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Task

Given a non-negative integer \$n\$, evaluate the factorial \$n!\$.

The factorial is defined as follows:

$$ n!=\begin{cases}1 & n=0\\n\times(n-1)!&n>0\end{cases} $$

Rules

  • All default I/O methods are allowed.
  • Standard loopholes are forbidden.
  • Built-ins are allowed.
  • There is no time or memory limit.
  • Giving imprecise or incorrect results for large inputs due to the limit of the native number format is fine, as long as the underlying algorithm is correct. Specifically, it is not allowed to abuse the native number type to trivialize the challenge, which is one of the standard loopholes.
  • This is . Shortest code in bytes wins, but feel free to participate in various esolangs (especially the ones hindered by the restrictions of the former challenge).

Test cases

0! = 1
1! = 1
2! = 2
3! = 6
4! = 24
5! = 120
6! = 720
7! = 5040
8! = 40320
9! = 362880
10! = 3628800
11! = 39916800
12! = 479001600

Note: We already have the old factorial challenge, but it has some restrictions on the domain, performance, and banning built-ins. As the consensus here was to create a separate challenge without those restrictions so that more esolangs can participate, here it goes.

Also, we discussed whether we should close the old one as a duplicate of this, and we decided to leave it open.

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147 Answers 147

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Pyramid Scheme, 510 bytes

    ^           ^     ^
   / \         / \   / \
  /set\       /   \ /out\
 ^-----^     /loop \-----^
/a\   /]\   ^-------^   /b\
---  ^---^ /a\      -^  ---
    / \ /#\---      /]\
   /set\---^       ^---^
  ^-----^ / \     / \  -^
 /b\   ^-/   \   /set\  -^
 ---  ^-/line \ ^-----^  -^
     /1\-------/b\   ^-  / \
     ---       ---  /*\ /set\
                   ^---^-----^
                  /b\ /a\   /-\
                  --- ---  ^---^
                          /a\ /1\
                          --- ---

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It works!

I spent way too long debugging this. There's a lot of empty space which I can't seem to get rid of.

This is basically the following pseudocode:

a = input
b = 1
while a
  b *= a
  a -= 1
print b
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JavaScript, 19 bytes

f=n=>n?f(n-1n)*n:1n

2 bytes longer than the other JS answers, but can safely calculate as high as 11190! instead of 18!, with anything higher exceeding the call stack size (tested on Node.js 18.6.0). the result has 40450 digits.

update: running with --stack-size=65536 allows me to calculate up to 95200!, resulting in 432625 digits. Trying to calculate any higher caused segmentation faults, no matter what I set the stack size to.

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2
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><>, 14 bytes

1$:@?!n$:1-@*!

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Explanation

1               # initialize product as 1
 $:@?!n         # if the current counter (starts as n) is 0, print the product
       $:1-@    # subtract 1 from the counter for the next iteration
            *   # multiply current counter with product
             !  # skip the product initialization for the next iteration
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2
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Uiua, 5 bytes

/×+1⇡

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    ⇡  # create range [0..input)
  +1   # amend it to [1..input]
/×     # reduce by multiplication
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2
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Vyxal 3L, 2 bytes

/*

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Fold / the input by multiplication *. Fold treats whole numbers as the range from 1 to n.

The factorial built-in is 4 bytes: fact

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1
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SNOBOL4 (CSNOBOL4), 65 bytes

 i =input
 p =1
i x =x + 1
 p =p * x
 output =p le(i,x) :f(i)
end

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1
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Desmos, 2 bytes

n!

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Finally a competitive Desmos answer that isn't graphical-output!

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1
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Wolfram Language (Mathematica), 9 3 bytes

-6 bytes thanks to @att

#!&

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  • 3
    \$\begingroup\$ 3 bytes \$\endgroup\$
    – att
    Aug 25, 2020 at 5:00
1
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MathGolf, 1 byte

!

