Product over a range

Question

Your task is simple: given two integers \$a\$ and \$b\$, output \$\Pi[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

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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.

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## Language Name, N bytes

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## Ruby, <s>104</s> <s>101</s> 96 bytes

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## Perl, 43 + 2 (-p flag) = 45 bytes

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

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Answer 1

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 2

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

Try it online!

Answer 3

F# (.NET Core), 27 bytes

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

Try it online!

Answer 4

Clam, 5 bytes

p;Brr

Explanation

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

Answer 5

Python 2, 30 bytes

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

Try it online!

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

Answer 6

Keg -hr, 3 bytes

ɧ∑*

Try it online!

Explanation:

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

Answer 7

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 8

Vyxal, 2 bytes

rΠ

Range of a to b, then product

Try it!

Answer 9

Pyth, 4 bytes

*F}F

Test suite

Range is inclusive.

Answer 10

Factor + math.unicode, 12 bytes

[ [a,b] Π ]

Try it online!

It's a quotation (anonymous function) that takes 2 integers from the data stack as input and leaves one integer on the data stack as output.

• [a,b] Create a range from two integers. As you may have guessed, this is inclusive of both. All of the other words are also available, such as [a,b).
• Π Find the product.

Answer 11

Pip, 6 bytes

$*Ya,b

Try it online!

-1 byte from DLosc.

Answer 12

HBL, 2.5 bytes

*(0.,

Uses inclusive range. Try it at HBL Online!

Explanation

After parentheses autocompletion, we have '(*(0.,)):

'(*(0.,))
'(      )  Define a function which returns
  *        the product of
   (0  )   the range from
     .     the first argument to
      ,    the second argument

Answer 13

Regex 🐇 (PCRE2 _{^{v10.35 or later}}), 128 bytes

^(?!-x*,(?!-))(-())?+(?=(x*),-?(x*))(?=(x*),?(?=\3)(?=\4)(x*,-?|[^,]*)(x*))\5,?+(?=(\2(xx)*(?11))?)((?*x+)x?+)*(?*x+)(\5\7(,|$))

Attempt This Online!

Takes its input in unary, as two strings of x characters whose lengths represent the numbers, with optional - signs on their left side, joined by a , delimiter, specifying an inclusive range. Returns the absolute value of its output as the number of ways the regex can match. (The rabbit emoji indicates this output method.)

Additionally, capture group \8 is set iff the result is negative. This value is taken from the last found possible match, if there is a conflict. But this regex always returns a consistent value in it within a run.

It takes two shortcuts:

• It assumes that the left argument is less than the right argument. This isn't necessary, and in fact removing this assumption wouldn't cost many bytes.
• If the two arguments have different signs, it doesn't even bother calculating anything, because the result will always be zero. This isn't a necessary shortcut (it worked without it), but it results in a much shorter regex.

^
(?!
    -x*,(?!-)           # If A is negative and B is nonnegative, let there
                        # be no possible matches, as the product would be 0.
)
(-())?+                 # \2 = set if A is negative
(?=
      (x*),             # \3 = abs(A); 
    -?(x*)              # \4 = abs(B)
)
(?=
    (x*),?(?=\3)(?=\4)  # \5 = what to skip so tail = the larger of A or B;
                        # tail = the larger of A or B
    (x*,-?|[^,]*)(x*)   # \7 = {the smaller of A or B} - \5
)
\5                      # tail = the larger of A or B
,?+                     # skip sign of B, if any
                        # First handle the range from larger inclusive to
                        # {smaller inclusive or 0 exclusive}
(?=
    (\2(xx)*(?11))?     # \8 = set iff the inclusive range is negative and has
                        #      an odd number of total members, i.e. the product
                        #      is negative
)
(
    (?*x+)              # Multiply the number of possibilities by tail.
    x?+                 # Advance by 1, but if already at the end, allow the
                        # loop to iterate once more so as to include 0 in
                        # the count.
)*                      # Iterate any number of times (cover the whole range)
(?*x+)                  # Multiply the number of possibilities by tail again
                        # at the end of the range.
(                       # Define subroutine (?11)
    \5\7                # Clamp the range at the smaller end.
    (,|$)               # Assert that we've reached the end.
)

As far as I know, this absolutely requires non-atomic lookaround (used here as (?*...) molecular lookahead) and 🐇-output, so it would not be portable to other regex engines currently – though it would be easily portable to 🐇 ECMAScript+(?*).

Answer 14

Fig, \$2\log_{256}(96)\approx\$ 1.646 bytes

rR

See the README to see how to run this

Calculates using the range [a, b] (because of bug). Explanation:

rR # Takes input as "a b"
 R # Range [a, b]
r  # Product

Answer 15

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 16

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 17

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 18

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 19

Python 56

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

Answer 20

𝔼𝕊𝕄𝕚𝕟, 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 21

Detour, 3 bytes (non-competing)

rP.

Try it online!

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 22

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 23

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 24

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 25

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 26

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 27

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 28

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 29

Perl 6 - 15 bytes

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

Usage:

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

Answer 30

Forth, 33 bytes

Pretty simple looping program.

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

Try it online