Product over a range

Question

Your task is simple: given two integers a and b, output ∏[a,b]; that is, the product of the range between a and b. You may take a and b in any reasonable format, whether that be arguments to a function, a list input, STDIN, et cetera. You may output in any reasonable format, such as a return value (for functions) or STDOUT. a will always be less than b.

Note that the end may be exclusive or inclusive of b. I'm not picky. ^_^

Test cases

[a,b) => result
[2,5) => 24
[5,10) => 15120
[-4,3) => 0
[0,3) => 0
[-4,0) => 24

[a,b] => result
[2,5] => 120
[5,10] => 151200
[-4,3] => 0
[0,3] => 0
[-4,-1] => 24

---

This is a code-golf, so the shortest program in bytes wins.

---

Leaderboard

The Stack Snippet at the bottom of this post generates the catalog 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:

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

var QUESTION_ID=66202,OVERRIDE_USER=44713;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>

Answer 1

Jelly, 2 bytes

rP

Takes two numbers as command line arguments. Try it online.

Note that this is inclusive range. For the cost of a byte (3 bytes), we can make this exclusive:

’rP

Try it online. Note that the arguments must be given in the order b a for this version.

Explanation

Inclusive

a rP b
  r   dyadic atom, creates inclusive range between a and b
   P  computes product of the list

Exclusive

b ’rP a
  ’   decrement b (by default, monadic atoms in dyadic chains operate on the left argument)
   r  range
    P product 

Answer 2

ArnoldC, 522 511 bytes

First post on codegolf !

I had fun doing this. Exclusive range.

LISTEN TO ME VERY CAREFULLY f
I NEED YOUR CLOTHES YOUR BOOTS AND YOUR MOTORCYCLE a
I NEED YOUR CLOTHES YOUR BOOTS AND YOUR MOTORCYCLE b
GIVE THESE PEOPLE AIR
HEY CHRISTMAS TREE r
YOU SET US UP 1
HEY CHRISTMAS TREE l
YOU SET US UP 1
STICK AROUND l
GET TO THE CHOPPER r
HERE IS MY INVITATION r
YOU'RE FIRED a
ENOUGH TALK
GET TO THE CHOPPER a
HERE IS MY INVITATION a
GET UP 1
ENOUGH TALK
GET TO THE CHOPPER l
HERE IS MY INVITATION b
LET OFF SOME STEAM BENNET a
ENOUGH TALK
CHILL
I'LL BE BACK r
HASTA LA VISTA, BABY

Explanations (Thanks Bijan):

DeclareMethod f
        MethodArguments a
        MethodArguments b
        NonVoidMethod
        DeclareInt r
        SetInitialValue 1
        DeclareInt l
        SetInitialValue 1
        WHILE l
                AssignVariable r
                        SetValue r
                        MultiplicationOperator a
                EndAssignVariable
                AssignVariable a
                        SetValue a
                        + 1
                EndAssignVariable
                AssignVariable l
                        SetValue b
                        > a
                EndAssignVariable
        EndWhile
        Return r
EndMethodDeclaration

Answer 3

Python, 30 bytes

f=lambda a,b:a>b or a*f(a+1,b)

Inclusive range. Repeatedly multiplies by and increments the left endpoint, until it is higher than the right endpoint, in which case it's the empty product of 1 (as True).

Answer 4

Minecraft 15w35a+, program size 456 total (see below)

This calculates PI [a,b). Input is given by using these two commands: /scoreboard players set A A {num} and /scoreboard players set B A {num}. Remember to use /scoreboard objectives add A dummy before input.

Scored using: {program size} + ( 2 * {input command} ) + {scoreboard command} = 356 + ( 2 * 33 ) + 34 = 456.

This code corresponds to the following psuedocode:

R = 1
loop:
  R *= A
  A += 1
  if A == B:
    print R
    end program

Download the world here.

Answer 5

Python 2, 44 38 bytes

lambda l:reduce(int.__mul__,range(*l))

Pretty much the obvious anonymous function answer.

EDIT: Thanks to xnor for saving 6 bytes with some features I didn't know.

Answer 6

TI-BASIC, 9 bytes

Input A
prod(randIntNoRep(A,Ans

Takes one number from Ans and another from a prompt.