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Explanation:

!  # Get the factorial (aka gamma(n+1)) of the (implicit) input-integer
   # (output the entire stack joined together implicitly as result)
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1
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Burlesque, 4 bytes

ri?!

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Explanation:

ri    # Read integer
  ?!  # Calculate factorial
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1
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Lua, 36 bytes

x=1 for i=1,...do x=x*i end print(x)

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On Lua <= 5.2 this will use double, allowing big inputs to work. On Lua 5.3+ (as currently on TIO) integers are used instead, making big inputs fail. This can be worked around at cost of one byte:

Lua, 37 bytes

x=1. for i=1,...do x=x*i end print(x)

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1
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cQuents, 5 bytes

$0:$!

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The first three bytes are for 0-indexing, since by default cQuents uses 1-indexing.

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1
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C# (Visual C# Interactive Compiler), 30 bytes

int f(int n)=>n==0?1:n*f(n-1);

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  • 1
    \$\begingroup\$ Suggest n<1 instead of n==0 \$\endgroup\$
    – ceilingcat
    Aug 31, 2020 at 9:36
1
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PHP, 38 bytes

Recursive...

function f($n){return$n?$n*f($n-1):1;}

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Or functional...

PHP, 39 bytes

fn($n)=>$n?array_product(range($n,1)):1

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1
  • \$\begingroup\$ If anyone knows if there's a way to do 7.4 arrow functions with recursion I'd be very interested to know! \$\endgroup\$
    – 640KB
    Aug 25, 2020 at 19:24
1
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Python 2, 46 bytes

lambda n:reduce(lambda a,b:a*b,range(1,n+1),1)

How it works:

the reduce function takes in the list that contains all the numbers from 1 to n and reduces it by the multiplication function (lambda a,b:a*b) to a single number. An optional initial parameter is set in the case that n is equal to 0.

Try it online (with all test cases)!

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  • 1
    \$\begingroup\$ You can do -~b to just do range(n). You've also got a trailing space in your submission \$\endgroup\$
    – Jo King ♦
    Aug 27, 2020 at 6:08
  • 1
    \$\begingroup\$ If you didn't notice yet, TIO has a Code Golf submission generator. \$\endgroup\$
    – Bubbler
    Aug 28, 2020 at 2:26
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C#, 42 bytes

Using the power of fresh and new C# 9 we can achieve a stunning 42 bytes!

int f(int n)=>n<2?1:n*f(n-1);return f(10);

for C# 8 and the online example we need to add 38 bytes for a total of 70 bytes

class P{static int Main(){int f(int n)=>n<2?1:n*f(n-1);return f(10);}}

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1
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ink, 50 bytes

==function f(n)
{
-!n:~return 1
}
~return n*f(n-1)

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Had to make it a proper function instead of just a stitch, since I actually have to use the return value (so outputting by printing is not an option - or at least not a good one).

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Python 3, 30 27 bytes

f=lambda n:1>>n or n*f(n-1)

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1
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Kotlin, 37 bytes

fun a(n:Int)=(1..n).fold(1){a,b->a*b}

Had to use fold(1){a,b->a*b}(surprisingly enough 1 less byte than something like fold(1,Int::times)) due to a lack of a product function in the stdlib.

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1
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BRASCA v0.4, 11 bytes

v0.4 was made after my initial solution and added the r operator, hence the seperate answer.

ig0$r[*$]xn

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Explanation

ig          - Convert to numbers and concatenate digits
  0$r       - Push range (i, 0) to stack
     [*$]   - Multiply it all
         xn - Remove the 0 and output the result

BRASCA, 15 14 bytes

ig[:1-]x[*$]xn

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Explanation

ig                   - Convert to numbers and concatenate
  [:1-]              - Generate a sequence from N to 0 on the stack
       x             - Remove the 0
        [*$]         - Multiply it all together
            $n       - Bring the number to the front and output it
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1
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Javascript (ES6),19 bytes

f=n=>n?f(n-1)*n:1
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  • 1
    \$\begingroup\$ I think there is a typo in your code. That will always return 1. \$\endgroup\$
    – EasyasPi
    Feb 12, 2021 at 14:13
  • \$\begingroup\$ @EasyasPi My bad, forgot to multiply. Works now. \$\endgroup\$
    – emanresu A
    Feb 12, 2021 at 21:33
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Tcl, 35 bytes

proc F x {expr $x?\[F $x-1]*($x):1}

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1
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Pxem, 26 bytes (filename) + 0 bytes (content) = 26 bytes.

