601
\$\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 2020 2021 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>

\$\endgroup\$
15
  • 15
    \$\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
  • 6
    \$\begingroup\$ Brainfuck isn't a valid language for this challenge. \$\endgroup\$ – Joe Z. Apr 1 '15 at 22:49
  • 6
    \$\begingroup\$ I know. That's why I posted it as a comment \$\endgroup\$ – Braden Best Apr 1 '15 at 22:51
  • 10
    \$\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

303 Answers 303

1
4 5
6
7 8
11
2
\$\begingroup\$

Python 3, 16 bytes (15 characters)

print(ord('ߞ'))

Try it online!

Explanation

ord returns the decimal Unicode of a character, and the decimal Unicode of ߞ happens to be 2014.

Python 2?

I tried doing the same thing in Python 2 (print ord('ߞ')), which would be 1 byte less, but this doesn't work. Why? Well, in Python 3, len('ߞ') returns 1, so everything is fine. However, in Python 2, it returns 2. And since ord only takes a string of length 1, Python 2 doesn't really like that: TypeError: ord() expected a character, but string of length 2 found

\$\endgroup\$
1
\$\begingroup\$

maybe not the shortest, but one of the more readable ones in Smalltalk:

Transcript show: Date today year.
\$\endgroup\$
2
  • 6
    \$\begingroup\$ see the comments. You can't use current Year. \$\endgroup\$ – Wasi Jan 1 '14 at 19:27
  • \$\begingroup\$ This doesn't work.. (lol) \$\endgroup\$ – Albert Renshaw Jan 30 '17 at 3:36
1
\$\begingroup\$

ANSI C - 95 47 52 characters

#include <stdio.h>
main() { printf("%i", (('a' + 'a')/'a') * ('\a' + '\f') * ('<' - '\a') ); }

This program uses characters to initialise integers and multiplies: 2 * 19 * 53.

#include main(){printf("%i",'\aÞ');}

This program initialises an integer using charaterbytes and prints it. '\aÞ' is the bitpattern 00000111 11011110 this is also the bitpattern of 2014.


Disclaimer: this was made on a windows system with visual studio. This code depends on a lot of things, including - How your compiler endodes the characters you input. Þ has an ascii value of 222 (or its negative equivalent), this may vary depending on your system. The notation int a = 'abcd'; is in itself evil and depends on how memory is handled on your system - this includes endian issues. int a = '\0A'; a is 65 on my system but may be 16640 on your system.

main(){printf("%i",('C'-'A')*('T'-'A')*('v'-'A'));}

I went back to Version one and multiplied 2 * 19 * 53. This version uses only one byte at a time so it is endian compatible. Also it uses only characters in the range of [0 - 127] to be compatible to all systems.

\$\endgroup\$
3
  • \$\begingroup\$ You don't actually need to include stdio.h, most compilers will give a warning but include it for you. \$\endgroup\$ – nyuszika7h Jan 1 '14 at 21:58
  • \$\begingroup\$ This code yields 7 for me with tcc and 508830 with gcc. clang gives an error: character too large for enclosing character literal type. \$\endgroup\$ – nyuszika7h Jan 1 '14 at 22:02
  • \$\begingroup\$ @nyuszika7h ... 7 is equivalent with \a - this could be a problem with passing to the printf function, it would be interesting for me to see if int a = '\aÞ' is 2014 using tcc. 508830 looks very strange to me - i would guess some endian thing but n * 256 + 7 can never be that number. So this illuminates how string this code depends on the system. \$\endgroup\$ – Johannes Jan 1 '14 at 22:58
1
\$\begingroup\$

Solution 1

Octave/Matlab (55 chars)

a=pi;b=a*a;disp(ceil(a^a^a/a/a/a-b*b*a-a^a*b+b*b-b-b));

Solution 2

PHP (9 chars without tags, 12 with them Actually 2022 because of the new lines involved)

<!--Comment
  previous
  2013 lines -->
<?=__LINE__; <!-- This should be on line 2014 -->
\$\endgroup\$
5
  • \$\begingroup\$ That would be 2013 newlines followed by <?=__LINE__ for 2024 chars, not 9. \$\endgroup\$ – Peter Taylor Jan 2 '14 at 9:28
  • \$\begingroup\$ @PeterTaylor I didn't write 9 chars to win the challenge (since there are lots of shortest answers!), but because the actual code is that <?=__LINE__;?>, which I thought would be funny :) Nvm, I'll edit that. \$\endgroup\$ – Vereos Jan 2 '14 at 9:31
  • \$\begingroup\$ If it doesn't work without the newlines, they're part of the "actual code". \$\endgroup\$ – Peter Taylor Jan 2 '14 at 9:32
  • \$\begingroup\$ I guess you're right, edited. \$\endgroup\$ – Vereos Jan 2 '14 at 9:37
  • 1
    \$\begingroup\$ In PHP, you can skip ?> at end of the program. But interesting idea with __LINE__, even if it's ridiculous for such huge number. \$\endgroup\$ – Konrad Borowski Jan 2 '14 at 12:15
1
\$\begingroup\$

Game Maker Language, 22

show_message(ord("ߞ"))
\$\endgroup\$
1
\$\begingroup\$

C - 44 characters (85 with headers)

What, no one is abusing strings yet?