Also 9 bytes, taking input as a list from Ans:

prod(randIntNoRep(min(Ans),max(Ans

Answer 7

Pyth, 5 bytes

*FrQE

Pyth doesn't have product, so we reduce * over the range.

Uses exclusive range.

Answer 8

R, 22 bytes

function(a,b)prod(a:b)

Answer 9

Mathematica, 15 bytes

1##&@@Range@##&

A shorter solution that only works for non-negative integers:

#2!/(#-1)!&

Answer 10

JavaScript (ES6), 34 bytes

(a,b)=>eval("for(c=a;a<b;)c*=++a")

Sometimes the simplest answer is the best! Just a for loop inside eval. Inclusive range.

Answer 11

Japt, 7 bytes

Easy challenges like this are always fun. :)

UoV r*1

Try it online!

Explanation

UoV r*1  // Implicit: U = first input, V = second input
UoV      // Generate range [U,V).
    r*1  // Reduce by multiplication, starting at 1.

Wow, this seems pathetic compared to the other answers so far. I need to work on Japt some more...

Answer 12

Seriously, 4 bytes

,ixπ

,         Read list [a,b] from stdin
 i        Flatten it to a b
  x       Pop a,b, push range(a,b)
   π      Pop the list and push its product.

Hex Dump:

2c6978e3

Try it online

Answer 13

Octave, 15 bytes

@(a,b)prod(a:b)

Straightforward. Uses the inclusive range.

Answer 14

Haskell, 19 17 bytes

a#b=product[a..b]

Usage example: 2#5-> 120.

Answer 15

CJam, 6 19 18 10 bytes

Thanks to Dennis and RetoKoradi for help with golfing!

q~1$-,f+:*

Try it online

Takes input as a b. Calculates PI [a,b).

Note: this program is 6 bytes long, and only works if a and b are positive.

q~,>:*

Try it online

Takes input as a b. Calculates PI [a,b).

Answer 16

Prolog, 45 bytes

Code:

p(A,B,C):-A=B,C=A;D is A+1,p(D,B,E),C is A*E.

Explained:

p(A,B,C):-A=B,      % A is unifiable with B
          C=A       % Unify C with A
          ;         % OR
          D is A+1, % D is the next number in the range
          p(D,B,E), % Recurse on the range after the first element
          C is A*E. % The result C is the product of the first element and the result 
                      of the recursion

Example:

p(5,10,X).
X = 151200

p(-4,-1,X).
X = 24

Answer 17

Bash + GNU utilities, 13

seq -s* $@|bc

Assumes there are no files in the current directory whose names start with -s. Start and end (inclusive) are passed as command-line parameters.

This simply produces the sequence from start to end, separated by *, then pipes to bc for arithmetic evaluation.

Answer 18

MATL (non-competing), 4 bytes

Inclusive Range

2$:p

Try it online!

Explanation

2$: % Implicitly grab two input arguments and create the array input1:input2
p   % Take the product of all array elements

Thanks to @Don Muesli for helping me get the hang of this whole MATL thing.

Answer 19

JavaScript (ES6), 22 bytes

I can't believe none of us JS golfers thought to use recursion...

a=>F=b=>a-b?b*F(b-1):a

Assign to a variable with e.g. var q = a=>F=b=>a-b?b*F(b-1):a, then call like q(2)(5).

Answer 20

Brachylog, 3 bytes

⟦₃×

Try it online!

Input is passed as [A,B]. The range is exclusive of B, but could be made inclusive by replacing the ₃ with ₂.

Answer 21

Jolf, 4 bytes

Try it here!

OrjJ
  jJ two inputs
 r   range between them [j,J)
O    product

Answer 22

Ruby, 22 bytes

->i,n{(i..n).reduce:*}

Ungolfed:

-> i,n {
  (i..n).reduce:* # Product of a range
}

Usage:

->i,n{(i..n).reduce:*}[5,10]
=> 151200

Answer 23

C, 32 bytes

For [a,b):

f(a,b){return a-b?a*f(a+1,b):1;}

For [a,b] (On Katenkyo's suggestions, 32 bytes again) :

f(a,b){return a<b?a*f(a+1,b):b;}

Answer 24

05AB1E, 2 bytes (non-competing)

Code:

ŸP

Explanation:

Ÿ   # Inclusive range [input, ..., input]
 P  # Total product of the list
    # Implicit printing top of the stack

Answer 25

J, 8 bytes

[:%/!@<:

Usage

>> f =: [:%/!@<:
>> f 10 5
<< 15120

where >> is STDIN and << is STDOUT.

