81
\$\begingroup\$

Given a number N, output the sign of N:

  • If N is positive, output 1
  • If N is negative, output -1
  • If N is 0, output 0

N will be an integer within the representable range of integers in your chosen language.

The Catalogue

The Stack Snippet at the bottom of this post generates the catalogue from the answers a) as a list of shortest solution per language and b) as an overall leaderboard.

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 snippet:

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

/* Configuration */

var QUESTION_ID = 103822; // Obtain this from the url
// It will be like https://XYZ.stackexchange.com/questions/QUESTION_ID/... on any question page
var ANSWER_FILTER = "!t)IWYnsLAZle2tQ3KqrVveCRJfxcRLe";
var COMMENT_FILTER = "!)Q2B_A2kjfAiU78X(md6BoYk";
var OVERRIDE_USER = 8478; // This should be the user ID of the challenge author.

/* App */

var answers = [], answers_hash, answer_ids, answer_page = 1, more_answers = true, comment_page;

function answersUrl(index) {
  return "https://api.stackexchange.com/2.2/questions/" +  QUESTION_ID + "/answers?page=" + index + "&pagesize=100&order=desc&sort=creation&site=codegolf&filter=" + ANSWER_FILTER;
}

function commentUrl(index, answers) {
  return "https://api.stackexchange.com/2.2/answers/" + answers.join(';') + "/comments?page=" + index + "&pagesize=100&order=desc&sort=creation&site=codegolf&filter=" + COMMENT_FILTER;
}

function getAnswers() {
  jQuery.ajax({
    url: answersUrl(answer_page++),
    method: "get",
    dataType: "jsonp",
    crossDomain: true,
    success: function (data) {
      answers.push.apply(answers, data.items);
      answers_hash = [];
      answer_ids = [];
      data.items.forEach(function(a) {
        a.comments = [];
        var id = +a.share_link.match(/\d+/);
        answer_ids.push(id);
        answers_hash[id] = a;
      });
      if (!data.has_more) more_answers = false;
      comment_page = 1;
      getComments();
    }
  });
}

function getComments() {
  jQuery.ajax({
    url: commentUrl(comment_page++, answer_ids),
    method: "get",
    dataType: "jsonp",
    crossDomain: true,
    success: function (data) {
      data.items.forEach(function(c) {
        if (c.owner.user_id === OVERRIDE_USER)
          answers_hash[c.post_id].comments.push(c);
      });
      if (data.has_more) getComments();
      else if (more_answers) getAnswers();
      else process();
    }
  });  
}

getAnswers();

var SCORE_REG = /<h\d>\s*([^\n,<]*(?:<(?:[^\n>]*>[^\n<]*<\/[^\n>]*>)[^\n,<]*)*),.*?(\d+)(?=[^\n\d<>]*(?:<(?:s>[^\n<>]*<\/s>|[^\n<>]+>)[^\n\d<>]*)*<\/h\d>)/;

var OVERRIDE_REG = /^Override\s*header:\s*/i;

function getAuthorName(a) {
  return a.owner.display_name;
}

function process() {
  var valid = [];
  
  answers.forEach(function(a) {
    var body = a.body;
    a.comments.forEach(function(c) {
      if(OVERRIDE_REG.test(c.body))
        body = '<h1>' + c.body.replace(OVERRIDE_REG, '') + '</h1>';
    });
    
    var match = body.match(SCORE_REG);
    if (match)
      valid.push({
        user: getAuthorName(a),
        size: +match[2],
        language: match[1],
        link: a.share_link,
      });
    else console.log(body);
  });
  
  valid.sort(function (a, b) {
    var aB = a.size,
        bB = b.size;
    return aB - bB
  });

  var languages = {};
  var place = 1;
  var lastSize = null;
  var lastPlace = 1;
  valid.forEach(function (a) {
    if (a.size != lastSize)
      lastPlace = place;
    lastSize = a.size;
    ++place;
    
