis Apr 8 '16 at 17:21
1
\$\begingroup\$

Mathematica, 10 bytes

N[E,E^E^E]

Prints the decimal expansion of the number e to over 3.8 million decimal places. The first occurrence of 2014 in that decimal expansion starts at the 3180th decimal place.

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1
\$\begingroup\$

VBA, 21 characters

?cells(,"BYL").column

Write and run the above code in the Immediate Window. Basically, the code converts column name BYL to its column index (2014).

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  • \$\begingroup\$ For future reference, this may be rewritten as ?[BYL:BYL].Column \$\endgroup\$ – Taylor Scott Sep 3 '17 at 21:04
1
\$\begingroup\$

Straw, 13 bytes (non-competing)

Non-competing because the language is 2 years newer than the question...

(…………………σ)«$>

« sum the codepoint of all characters in a string, $ convert from unary to decimal and > is the print operator.

Try it online

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1
\$\begingroup\$

Actually, 15 bytes

This language was created long after this challenge was made, but I thought I'd still try my hand at it. This answer avoids all numerals, including Actually's ² for a*a. Golfing suggestions welcome. Try it online!

╜⌐u;*⌐úl¬¬τu;*-

Ungolfing

╜    Push register 0 (initialized to 0).
⌐u   Add 2 and increment. Returns 3.
;*   Duplicate and multiply. Equivalent to squaring. Returns 9.
⌐    Add 2 again. Returns 11.
úl   Pushes the lowercase alphabet and gets its length. Returns 26.
¬¬   Subtracts 2 twice. Returns 22.
τ    double(). Returns 44.
u    Increment. Returns 45.
;*   Square. Returns 2025.
-    Subtract. Returns 2025 - 11 == 2014.
\$\endgroup\$
1
\$\begingroup\$

Vim 8.0, 15 bytes

:h u
ggf:wywZZp

I didn't see a vim answer yet, so I figured I'd add one. This opens up a helpfile, so it is specifically vim 8.0, since it might not work with a future version that updates that file.

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  • \$\begingroup\$ Clever! In my case :h followed by 9wywZZp works. So down to 10 symbols/bytes \$\endgroup\$ – defhlt Dec 5 '16 at 17:22
  • \$\begingroup\$ @defhlt OK, good to know! Although technically that doesn't work because it uses the digit 9. You could do f:w in place of 9w for only one byte more. Which version of vim do you have? \$\endgroup\$ – DJMcMayhem Dec 5 '16 at 17:25
  • \$\begingroup\$ Also, !!date +\%Y is 12 symbols (assuming you are reading this from 2014) \$\endgroup\$ – defhlt Dec 5 '16 at 17:26
  • \$\begingroup\$ You are totally right! I use nvim 0.1.6-dev. \$\endgroup\$ – defhlt Dec 5 '16 at 17:28
1
\$\begingroup\$

SmileBASIC, 10 bytes

?&HFBC>>!.

&HFBC is hexadecimal for 2014*2, which is right shifted by not(0.0)

?ASC("ߞ") looks shorter, but it's actually the same length when saved in UTF-8, and about 100000x more boring.

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1
\$\begingroup\$

Braingolf, 3 bytes

Try it online!

The ordinal of ߞ is 2014, # pushes the ordinal of the next character to the stack, and Braingolf implicitly outputs the last item on the stack.

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  • \$\begingroup\$ Braingolf is your language, right? \$\endgroup\$ – MD XF Jun 6 '17 at 19:11
  • \$\begingroup\$ @MDXF Indeed it is \$\endgroup\$ – Skidsdev Jun 6 '17 at 19:46

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