Explanation

It computes ∏[a,b] as (b-1)!/(a-1)!.

minus_one =: <:
factorial =: !
of        =: @
monadic   =: [:
division  =: %/
f =: monadic division factorial of minus_one

Previous 13-byte version

Written when I had no idea what J even is :p

*/(}.[:>:i.)/

Usage:

   */(}.[:>:i.)/ 5 10
30240

Explanation:

*/            NB. multiply over
  (
   }.         NB. remove [the first x items] from
     [:>:     NB. increment all of
         i.   NB. the numbers from 0 to [y-1]
           )
            / NB. insert the above code into the following numbers

Detailed explanation:

i.10 would produce 0 1 2 3 4 5 6 7 8 9

>:i.10 would make it 1 2 3 4 5 6 6 7 8 9 10

the [: is used to make the ">:" take only one argument (a monad)
because if it takes two arguments, it is a different function.
so [:>:i.10 becomes 1 2 3 4 5 6 7 8 9 10

}. means take away the first [x] items from the following list,
so 5}.1 2 3 4 5 6 7 8 9 10 becomes 6 7 8 9 10

the two slashes "/" in the code are actually the same
for example, */6 7 8 9 10 becomes 6*7*8*9*10

Answer 26

Minkolang 0.14, 7 bytes

nnL$*N.

Try it here.

Explanation

nn         Takes two numbers from input
  L        Pops b,a and pushes a..b
   $*      Product the whole stack
     N.    Output as number and stop.

Answer 27

Python, 52 bytes

Very simple code; a bit too long.

def p(a,b):
 t=1
 for i in range(a,b):t*=i
 return t

Answer 28

JavaScript (ES6), 45 41 bytes

Saved 4 bytes thanks to @Cᴏɴᴏʀ O'Bʀɪᴇɴ

(a,b)=>[...Array(b-a)].reduce(x=>x*a++,1)

Seems a little too long...

(a,b)=>           // Define an anonymous function that takes parameters a and b, and returns:
[...Array(b-a)]   // An array of b-a items,
.reduce(          // Reduced by
x=>x*a++          //  multiplying each item with the previous,
,1)               //  starting at 1.

Answer 29

Julia, 16 bytes

f(a,b)=prod(a:b)

Note: if the range object a:b (which is literally stored as a start value and a stop value, and internally includes a "increment by 1 on each step" value) is permitted as the input, then just 4 bytes are required: prod.

Answer 30

Perl 6, 14 bytes

{[*] $^a..$^b}

usage:

my &code = {[*] $^a..$^b}
say code |$_ for (2,5),(5,10),(-4,3),(0,3),(-4,-1);
# 120
# 151200
# 0
# 0
# 24

say chars code 1,10000;
# 35660

If you wanted to exclude the last element use ..^ instead of ..

Answer 31

gs2, 4 bytes

Hex: 57 0e 4f 65

Mnemonics: read-nums dump crange product

Try it online!

dump pops a list from the stack and pushes its contents (like ~ in GolfScript), and crange gives [a,b).

Answer 32

PowerShell, 30 Bytes

param($a,$b)$a..$b-join'*'|iex

Takes input as two integers, creates a range with .., then -joins that with asterisks, pipes it into Invoke-Expression (similar to eval). The range operator in PowerShell is inclusive.

Pretty competitive with non-golfing languages.

Answer 33

Java 7, 54 characters

int p(int a,int b){int p=a;while(++a<b)p*=a;return p;}

We cheat the first multiply by assigning it to the temp-product. Then we pre-increment the counter (shamelessly stolen from the input parameter) to skip the first one.

Compare < b instead of <= b - both are to spec, but this saves a character.