    var answer = jQuery("#answer-template").html();
    answer = answer.replace("{{PLACE}}", lastPlace + ".")
                   .replace("{{NAME}}", a.user)
                   .replace("{{LANGUAGE}}", a.language)
                   .replace("{{SIZE}}", a.size)
                   .replace("{{LINK}}", a.link);
    answer = jQuery(answer);
    jQuery("#answers").append(answer);

    var lang = a.language;
    lang = jQuery('<a>'+lang+'</a>').text();
    
    languages[lang] = languages[lang] || {lang: a.language, lang_raw: lang, user: a.user, size: a.size, link: a.link};
  });

  var langs = [];
  for (var lang in languages)
    if (languages.hasOwnProperty(lang))
      langs.push(languages[lang]);

  langs.sort(function (a, b) {
    if (a.lang_raw.toLowerCase() > b.lang_raw.toLowerCase()) return 1;
    if (a.lang_raw.toLowerCase() < b.lang_raw.toLowerCase()) return -1;
    return 0;
  });

  for (var i = 0; i < langs.length; ++i)
  {
    var language = jQuery("#language-template").html();
    var lang = langs[i];
    language = language.replace("{{LANGUAGE}}", lang.lang)
                       .replace("{{NAME}}", lang.user)
                       .replace("{{SIZE}}", lang.size)
                       .replace("{{LINK}}", lang.link);
    language = jQuery(language);
    jQuery("#languages").append(language);
  }

}
body {
  text-align: left !important;
  display: block !important;
}

#answer-list {
  padding: 10px;
  width: 290px;
  float: left;
}

#language-list {
  padding: 10px;
  width: 500px;
  float: left;
}

table thead {
  font-weight: bold;
}

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="https://cdn.sstatic.net/Sites/codegolf/all.css?v=ffb5d0584c5f">
<div id="language-list">
  <h2>Shortest Solution 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>
<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>
<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\$
12
  • 49
    \$\begingroup\$ This is a trivial challenge with a lot of trivial solutions. There are however some non-trivial solutions too. To voters: Please read the first sentence of this meta post before upvoting builtin functions. \$\endgroup\$ Dec 20 '16 at 10:10
  • 8
    \$\begingroup\$ This could probably use a leaderboard. \$\endgroup\$ Dec 20 '16 at 11:13
  • 2
    \$\begingroup\$ With a challenge such as this, I'd only be impressed if there was an answer that used less than 1 byte. Say, 5 bits or something. That I would upvote. \$\endgroup\$
    – Mr Lister
    Dec 22 '16 at 13:13
  • 3
    \$\begingroup\$ @FlipTack Oh, I thought it was codegolf. \$\endgroup\$
    – Mr Lister
    Dec 28 '16 at 20:23
  • 4
    \$\begingroup\$ @MrLister that's the objective winning criterion. but does it really take more effort to type s for a sign builtin, or use some clever bitshifting/maths to work it out? Have a look at this meta post \$\endgroup\$
    – FlipTack
    Dec 28 '16 at 20:25

176 Answers 176

1
\$\begingroup\$

QBIC, 18 8 bytes

:?sgn(a)

This utilizes Qbasic's SGN() function. : gets the input in variable a, ? prints.


Original version, before I learnt that QBasic has a SGN() function:

:~a=b|?a\?a/abs(a)

18 bytes. Explanation

:           Get 'a' from the command line
~a=b        If a == b (and b==0 by default)
|?a         Then print a
\?a/abs(a)  Else, print a / abs(a) --> -2/2 leaves the req. -1, 4/4 = 1
\$\endgroup\$
1
\$\begingroup\$

Javascript, 37 bytes

function s(n){return n>0?1:n<0?-1:0}
\$\endgroup\$
2
  • \$\begingroup\$ Sorry, updated! \$\endgroup\$
    – Ostbullen
    Dec 21 '16 at 15:35
  • 2
    \$\begingroup\$ You could use a lambda: n=>n>0?1:n<0?-1:0 \$\endgroup\$
    – user45941
    Dec 23 '16 at 14:45
1
\$\begingroup\$

brainfuck, 23 bytes

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

This takes a single byte from stdin and interprets it as a signed char, printing \xff for negative, \x00 for zero, and \x01 for positive.

