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

175 Answers 175

1
\$\begingroup\$

Io, 28 bytes

Does a conditional checking over x/abs(x).

method(x,if(x!=0,x/x abs,0))

Try it online!

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

Symja, 29 26 bytes

f(x_):=If(x==0,0,x/Abs(x))

Try It Online!

-3 bytes to due porting the idea behind the Io answer.

Answer History

29 bytes

f(x_):=If(x<0,-1,If(x>0,1,0))

For some reason, my edit history didn't save this old approach.

\$\endgroup\$
3
  • \$\begingroup\$ I'd be interested to know what approach you used for the 29 bytes solution \$\endgroup\$
    – user92069
    Mar 26 '20 at 2:29
  • \$\begingroup\$ Wait, why is my edit history not showing? \$\endgroup\$
    – lyxal
    Mar 26 '20 at 2:30
  • \$\begingroup\$ @a'_' I added the 29 byter \$\endgroup\$
    – lyxal
    Mar 26 '20 at 2:31
1
\$\begingroup\$

Python 3, 25 bytes

lambda x:x and(1,-1)[x<0]

Try it online!

Uses a different approach from other python answers

\$\endgroup\$
1
  • \$\begingroup\$ You could use (x>0)*2-1 instead of (1,-1)[x<0]. \$\endgroup\$ Mar 26 '20 at 11:45
1
\$\begingroup\$

Desmos, 12 bytes

f(x)=sign(x)

or $$f\left(x\right)=\operatorname{sign}\left(x\right)$$ Try It On Desmos!

\$\endgroup\$
1
  • \$\begingroup\$ I would change "TIO" to something like "View the plot on desmos.com" not to cause confusion with tio.run. \$\endgroup\$ Mar 26 '20 at 11:43
1
\$\begingroup\$

ArnoldC, 498 bytes

IT'S SHOWTIME
HEY CHRISTMAS TREE n
YOU SET US UP 0
GET YOUR ASS TO MARS n
DO IT NOW
I WANT TO ASK YOU A BUNCH OF QUESTIONS AND I WANT TO HAVE THEM ANSWERED IMMEDIATELY
HEY CHRISTMAS TREE m
YOU SET US UP 0
GET TO THE CHOPPER m
HERE IS MY INVITATION n
GET UP n
GET UP 1
ENOUGH TALK
GET TO THE CHOPPER m
HERE IS MY INVITATION m
GET DOWN 1
I LET HIM GO m
GET DOWN 1
ENOUGH TALK
GET TO THE CHOPPER m
HERE IS MY INVITATION m
GET UP 1
I LET HIM GO m
ENOUGH TALK
TALK TO THE HAND m
YOU HAVE BEEN TERMINATED

Try it online!

Explanation

IT'S SHOWTIME

# n = 0
HEY CHRISTMAS TREE n
YOU SET US UP 0

# n = input
GET YOUR ASS TO MARS n
DO IT NOW
I WANT TO ASK YOU A BUNCH OF QUESTIONS AND I WANT TO HAVE THEM ANSWERED IMMEDIATELY

# m = 0
HEY CHRISTMAS TREE m
YOU SET US UP 0

# m = n + n + 1
GET TO THE CHOPPER m
HERE IS MY INVITATION n
GET UP n
GET UP 1
ENOUGH TALK

# m = ((m - 1) % m) - 1
GET TO THE CHOPPER m
HERE IS MY INVITATION m
GET DOWN 1
I LET HIM GO m
GET DOWN 1
ENOUGH TALK

# m = (m + 1) % m
GET TO THE CHOPPER m
HERE IS MY INVITATION m
GET UP 1
I LET HIM GO m
ENOUGH TALK

# print m
TALK TO THE HAND m

YOU HAVE BEEN TERMINATED
\$\endgroup\$
1
  • \$\begingroup\$ Sorry, skynet got my computer before I could read the explanation. \$\endgroup\$
    – Razetime
    Sep 2 '20 at 2:26
1
\$\begingroup\$

Haskell, 6 bytes

signum

It's a function belonging to the Num typeclass, so every number works.