#include<stdio.h>
#include<netinet/in.h>
main(){printf("%u",ntohs(*(int*)"\a\xde"));}

Interestingly, this is a special case where neither character is printable, but their special code doesn't involve a number.

If we want no warnings, it needs to become 55 (96) characters:

#include<stdio.h>
#include<netinet/in.h>
int main(){return!printf("%u",ntohs(*(int*)"\a\xde"));}
\$\endgroup\$
1
\$\begingroup\$

Python, 14 characters

print ord('ߞ')

Short way ;)

\$\endgroup\$
4
  • \$\begingroup\$ Does it work without the u before the string? \$\endgroup\$ – Joe Z. Jan 7 '14 at 2:16
  • \$\begingroup\$ TypeError: ord() expected a character, but string of length 2 found \$\endgroup\$ – boothby Jan 7 '14 at 5:41
  • 1
    \$\begingroup\$ @JoeZ. this is valid Python 3 \$\endgroup\$ – Peter Gibson Jan 7 '14 at 6:21
  • 2
    \$\begingroup\$ In Python 3: SyntaxError: invalid syntax (Python 3 needs brackets around stuff being printed). \$\endgroup\$ – Joe Z. Jan 7 '14 at 7:41
1
\$\begingroup\$

C/C++ 39

main(){printf("%d%d",':'-'&',':'-',');}

ASCII for: ':' = 58, '&' = 38, ',' = 44. Using that, 58-38 = 20 and 58-44 = 14.

\$\endgroup\$
1
\$\begingroup\$

120 characters in Squeak Smalltalk trunk (4.5).
I did not search the shortest, but kind of graphical solution:

((Text string:'Happy\New year'withCRs attribute:TextEmphasis narrow)asMorph borderWidth:Float one+Float one)bounds area

It depends on font, margins, and so is quite fragile, but at least for me it worked.
In Squeak 4.4, it works with lowercase 'happy\new year'.

\$\endgroup\$
1
\$\begingroup\$

JavaScript, 49 Chars

A mathematical JavaScript version making use of only PI and E as source numbers.

(m=Math).pow(e=m.E,e*(p=~~m.PI))/m.sqrt(p)+e+e|""

... mmmm PIE.

Oh and just in case implicit returns are vetoed (56 Chars with alert):

alert((m=Math).pow(e=m.E,e*(p=~~m.PI))/m.sqrt(p)+e+e|"")
\$\endgroup\$
1
\$\begingroup\$

Clojure - 22

(apply *(map int"j#"))

(note: the # is ASCII character 19, Stack Overflow doesn't seem to like this but it's valid Clojure source...)

Clojure - 36

(dec(reduce +(nnext(range(int\@)))))
\$\endgroup\$
1
\$\begingroup\$

C++ - 63 bytes

I'm not sure if this method has been used, but I designed this myself anyway:

#include<iostream>
int main(){std::cout<<int('&'*(','+'\t'));}
\$\endgroup\$
10
  • \$\begingroup\$ You can use int instead of toascii for all of those to save a lot of characters. \$\endgroup\$ – Joe Z. Jan 7 '14 at 2:58
  • \$\begingroup\$ @JoeZ. OK, I didn't think of that. How can I calculate the bytes? \$\endgroup\$ – user10766 Jan 7 '14 at 2:59
  • \$\begingroup\$ Save your program as a text file, and then view how many bytes it is in the file manager. \$\endgroup\$ – Joe Z. Jan 7 '14 at 3:00
  • \$\begingroup\$ 2 is not allowed. \$\endgroup\$ – Danko Durbić Jan 7 '14 at 7:46
  • \$\begingroup\$ Oops, I'll fix that. \$\endgroup\$ – user10766 Jan 7 '14 at 16:08
1
\$\begingroup\$

Ruby 1.9, 10 bytes 

p 'ߞ'.ord
\$\endgroup\$
3
  • \$\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

\$\endgroup\$
5
  • \$\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)))

\$\endgroup\$
1
\$\begingroup\$

Python, 23

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

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

\$\endgroup\$
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.

\$\endgroup\$
1
\$\begingroup\$

JavaScript 45

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

\$\endgroup\$
4
  • 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.

\$\endgroup\$
1
  • \$\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.

\$\endgroup\$
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');}
\$\endgroup\$
8
  • \$\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());
    }
}
\$\endgroup\$
3
  • 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))
\$\endgroup\$
0
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))
\$\endgroup\$
5
  • \$\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($"#&,.)
\$\endgroup\$
1
\$\begingroup\$

Lua - 30 bytes

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

# is the length operator, so b = 2.

\$\endgroup\$
2
  • 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<<('&')*(']'-'(');}
\$\endgroup\$
1
  • \$\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

\$\endgroup\$
2
  • 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.

\$\endgroup\$
1
4 5
6
7 8
11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.