Sample:

C:\files>type BadProd.java
class BadProd {
    public static void main(String...args) {
        System.out.println(p(2,5));
        System.out.println(p(5,10));
        System.out.println(p(-4,3));
        System.out.println(p(0,3));
        System.out.println(p(-4,0));

    }
    static int p(int a,int b){int p=a;while(++a<b)p*=a;return p;}
}
C:\files>java BadProd
24
15120
0
0
24

Answer 34

APL (Dyalog Extended), 8 3 bytes^SBCS

×/…

Try it online!

… range

×/ multiplication across

This is an "atop" so a(×/…)b is ×/a…b.

Answer 35

dc, 41 38 37 bytes

This is my first code golf, so bear with me

[lad1+sa*dstlalb>d]sd?sbdst1+saltldxf

It must be saved as a file, and invoked (with bash, so UNIX only) as

dc -f [filename]

It will proceed to wait for input, in the form of a b integers. Note that negative numbers must come as _number.

---

Alternatively,

echo "X Y [lad1+sa*dstlalb>d]sd?sbdst1+saltdxf" | dc

with X and Y replaced with the two numbers you'd like to product-over-range. This one won't wait unless you don't pipe two numbers.

---

Explanation:

The recursive macro:
[lad    # load the contents of register a onto the stack, and duplicate it
 1+sa   # Add 1 to the top of the stack (a) and store it in register a
 *dst   # Multiply two off the stack, duplicate and store in register t (stack contains t)
 lalb>d # load a and b, as long as b>a execute d (which is this macro)
 ]sd    # Store that macro in register d
 ?      # Wait for input, hope it is [a b] without braces 
 sbd    # store b, duplicate a (stack == a a) (old: sbddsa)
 st     # store a => t (stack == a)
 1+sa   # add 1 to a and store it in a
 lt     # load t since d expects the total to be on the stack
 ldx    # load d and execute it as a macro
 f      # after d is done recursively calling, t will still be on the stack.
        # Print the stack and close (close is implicit)

There was a removal of a spare duplication and store (dsa)

Answer 36

Keg -hr, 3 bytes

ɧ∑*

Try it online!

Explanation:

ɧ   # Generate inclusive range from two integer inputs
 ∑* # Calculate product of whole stack
    # Output

Answer 37

W, 3 bytes (inclusive)

.*r

Explanation

ab.   % Generate inclusive range from a to b
   *  % Multiplication
    r % Preserve the operation's action & reduce by it

Answer 38

Mouse, 36 bytes

?A:?B:A.I:(A.B.<^A.1+A:A.I.*I:)I.!$

This reads two integers from STDIN and prints an integer to STDOUT. This computes the sum over the closed interval, i.e. [a,b] rather than [a,b).

Ungolfed:

? A:             ~ Read an integer from STDIN, assign to A
? B:             ~ Read STDIN, assign B
A. I:            ~ Begin an accumulator at A
( A. B. < ^      ~ While A < B...
  A. 1 + A:      ~ Increment A
  A. I. * I:     ~ I *= A
)
I. !             ~ Print I to STDOUT
$

Answer 39

LabVIEW, 14 LabVIEW Primitives

Range is exclusive. Creates an array from a to b then multiplies all elements.

Answer 40

JavaScript (ES6), 39

Inclusive range.

(a,b)=>[...Array(b-a)].map(_=>b*=a++)|b

Note: map beats reduce once again

Answer 41

zsh, 30 bytes

f=({$1..$2})
echo $[${f// /*}]

Sample run:

manatwork% zsh product.zsh 5 10
151200

Answer 42

jq, 36 bytes

(35 characters code + 1 character command line option)

reduce range(.[0];.[1])as$i(1;.*$i)

Sample run:

bash-4.3$ bin/jq -s 'reduce range(.[0];.[1])as$i(1;.*$i)' <<< '5 10'
15120

On-line test (Passing -s through URL is not supported – so input passed as [5, 10].)

Answer 43

TI-Basic, 14 11 bytes

Prompt A,B:prod(randIntNoRep(A,B

Answer 44

Fortran, 44 bytes

Finally a use for those pesky implicit types.

function j(k,l)
j=1
do i=k,l
j=j*i
enddo
end

Test program:

program testProduct
  integer :: a, k, l
  k = -4
  l = -1
  a = j(k,l)
  print*,a
end program testProduct

Answer 45

Vitsy, 14 bytes

_{Note to self: add range}

Expects input as b a.