  • Filename (escaped): \001._.c.w.t.m.!.m\001.-.c.a.s.n
  • Content: empty

Usage

  • Input decimal integer from STDIN
  • Output to STDOUT

How it works

XX.z
.a\001XX.z # push one
.a._.c.wXX.z # push getint; dup; while pop!=0; do
  .a.t.m.!XX.z # heap=pop; push heap; push pop*pop
  .a.m\001.-XX.z # push heap; push 1; push abs(pop-pop)
  .a.cXX.z # dup
.a.a.s.nXX.z # done; pop; printf "%d" pop
.a

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1
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M4, 49 bytes

define(f,`ifelse($1,0,1,`eval($1*f(decr($1)))')')

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Usage

f(n)dnl where n is an integer.
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1
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Python 3, 56 bytes

def f(n):
 k=1
 for i in range(2,n+1):
  k=k*i
 return k

Try it online!

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1
  • \$\begingroup\$ Welcome to Code Golf! \$\endgroup\$
    – Makonede
    Apr 16, 2021 at 16:25
1
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Pinecone, 45 bytes

f::{Int}:(k:1;i:1|i<=in|i:i+1@k:k*i;print:k)
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1
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Scala 3, 79 bytes

import compiletime.ops.int._
type F[N<:Int]=N match{case 0=>1 case _=>N*F[N-1]}

Try it in Scastie

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1
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Red, 40 bytes

f: func[n][either n = 0[1][n * f n - 1]]

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I'm surprised there are no Red answers here yet. I'm still pretty new to Red, so there may be golf potential here. But I condensed it as much as I could (all the whitespace is necessary). It's the simple recursive definition: if n = 0, return 1, else return n * f(n-1).

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Wren, 36 bytes

var F=Fn.new{|n|n<1?1:n*F.call(n-1)}

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1
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Cubix, 24 bytes

I:;.^!^u.>($.sr*u..;uO;@

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And here's a link to the Cubix online interpreter you can run in "debug" mode if you want to see the IP as it moves around the cube the code on.

http://ethproductions.github.io/cubix/?code=STo7Ll4hXnUuPigkLnNyKnUuLjt1TztA&input=NgoK&speed=6

I find it hard to explain code in this language, since the directions the code takes a confusing (since it's running through code wrapped around a cube). The size of the cube used varies based on the length of the code. In this case each side has 4 characters.

Here goes...

I:;.^!^u.>($.sr*u..;uO;@
    ^                    - code starts here, change IP to "up"
I:                       - read a number from STDIN, duplicate it on stack
         >               - change IP to "right" (start of loop)
          (              - decrement top of stack
           $             - skip the next instruction
    ^                    - SKIPPED
     !                   - if top of stack not 0, skip next instruction
      ^                  - TRUE, change IP to "up" (exit path 0)
       u                 - FALSE, IP does a "u turn"
               *         - multiple top two entries on the stack
              r          - rotate top three stack entries twice
             s           - flip the top two entries on the stack
                   ;     - delete the top of stack
                u        - IP does a "u turn", (end of loop)
  ;                      - exit path 1, delete top of stack
I                        - exit path 2, read STDIN as number onto stack
           $             - exit path 3, skip next instruction
                   ;     - exit path 4, SKIPPED
                      ;  - exit path 5, delete top of stack
                    u    - exit path 6, IP makes a "u turn"
                     O   - exit path 7, output top of stack as a number
                       @ - exit path 8, halt program
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