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

8th, 40 3 bytes

With 8th is quite simple to get the sign of N, which is left on TOS

sgn

Testing and Output

ok> 42 sgn .
1
ok> -42 sgn .
-1
ok> 0 sgn .
0

The following code, as an alternative, has the same behaviour of 8th's builtin word n:sgn

: f dup 0; 0 n:> if 1 else -1 then nip ;

Explanation of word f

: f \ n -- -1|0|1
  dup     \ Duplicate input
  0;      \ Check if number is 0. If true, leave 0 on TOS and exit from word
  0 n:>   \ Check if positive
  if 1    \ Return 1 if positive
  else -1 \ Return -1 if negative
  then
  nip     \ Get rid of input
; 

Testing and Output

ok> 42 f .
1
ok> -42 f .
-1
ok> 0 f .
0
\$\endgroup\$
1
\$\begingroup\$

COW, 153 75 bytes

oomMMMmoOMoOmoOMMMMOOmoOOOOMOomOoMoOmOomOoMOOMOomoOmoOMMMOOOmooMMMmoomoOOOM

Try it online!

COW can't compare numbers so N is stored in two cells, (if N is not 0) in turns one will be increased and the other decreased. When eventually one of them reaches 0, the loop will stop.
The sign will be in the next cell.

Detailed Explanation:

[0]: N to be incr   [1] = -1   [2]: N to be decr   [3] = 1


i=>+>=          ;   Read N in [0], set [1] = 1 and copy it to [2]
[               ;       Loop while [2] is non-zero
    >°-<+<<     ;       Set [3] = -1, increase [2] then point to [0]
    [           ;           Loop while [0] is non-zero
        ->>=°   ;           Decrease [0], point to [2] copy in register and set it to 0 (to exit)
    ]           ;
    =           ;       Paste from register to [2]
]               ;   When exiting the poited cell could be [0] or [1], in either case...
>o              ;   ...move one right and it would contain the correct output


moo ]    mOo <    MOo -    OOO °    OOM i
MOO [    moO >    MoO +    MMM =    oom o

I've tried other permutations of memory cells but this seems the best.
Setting [3] (a constant) inside the loop is needed to manage the case N=0.

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

Stax, 2 bytes

:+

Run and debug it

Stax has a sign builtin. Kind of boring. Here's a solution that uses clamp instead

Stax, 4 bytes

U1:c

Run and debug it

U    push -1 to stack
 1   push 1 to stack
  :c clamp input between -1 and 1
\$\endgroup\$
1
\$\begingroup\$

Zsh, 17 bytes

<<<$[(n>0)-(n<0)]

Idea from stackoverflow (not copypasta, I still had to use my brain somewhat!)

Try it Online!

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

Python - 23 bytes

print((n>0)*2+(n==0)-1)

This checks if n is greater than 0 (in which case it outputs 2 for the next step), then if it is equal to 0 (in which case it outputs 1) and then subtracts 1, leaving us with -1 if n < 0, 0 if n == 0, and 1 otherwise.

(Yes, the double brackets are necessary, I have checked.)

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

Pxem, Contents: 0 bytes + Filename: 33 25 bytes

  • Filename(unprintables are escaped): 1._.c.w\001.r.yXX-.a.p.d.a.n
  • Content is empty.

Try it online!

Usage

Give an integer from STDIN. The result is output from stdout, with no LF termination.

How it works

XX.z
# Prepare character 1 at this point
.aXX1.z
# integer input is pushed
.a._XX.z
# duplicate, and is it non-zero?
.a.c.wXX.z
  # then come here
  # also \001.r is an idiom to push zero:
  # actually "n=pop;push(int(random()*n))"
  # where 0<=random()<1
  # because filename cannot have null character
  # don't be afraid of such binary for code golf
  .a\001.rXX.z
  # is input less than zero?
  # DOT-y to DOT-a is:
  # while size<2 OR pop>pop; do something; done
  .a.yXX.z
    # if so, push hyphen so string "-1" is made
    .aXXXX-.z
  .a.aXX.z
  # at this point DOT-y popped two items: zero and input integer
  # so no worry about garbages
  # print the string and exit
  # .p is actually: while !empty; do printf "%c", pop; done
  .a.p.dXX.z
# if input is zero, come here
# .n is printf "%d", pop
# OBTW the "1" will on stack, but it's not matter,
# because it is ignored when program ends
.a.a.n

Previously

  • 33bytes of filename: ._.c.w.c00.-.-.z-1.p.d.a1.o.d.a.n
\$\endgroup\$
1
  • \$\begingroup\$ this language is really cool \$\endgroup\$
    – wasif
    Jun 1 at 8:30
1
\$\begingroup\$

Japt, 1 byte

g

Try it online!