Try it online.

Viewed as an unsigned char, we are checking whether it is greater than 127, with 0 as a special case. We can do the comparison by decrementing from 255 twice at a time.

\$\endgroup\$
6
  • \$\begingroup\$ Doing I/O as decimal strings could be fun too. \$\endgroup\$ Dec 21 '16 at 16:48
  • \$\begingroup\$ You could argue that all numbers in BF are positive, so ,>+. is a valid answer. \$\endgroup\$ Dec 22 '16 at 4:38
  • 1
    \$\begingroup\$ @Challenger5 Even if that were modified to handle zero, e.g. with ,[>+>]<., I don't think that could be argued as anything other than a loophole. \$\endgroup\$ Dec 22 '16 at 5:49
  • \$\begingroup\$ Definitely, but it's funny. \$\endgroup\$ Dec 22 '16 at 5:50
  • 1
    \$\begingroup\$ @Challenger5 That would be abusing native number types to trivialize the challenge. The solution Mitch has used (interpreting the input as an 8-bit two's complement integer) is valid, and the only valid way to do this in brainfuck other than interpreting decimal strings (unless there's another way and I'm not seeing it). \$\endgroup\$
    – user45941
    Dec 23 '16 at 14:44
1
\$\begingroup\$

MarioLANG, 868 bytes

;                                                    
=
  [@:
 =======================================================
                                                 ))    <
 ======================================================"
@ ((((++
=======            <
        ==========="
      @ -)+)+)+((([!))+((
      =============#====    (<
                         ===="
                       @ +)-[!)))
                       ======#==      )  <
                                 ========"
                                      >([!) 
                                      "==#        ))  <
                               @ -(-)[!+   ==========="
                               =======#:          >(([!!
                                                  "===##
                                         @ +((-))[!-
                                         =========#:

Ungolfed, with comments

;[>                                                        
==                                                                                                   output zero
    [@                                                                                            :                                                                      * start
  ==================================================================================================                  memory: [limit|limit_copy|counter_add|counter_sub|arg]
                                                    move pointer back to arg  ))    <
  =================================================================================="===
       increase counter limit
@   ((((++
=================    set counters
                              <
                  ============"     increase counter_add by one
                @ -)+)+)+ ((([! ))+((
                ==============#======
                                        reset limit
                                            (<
                                      ======="
                                    @ +)-   [! )))
                                    =========#====   try subtraction
                                                           )  <
                                                   ==========="
                                                         > ( [! ) 
                                                         "====#==
                                                 @ -(-) [!               try addition
                                                 ========#                   ))  <
                                                                  ==============="
                                                                           > (( [!  !
                                                                           "=====#==#
                                                                @  +((-)) [!
                                                                ===========#



                                                             output one       output minus one
                                                           +:                -:
                                                          ===               ===

What's going on?

The program maintains 5 memory fields (right to left):

  • Input value, continuously edited in search of zero
  • Counter for search in negative direction (subtraction)
  • Counter for search in positive direction (addition)
  • Helper to reset search radius (limit)
  • Current search interval radius (limit, k)

The algorithm keeps on searching for zero in both (+ and -) directions, starting at the input value. It does k negative and k+1 positive steps on each iteration, then increases k by 2. Once zero has been found, 1 or -1 is output, depending from which side it was reached.

Detection of zero as input is a special case, handled right at the beginning.