D{-\[D1+]l1-\*
D               Duplicate the top input.
 {-\[   ]       Repeat the stuff in the brackets by the difference between the 
                inputs.
     D1+        Duplicate, add one.
                I now have the range [a,b].
         l1-\*  Multiply all the items together.

This is a function in Vitsy that will leave the product of the range [a,b] on the stack. "But how do I test it?!?" I hear you ask?

Concatenate N. c:

Try it online!

Answer 46

dc, 27 bytes

Input assumes a is on top of the main stack and b is under it

dsar[dsc*lc1-dla<A]sAlAxrp

Ungolfed:

dsa        # duplicate 'a', store a copy in register a
r          # reverse top two items on stack
[          # start macro definition 
  dsc      # duplicate counter, store in register c 
           #  ('b' is the initial counter value)
  *        # pop top two elements of stack, multiply and push result
  lc       # copy the stored counter to the stack
  1- d     # decrement and duplicate it
  la       # copy 'a' to the stack
  <A       # if 'a' is < the counter, run macro A
]sA        # store this macro as A
lAx        # execute macro A
r          # reverse top two elements of stack
p          # pop/print the result

Answer 47

Milky Way 1.5.17, 28 bytes

1%{¢¢L§{?{¢1-e__^}}£*}!

---

Explanation

1          1             # push integer to the stack
 %{                £ }   # for loop
   ¢¢     ¢              # read a single line of input from the command line
     L                   # push a Pythonic range(0, TOS + 1)
      §{          }      # mapping
        ?{    __ }       # if-else statement
            -            # subtract the STOS from the TOS
             e           # order the TOS and STOS (greater-than)
                ^        # pop the TOS
                      !  # output the TOS

---

Usage

$ ./mw <path-to-code> -i <input>

Example:

$ ./mw test.mwg -i "2
5"

Milky Way takes line breaks literally. \n is not interpreted as a line break. In future versions, this will be changed.

Answer 48

Arcyóu, 17 bytes

(F(a b)(r *(_ a b

This includes a but excludes b.

Explanation:

(F(a b)    ; Anonymous function F(a, b)
  (r *     ; Reduce by multiplication
    (_ a b ; Range from a to b

Answer 49

PlatyPar, 4 bytes

_XF*

_ gets the range [a,b), X expands it onto the stack, and F* folds multiplication over the range. Try it online!

Answer 50

Rust, 52 51 bytes

fn p(mut a:i32,b:i32)->i32{for i in a+1..b{a*=i;}a}

Usage (+ ungolfed version) as follows:

fn p(mut a:i32, b:i32) -> i32 {
    for i in a+1..b {
        a *= i;
    }
    a
}

fn main() {
    println!("{}", p(2, 5));
    println!("{}", p(5, 10));
    println!("{}", p(-4, 3));
    println!("{}", p(-4, 0));
}

Uses the current stable version of Rust (1.5.0)

Answer 51

PHP, 42 Bytes

<?=array_product(range($ARGV[1],$ARGV[2]))

<?=                         // Open file, '=' is shorthand for 'echo'
array_product(              // Get product of array
range($argv[1],$argv[2])    // array is range from a to b
)

Answer 52

Swift, 26 58 bytes

func p(a:Int,b:Int)->Int{return(a+1..<b).reduce(a){$0*$1}}

Answer 53

Pushy, 8 bytes

Non-competing as the language postdates the challenge.

w-:&h;P#

Try it online!

Most of the code is building the range, as there is a builtin for finding the product (P).

w          \ Mirror stack, yielding [a, b, a]
 -         \ Pop (b, a) and push b - a, the difference
  :  ;     \ That many times do:
   &h      \   Push last item +1
      P#   \ Print product

The w-:&h; is essentially a long-winded binary range function, as Pushy only has unary range commands.

Answer 54

tcl, 55

incr p
while \$a<$b {set p [expr $p*$a];incr a}
puts $p

demo

Answer 55

Ahead, 7 bytes

Input is taken as two numbers on standard in. Range is [a,b].