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

Alice, 19 13 bytes

-6 bytes thanks to Martin Ender

1i4/o
-% \7@/

Explanation

1                     - Push 1                          [1]
 i                    - Take input from STDIN           [1,x]
  4/7                 - Push 47 while going to Ordinal  [1,n,47]
     /                - Back to Cardinal, wrap around.  [1,n,47]
      -               - Subtract 47 from n              [1,n]
       %              - 1 mod n                         [n]
        \o@           - Go to Ordinal, print and terminate.

Try it online!

\$\endgroup\$
2
  • \$\begingroup\$ It's normally not necessary to use " to push numbers onto the stack. A simpler way is to push the 4 in Cardinal mode and then append the 7 in Ordinal mode. You can then also save a few bytes by interleaving the two segments in Ordinal mode: tio.run/##S8zJTE79/98w00Q/n0tXVSHG3EH//39dQyNjAA (I also push the 1 at the start to avoid the ~ but it doesn't really make a difference for the byte count) \$\endgroup\$ Jun 4 at 13:15
  • \$\begingroup\$ Ah, I somehow didn't see appending numbers in Ordinal while checking the command list. Thanks for the advice! \$\endgroup\$ Jun 4 at 13:25
1
\$\begingroup\$

BitCycle -U, 8 bytes

?v
/=
>!

Try it here!

Explanation

BitCycle's -U flag represents signed integer inputs as follows:

  • Positive integer \$N\$: a stream of \$N\$ 1's
  • Negative integer \$-N\$: 0 followed by a stream of \$N\$ 1's
  • Zero: 0

So all we need to do is output the leading 0 (if any) and the first 1 (if any), and discard the rest of the 1's.

The switch device (=) is very convenient for this task. The first bit that hits a switch passes straight through, but it also activates the switch in one of two directions: a 0 bit turns the switch into {, which redirects all subsequent bits westward, while a 1 bit turns the switch into }, which redirects all subsequent bits eastward.

In our program, we direct the input from the source ? into the switch =, moving southward. The first bit (either the leading 0 or a 1 if there is no leading 0) goes straight through to the sink ! and is output. Then:

  • If the first bit was a 1 (positive number), the switch activates as }, which redirects the remaining bits off the playfield.
  • If the first bit was a 0 (non-positive number), the switch activates as {, which redirects the remaining bits westward into the splitter /. This device sends the first bit that hits it southward, where it is redirected into the sink and output. All subsequent bits that reach the splitter pass through it and off the playfield.
\$\endgroup\$
1
\$\begingroup\$

Risky, built-in, 1 byte

1?

Try it online!

Risky, non-built-in, 3 bytes

_?/[?

Try it online!

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

Pyke, 5 bytes

0'<>-

Try it here!

0'<>  - 0<input, 0>input
    - - ^-^
\$\endgroup\$
0
\$\begingroup\$

Loader, 61+44+1=106 bytes

Main module, 61 bytes

readline 0
set B =48
load a
set B =45
set K -1
load a
print 1

Module a, 44 bytes

decr B
decr *0
B:load a
~*0:print K
~*0:exit
\$\endgroup\$
0
\$\begingroup\$

PARI/GP, 4 bytes

sign

Another boring built-in.

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

GameMaker Language, 22 bytes

return sign(argument0)

Alternative solutions:

return median(1,argument0,-1)
a=argument0 return (a>0)-(a<0)
a=argument0 return max(-1,min(a,1))
a=argument0 if a return a/abs(a)return 0
\$\endgroup\$
0
\$\begingroup\$

Clojure, 24 bytes

#(cond(> % 0)1(< % 0)-1 0 0)

An anonymous function that checks for being greater than 0, less than 0, then defaults on 0 if neither is true. cond acts as (and is a macro resulting in) an if-else tree. Unfortunately, it requires an even number of arguments, so I had to waste 2 bytes by adding an extra number to mean true to default on.