Try it online (commented, ungolfed version)

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

Python 3, 13 bytes

n//abs(n-.1)
\$\endgroup\$
1
  • 7
    \$\begingroup\$ Welcome to the site, and nice first answer! Just so you know, this is only a REPL snippet, which aren't a default valid form of output. You could wrap this in a lambda to make a function submission though. lambda n:n//abs(n-.1) \$\endgroup\$
    – DJMcMayhem
    Dec 24 '16 at 7:27
1
\$\begingroup\$

C, 23 bytes

A more portable (I think) 23-byte solution in C:

f(n){return(n|1)%2-!n;}
\$\endgroup\$
1
\$\begingroup\$

C#, 40 bytes

b=>System.Console.Write(b>0?1:b<0?-1:0);

Or with a built-in:

C#, 44 bytes

using System;b=>Console.Write(Math.Sign(b));

Unfortunately it's longer, then the first solution.

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

MATLAB + Octave, 15bytes

There are a few other Octave/MATLAB answers, but two of the others are simply using a built in, and the other is significantly longer.

The anonymous function:

@(a)(a>0)-(a<0)

Quite simple. If a>0, the answer will be (1-0)=1. If a<0, the answer will be (0-1)=-1. If a==0 the answer will be (0-0)=0.

You can try online here. Simply run the above code and then try with ans(input).

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

Java 8, 33, 17, 14 bytes

i->i>0?1:i>>31

Does not rely on any questionable code constructs or fragments. This is a complete functional interface implementation.

\$\endgroup\$
3
  • \$\begingroup\$ You can reduce to (int i)->i<0?-1:i>0?1:0 (or even further to i->i<0?-1:i>0?1:0) which are all accepted solutions here on codegolf (codegolf.stackexchange.com/a/16100/16236). \$\endgroup\$ Dec 24 '16 at 0:23
  • \$\begingroup\$ @OlivierGrégoire thanks, for some reason I was not thinking "this is a single expression, so the lambda can be reduced." \$\endgroup\$
    – user18932
    Dec 24 '16 at 19:12
  • \$\begingroup\$ It can be reduced to this: i->i>0?1:i>>31 (14 bytes) \$\endgroup\$ Jan 10 '17 at 10:49
1
\$\begingroup\$

Scala, 16 bytes

n=>n compareTo 0
\$\endgroup\$
1
\$\begingroup\$

Common Lisp, 21 6 bytes

signum

Try it online!

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

dc, 22 bytes

[pq]sqd0=qdd*vr1-d*v-p

Try it online!

I don't like that 10 bytes of this is eaten up testing for zero, will continue to mull over that. The second half, dd*vr1-d*v-p uses the square root of the square to calculate the absolute value of both our value to test and that value less one. Subtracting the latter from the former yields 1 for a positive value, -1 for a negative.

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

Aceto, 4 bytes

riyp
ri reads input as integer
y puts sign on stack
p prints it

Try it online!

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

dc, 15 bytes

[_1*]sa?dd0>a/p

Try it online!

Explanation

This (ab)uses the fact that dc keeps the stack as is, if you're trying to divide by 0 which really helps for this challenge:

[_1*]sa          # store the macro in register a (macro negates the top)
       ?         # push input to stack
        d        # duplicate top
         d0>a    # if top is negative, execute macro
             /   # divides top two values
              p  # print the top of the stack
\$\endgroup\$
1
\$\begingroup\$

Functoid, 30 28 bytes

In functoid there is no support for negative integers since numbers are handled as Church numerals. This solution redefines numbers in a way that supports negative numbers (please refer to the explanation):

<|12K0K$BbB
r<@.1,"-"
)<@.([

Try it online! (Here is an input generator)

Explanation

The definition of negative numbers extends the Church numerals by simply introducing another lambda abstraction that encodes the sign of a number (the first argument, ie. x3). Here is how the numbers -2,-1,0,1,2 would look like in this notation:

