Output the sign

Question

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) {
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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;
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    var answer = jQuery("#answer-template").html();
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                   .replace("{{LINK}}", a.link);
    answer = jQuery(answer);
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  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;
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    language = jQuery(language);
    jQuery("#languages").append(language);
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body {
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  display: block !important;
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#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">

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<div id="answer-list">
  <h2>Leaderboard</h2>
  <table class="answer-list">
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      <tr><td></td><td>Author</td><td>Language</td><td>Size</td></tr>
    </thead>
    <tbody id="answers">

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<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>
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<table style="display: none">
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    <tr><td>{{LANGUAGE}}</td><td>{{NAME}}</td><td>{{SIZE}}</td><td><a href="{{LINK}}">Link</a></td></tr>
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Answer 1

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.

Answer 2

Yabasic, 16 bytes

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

input""n
?sig(n)

Try it online!

Answer 3

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

$_=$_<=>0

Try it online!

Answer 4

Pyt, 1 byte

±

Try it online!

Boring built-in answer (padding for body length)

Answer 5

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

Answer 6

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).

Answer 7

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

Answer 8

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!

Answer 9

Brain-Flueue, 40 bytes

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

Try it online!

Readable version

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

Answer 10

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.

Answer 11

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

Answer 12

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?

Answer 13

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

Answer 14

Keg, -hr, 13 bytes

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

Try it online!

A very simple switch like comparison happening here.

Answer 15

05AB1E, 2 bytes

A signum built-in.

.±

Try it online!

Answer 16

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

Try it online!

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

Answer 17

PowerShell, 22 21 bytes

[math]::sign("$args")

Boring built-in, calls the .NET function that does exactly what it says on the tin. Ho-hum.
Try it online!

-1 byte thanks to Veskah.

---

For 26 bytes however, we get the classic greater-than less-than equation

param($b)($b-gt0)-($b-lt0)

This, at least, has a little bit of logic and thought put into it. Try it online!

---

Best yet, though is 44 bytes, where we roll our own solution.

param($b)if("$b".indexof('-')){+!!$b;exit}-1

Here we take input $b, stringify it, take the .IndexOf('-') on it, and use it in an if clause. If the negative sign isn't found, this returns -1, which is truthy in PowerShell, so we turn $b into a Boolean with !, invert the Boolean with another !, cast it as an int with +, leave it on the pipeline, and exit. This turns a positive integer (which is truthy) into $false, then $true, then 1, while turning 0 into $true, then $false, then 0. Otherwise, the .IndexOf returned 0 (meaning it was the first character in the string), which is falsey, so we skip the if and just place a -1 on the pipeline. In either case, output via implicit Write-Output happens at program completion. Try it online!

Answer 18

Scratch 3.0, 9 blocks/44 bytes

For those who are tired of the old way of introducing these answers [I'm looking at you a'_' ...(ಠ ͟ʖಠ)], I'll put the ScratchBlocks Syntax first:

define f(n
set[r v]to(<(n)>(0)>-<(n)<(0
say(r

I decided to use a function instead of the old when gf clicked approach because that means I don't have to deal with having an ask() and wait next.

Try it not online but on Scratch

Answer 19

Io, 28 bytes

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

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

Try it online!

Answer 20

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.

Answer 21

Python 3, 25 bytes

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

Try it online!

Uses a different approach from other python answers

Answer 22

dc, 7 bytes

?dd*v/p

Try it online!

Input on stdin, output on stdout. (There may also be spurious output to stderr, which should be ignored, as usual.)

This uses the formula $$\frac{\left\lfloor\sqrt{n^2}\right\rfloor}n$$

plus the fact that if you divide by 0 in dc, it leaves the 0 at the top of the stack.

(If you try this out, note that negative numbers are written in dc with an initial _ instead of -, since - is only used for the two-argument subtraction operation.)

Answer 23

Desmos, 12 bytes

f(x)=sign(x)

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

Answer 24

Haskell, 6 bytes

signum

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

Answer 25

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

Answer 26

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.)

Answer 27

R, 21 bytes, no builtin

x=scan();x/abs(x^!!x)

Try it online!

Takis input from STDIN. Avoids the builtin sign. This comes out 4 bytes shorter than JAD's solution with if. We cannot use directly x/abs(x) since we would divide by 0 when x=0.

After coercion to integer, !!x will be equal to 1 for all input except 0 (when it is equal to 0). Since 0^0=1, this trick allows to avoid dividing by 0:

• For nonzero input, it is equivalent to x/abs(x^1), giving +1 or -1
• For x=0, it is equivalent to x/1, giving 0

Answer 28

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

Answer 29

Japt, 1 byte

g

Try it online!

Answer 30

Python 3.8, 47 bytes

print(1if(n:=int(input()))>0else-1if n<0else 0)