Ungolfed:

(defn sign [n]
  (cond
    (> n 0) 1
    (< n 0) -1
    :else 0))
\$\endgroup\$
0
\$\begingroup\$

BotEngine, 7x17=119

v #  #0123456789
>Ie~>SSSSSSSSSSS
    ^ <<<<<<<<<<P
  PeeS-        1e
 Pe1-S0123456789
  0  > SSSSSSSSS
       >>>>>>>>>^
\$\endgroup\$
0
\$\begingroup\$

F#, 31 9 bytes

Math.Sign

(Saved 22 bytes thanks to @pmbanka's advice :))

fun(i:int)->System.Math.Sign(i)

I was about to post a C# solution.. but it already exists :(

So I decided to have some fun with F#!!

//a solution with 24 bytes, but you have to put `open System` into the file. 
//Not sure if that would be valid...

fun(i:int)->Math.Sign(i)
\$\endgroup\$
2
  • \$\begingroup\$ I think you don't need to wrap a function inside a lambda. The answer could be as simple as System.Math.Sign, with example usage System.Math.Sign 2 BTW, in F# you don't need to wrap a single argument in a parentheses, saving a byte. And as a side note, I think that C# not counting using System; is a bit weird, but maybe these are the rules here... In such case, assume open System is there as well. \$\endgroup\$
    – pmbanka
    Dec 21 '16 at 11:19
  • \$\begingroup\$ I'm not very familiar with F#, but unless System is imported by default, you have to use the fully-qualified name System.Math.Sign. \$\endgroup\$
    – user45941
    Dec 23 '16 at 14:47
0
\$\begingroup\$

Go, 49 bytes

func s(x int)int{return(x>>63)|int(uint(-x)>>63)}

Note: this won't work on the playground until Go 1.8 is released due to this bug: https://github.com/golang/go/issues/16203

\$\endgroup\$
3
  • \$\begingroup\$ Possibly doesn't work for the minimum integer (-1<<63)? \$\endgroup\$
    – Neil
    Dec 21 '16 at 10:30
  • \$\begingroup\$ @Neil nope! Should work for all 64 bit integers. \$\endgroup\$ Dec 21 '16 at 16:12
  • \$\begingroup\$ Oh, I see now, you just get -1|1 in that case, which is still -1. \$\endgroup\$
    – Neil
    Dec 21 '16 at 20:13
0
\$\begingroup\$

OCaml, 4 bytes

sign

Didn't even know there was a built-in for this. Alternative 18 bytes solution:

fun x->compare x 0
\$\endgroup\$
0
\$\begingroup\$

Stacked, 4 bytes

sign

Takes input from top of stack. Bonafide function:

$sign

Shortest non-builtin so far:

:0>\0<-

Which does:

:0>\0<-                        stack: [..., n]
:        duplicate TOS         stack: [..., n, n]
 0       push 0                stack: [..., n, n, 0]
  >      pop n,0 and push n>0  stack: [..., n, n>0]
   \     switch top two        stack: [..., n>0, n]
    0    push 0                stack: [..., n>0, n, 0]
     <   pop n,0 and push n<0  stack: [..., n>0, n<0]
      -  subtract top two      stack: [..., (n>0)-(n<0)]

Surround with [] to be a function.

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

TI-Basic, 11 bytes

:Input A
:A>0
:Ans-A<0

Reads from variable A.

:A>0            // Reads from A, returns 0 or 1
:Ans-A<0.   // Computes A<0 first, then subtracts from previous result. Implicit print.
\$\endgroup\$
1
  • \$\begingroup\$ To be fair, the other TI-Basic Answer used Ans for input, which decreases his code by 3 bytes. You could add 3 bytes to both of ours through Input A: then use the same metric. \$\endgroup\$ Dec 23 '16 at 13:30
0
\$\begingroup\$

JavaScript (ES6), 14 bytes

No conditionals, no Math lib usage

x=>(x>0)-(x<0)

Explanation

x > 0 is true/false for positives/non-positives and cast to number (1/0) in case for subtraction. x < 0 is analogous for negatives, so we end up with 1-0 for positives, 0-0 for zero and 0-1 for negatives.

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0
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Math++, 28 bytes

?>a
3+2*!a>$
a/sqrta*a
0>$
0
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1
  • \$\begingroup\$ Could you link to an interpreter? \$\endgroup\$
    – user45941
    Dec 23 '16 at 15:01
0
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Python 2, 26 Bytes

b=input()
print(b>0)-(b<0)
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