Decimal number | extended Church numerals | absolute value as Church numeral
---------------+--------------------------+---------------------------------
            -2 |     λλλ(x3 (x2 (x2 x1))) |                   λλ(x2 (x2 x1))
            -1 |          λλλ(x3 (x2 x1)) |                        λλ(x2 x1)
             0 |                  λλλ(x1) |                           λλ(x1)
             1 |               λλλ(x2 x1) |                        λλ(x2 x1)
             2 |          λλλ(x2 (x2 x1)) |                   λλ(x2 (x2 x1))

Now for the explanation of the above code: The characters < will change the direction to left, this saves us four bytes but makes it more difficult to read.. Here's how the code would look like without doing this:

BbB$K0K21|
 @.1,"-"r<
    @.([)<

The combinator BbB ensures that K0 and K2 get grouped as one expression each, here's the more verbose code:

$(K0)(K2)1|
  @.1,"-"r<
     @.([)<

Okay now it's quite readable: First $ will apply the input so we'll have a sign-extended Church numeral as current expression. Applying K0 (constant 0) will evaluate to KK0 (which will "swallow" the next two arguments) in the case of a negative number, in the case of a non-negative number it simply vanishes. In the case of a positive number we'll have K2 which will swallow the argument 1 and else we'll finally get 1.

Summing up the function $(K0)(K2)1 gives the following numbers (regular Church numerals):

if input < 0:
    return 0
elif input > 0:
    return 2
else:
    return 1

The command | will set the direction down iff the current expression evaluates to 0 and up in the other case.

  • The second line is thus only relevant if the input was negative and it simply prints -1 and terminates
  • In the other case (ie. hitting the third line) it applies the current expression to [ which decrements it by 1, prints the expression (as a numeral) and terminates

Disclaimer

It might look like the above extension was done just to make this task easier, however this is not the case. In fact most definitions in the untyped (or rather uni-typed) lambda calculus are made because they simplify a lot of operations that one could be interested (such as Booleans or the Church numerals themselves).

As an example the function abs could be "implemented" by just applying the identity function to such an extended numeral which will get rid of one lambda abstraction (and in case of a negative number, keep the absolute value as is).

If you're interested, here is an article that talks more about this definition which is worth a read.

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

Yabasic, 16 bytes

Anonymous function that takes input as a number, n, and output to STDOUT

input""n
?sig(n)

Try it online!

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

Perl 5, 9 + 1 (-p) = 10 bytes

$_=$_<=>0

Try it online!

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

Pyt, 1 byte

±

Try it online!

Boring built-in answer (padding for body length)

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

Lua (51 Bytes)

n=tonumber(...) print(n<0 and -1 or n>0 and 1 or 0)

... is the variable that represents arguments in lua, tonumber(...) converts the first argument to a number.

also in lua a and b or c is a structure that returns b if a is truthy or c if a is falsy, you can nest this structure as well

\$\endgroup\$
1
  • \$\begingroup\$ Answers should be either functions or full programs. This doesn't seem to be able to take input. N is also guaranteed to be an integer \$\endgroup\$
    – Jo King
    Feb 17 '18 at 2:01
1
\$\begingroup\$

Stax, 2 bytes

:+

Run and debug online!

Explanation

Added for completeness. It is a builtin, but all built-ins starting with a colon in Stax is actually a macro defined using other Stax operations.

Internally it is defined as c0>s0<-, which is simply (x>0)-(x<0).

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

SHELL ( 20 Bytes )

sed s/[1-9][0-9]*/1/

tests:

echo "789" | sed s/[1-9][0-9]*/1/
1

echo "-789" | sed s/[1-9][0-9]*/1/
-1

echo "0" | sed s/[1-9][0-9]*/1/
0
\$\endgroup\$
1
\$\begingroup\$

Swift, 44 bytes

func s(i:Int)->Int{return i>0 ?1:i<0 ? -1:0}

Strange spacing around the ternary options, I know, but this was the shortest way.

Try it online!

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

Brain-Flueue, 40 bytes

{([({})]){({}())<>([{}()()])<>}}(<>[]{})

Try it online!