IIEK*O@

II       read two ints and push to stack
  E      expand range between top two stack items
   K     reduce stack by...
    *    multiplying
     O   pop and print number
      @  terminate

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Answer 56

F# (.NET Core), 27 bytes

fun a b->Seq.fold(*)1[a..b]

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Answer 57

Clam, 5 bytes

p;Brr

Explanation

p     - Print...
 ;    - Product of...
  B   - Range...
   r  - Next input
    r - Next input

Answer 58

Python 2, 30 bytes

f=lambda a,b:b<a or a*f(a+1,b)

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Range is inclusive (e.g., f(2,5) => 120).

Answer 59

GolfScript, 8 bytes

Idea copied from the gs2 solution.

~,\>{*}*

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Explanation

~        # Dump the input,            e.g. 2 4
 ,       # Generate range from 0 to input: 2 [0 1 2 3 4]
  \      # Swap the stack                : [0 1 2 3 4] 2
   >     # Keep all those that are larger: [2 3 4]
    {*}* # Reduce by multiplication      : 24

Answer 60

golflua, 26 characters

\p(f,t)~@i=f+1,t f=f*i$~f$

Sample run:

Lua 5.2.2  Copyright (C) 1994-2013 Lua.org, PUC-Rio
> \p(f,t)~@i=f+1,t f=f*i$~f$
> w(p(5, 10))
151200

Answer 61

C#, 66 bytes

Not great, but not too bad.

int d(int a,int b)=>Enumerable.Range(a,b-a).Aggregate((x,y)=>x*y);

Answer 62

Hassium, 49 Bytes

func f(a,b){t=a;for(x=a+1;x<b;x++)t=t*x;print(t)}

See expanded and run online with test case here

Answer 63

Befunge, 51 bytes

 v          @.$<
v>&20p&30p1
>20g*20g:30g-!#^_1+20p

This can probably be optimised with some stack manipulation wizardry.
I went for put and get instructions instead. Might get back to it later on.
Also fails with negative numbers, since AFAIK Befunge doesn't handle them, instead looping back to maxint.

Answer 64

, 4 chars / 11 bytes

⨴⩥…ï

Try it here (Firefox only).

Creates an exclusive range from the array input, and multiplies everything in the resulting range. (Note that although the interpreter is using v2 of the language, this code still works in v1.)

Answer 65

Python 56

a=input()
b=input()
print reduce(int.__mul__,range(a,b))

Answer 66

Detour, 3 bytes (non-competing)

rP.

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Same as @quartata's Jelly answer, except . is required to output the result so it doesn't wrap back around to r.
This was not intended to be a golfing language, but the 1-character commands necessary to operate on a 2d grid make small challenges like this really short.

Answer 67

CoffeeScript, 46 45 44 29 bytes

Inclusive range

29 bytes version (anonymous function):

(a,b)->c=a;c*=++a while a<b;c

Equivalent to next JS code:

(function(a, b) {
    var c;
    c = a;
    while (a < b) {
        c *= ++a;
    }
    return c;
});

44 bytes version (prompt()):

alert [prompt()..prompt()].reduce (a,b)->a*b

45 bytes version:

alert [prompt()..prompt()].reduce((a,b)->a*b)

46 bytes version:

alert([prompt()..prompt()].reduce((a,b)->a*b))

Answer 68

Reng v.3, 25 bytes

This is encoded in ISO 8859-1, and you can try it out at the new IO page!

ii#x:1+::xe)2(¿Bh$1¶a*¡n~

Is an inclusive range.

ii takes 2 inputs, and stores the second one (the maximum) in x. :1+:: sets a to a (a+1) (a+1) (a+1). xe sets the TOS to the equality of the counter (a+1) and )2( sets the STOS to 2. ¿ pops two numbers and, if the TOS is true, the pointer moves STOS times. Otherwise, we just go forward. In the event that we have not reached the maximum, Bh goes back 11 units. $ drops the excess counter from the TOS once we are done looping. 1¶ sets the default pop from an empty stack to 1, so we can use a double-sided mirror loop to multiply everything. a is a one-sided mirror, and * multiplies values. ¡ mirrors until the stack's length is 1. Once the length is 1, n~ is met and outputs the result and exits the program.

Answer 69

Factor, 79 43 29 bytes

Hooray for auto use!

It's quite elegant, really. A lambda, that goes on the stack.

[ 1 - [a,b] 1 [ * ] reduce ]

Answer 70

Mathcad, 4 "bytes"

Uses the built-in Iterated Product Operator, which is entered from the keyboard by ctl-#. Type k into the iterator and expression placeholders and then type the evaluation operator (=), giving a total of 4 "bytes" where one byte is taken to be the number of characters needed to enter an expression.

Answer 71

Matricks, 24 bytes

Matricks is a new esolang I invented to deal with matrices. Run like: python matricks.py [[<num1>,<num2>]]

mg:;+c:1:g:1;-g:;;kp{};;

This is a simple answer. Matricks has a built-in for taking the product of an array, p, so all I need to do is make the range with the constructor, m

Answer 72

Java 7, 49 48 bytes

int c(int a,int b){return-~a!=b?a*c(a+1,b):a*b;}

corsiKa's answer is exclusive, mine is inclusive (and slightly shorter by using recursion instead of a loop).

Explanation:

Try it here.

int c(int a,int b){  // Method with two integer parameters and integer return-type
  return-~a!=b?      //  If `a+1` is not equal to `b`
         a*c(a+1,b)  //   Return `a` multiplied by the recursive call with `a+1,b`
        :            //  Else:
         a*b;        //   Return `a*b`
}                    // End of method

Answer 73

k/kona, 8 bytes

*/a+!b-a

Let's break this down: k reads right-to-left, so:

b-a finds the difference between b and a, to give the number of numbers required

!(b-a) generates a list from 0 to (b-a), giving the range [0, (b-a))

a+(!b-a) adds a to each element of the list, to get the range [a, b)

*/(a+!b-a) applies multiplication over all elements of the list; * is the multiply operator, and / is the over adverb, which is like a map/reduce function, applying its left-hand argument over all elements of its right-hand argument.

Answer 74

Racket (scheme), 29 bytes

Functional programming woot

Apply applies a function to all elements of a list, and range creates a list ranging from a to b. The function being applied is *. Exclusive.

(λ(a b)(apply *(range a b)))

Answer 75

Fith, 17 bytes (non-competing)

This language was created after the challenge.

{ range product }

Anonymous function. Excludes b. Pretty self-explanatory.

Stack effect: a b -- *[a,b)

Answer 76

><>, 33 bytes

 1&:@$:@)?\&n;
:$@:+1&*&:/?@@$@)@

My first fish attempt.

Inclusive range
Assumes a and b are on the stack.

Answer 77

Perl 6 - 15 bytes

{[*] $^a..$^b}

Usage:

> {[*] $^a..$^b}(2,5)
120

Answer 78

Forth, 33 bytes

Pretty simple looping program.

: f DUP ROT DO DUP I * LOOP ; f .

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Answer 79

JavaScript (using external library) (31 bytes)

 (a,b)=>_.RangeTo(a,b).Product()

Link to lib: https://github.com/mvegh1/Enumerable/

Explanation of code: Anonymous method accepts low bound and high bound for range, and uses built in .Product method to produce the product for that range

Answer 80

C#, 56 Bytes

int f(int a,int b){var r=1;for(;a<b;a++)r*=a;return r;}

Answer 81

Excel, 37 bytes

=PRODUCT(A2-ROW(OFFSET(A1,,,A2-A1)))

Data will be a in A1 and b in A2.

The above should be entered as an array formula (ctrl_shift_enter) in any other cell. The result is exclusive of b.

Answer 82

Triangular, 21 20 bytes

$\S=t$iUprP%..%/*U:<

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Takes input with order b, a from standard in.

Ungolfed: (Explanation to be added later)

      $ 
     \ S 
    = t $ 
   i U p r 
  P % . . % 
 / * U : <

---

Previous Version (21 bytes):

1\P\p$?Ud$=0%)(/P*U:<

Answer 83

JavaScript (ES6), 25 bytes

f=(a,b)=>a-b?a*f(a+1,b):b

Inclusive.

Answer 84

Perl 5 -p, 22 bytes

$_=eval join'*',$_..<>

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Answer 85

Perl 6, 14 bytes

{[*] [...] @_}

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Reduce the two inputs by range, then reduces by multiplication.