Readable version

{
 ([({})])
 {
  ({}())<>
  ([{}()()])<>
 }
}
(<>[]{})
\$\endgroup\$
1
\$\begingroup\$

MachineCode, 26 bytes

31c085ff0f9fc0c1ef1f29f8c3

Requires the i flag for integer output. Input must be appended to the code on a separate line. This is equivalent to the C code (n>0)-(n<0). Try it online!

Alternatives with the same bytecount:

  • 31c085ff0f95c0c1ff1f09f8c3 - Try it online! Equivalent to the highest voted C submission.
  • 89f8c1ff1ff7d8c1e81f01f8c3 - Try it online! Equivalent to this C code.
\$\endgroup\$
1
\$\begingroup\$

Canvas, 3 2 bytes

⤢÷

Try it here!

This might be the shortest solution without a builtin that does the problem for you (I'm looking at you, Jelly.)

Explanation:
⤢÷ | *Full code*
 ÷ | Divide
   | The input (implicit)
⤢  | By its absolute value
\$\endgroup\$
1
  • 1
    \$\begingroup\$ Canvas doesn't interpret 0/0 as 0, it actually errors & removes the 2 items without pushing anything (though I'm about to change that as 0 makes much more sense than nothing). The reason that 0 is outputted is that the stack contains an infinite amount of the inputs. And because of that you can actually remove the for 2 bytes \$\endgroup\$
    – dzaima
    May 5 '18 at 13:46
1
\$\begingroup\$

F#, 140 bytes

open Checked
let s n=
 if n=0 then 0
 else
  try
   let mutable p=n
   while p<>0 do
    p<-p+1
   -1
  with| :? System.OverflowException->1

Try it online!

Basically, if the number is non-zero, keep adding 1 to it until you get either 0 (so the original value was negative) or an overflow (so the original value was positive).

That's all right, isn't it?

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

naz, 64 bytes

9a5m2x1v3a2x2v1x1f1o0m1a1o0x1x2f1o0x1x3f1r3x1v1e3x2v2e0m1a1o0x3f

Explanation (with 0x commands removed)

9a5m2x1v                 # Set variable 1 equal to 45 ("-")
3a2x2v                   # Set variable 2 equal to 48 ("0")
1x1f1o0m1a1o             # Function 1
                         # Output once, set the register equal to 1, and output again
1x2f1o                   # Function 2
                         # Output once
1x3f                     # Function 3
    1r                   # Read a byte of input
      3x1v1e             # Jump to function 1 if the register equals variable 1
            3x2v2e       # Jump to function 2 if the register equals variable 2
                  0m1a1o # Otherwise, set the register equal to 1 and output
3f                       # Call function 3
\$\endgroup\$
1
\$\begingroup\$

Keg, -hr, 13 bytes

:0<[0;|0>[1|0

Try it online!

A very simple switch like comparison happening here.

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

05AB1E, 2 bytes

A signum built-in.

Try it online!

\$\endgroup\$
0
1
\$\begingroup\$

Gol><>, 8 bytes

I:SA:?,h

Try it online!

How it works

I:SA:?,h
I         Input as number; [n]
 :        Duplicate; [n n]
  SA      Absolute value; [n abs(n)]
    :     Duplicate again; [n abs(n) abs(n)]
     ?    Pop one; if it is zero, skip next command
      ,   Nonzero n: Divide n by abs(n)
       h  Print top as number and exit

Gol><>, 11 bytes

I:0(qm$0)+h

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How it works

I:0(qm$0)+h

I            Input as number
 :           Duplicate top
  0(         Change top to "top < 0"
    q        If top is true...
     m$      Push -1 and swap top two; the stack is [-1 x]
             Otherwise, skip two commands (m$); the stack is [x]
       0)    Change top to "top > 0"
         +   Add top two
          h  Print top as number and halt
\$\endgroup\$
1
  • \$\begingroup\$ How about I:0)$0(-h for 9 bytes \$\endgroup\$
    – Jo King
    May 6 '18 at 20:21

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