Multiply two numbers

Question

Input: Two decimal integers. These can be given to the code in standard input, as arguments to the program or function, or as a list.

Output: Their product, as a decimal integer. For example, the input 5 16 would lead to the output 80.

Restrictions: No standard loopholes please. This is code-golf, answer in lowest amount of bytes wins.

Notes: Layout stolen from my earlier challenge, Add two numbers.

Test cases:

1 2   -> 2
4 5   -> 20
7 9   -> 63
-2 8  -> -16
8 -9  -> -72
-8 -9 -> 72
0 8   -> 0
0 -8  -> 0
8 0   -> 0
-8 0  -> 0
0 0   -> 0

Or as CSV:

a,b,c
1,2,2
4,5,20
7,9,63
-2,8,-16
8,-9,-72
-8,-9,72
0,8,0
0,-8,0
8,0,0
-8,0,0
0,0,0

Leaderboard

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Answer 1

dc, 3

?*p

Try it online (wrapped in a shell script to run all given tests).

Note dc thinks underscores (_) are -ve signs.

Answer 2

Befunge-93, 5 bytes

&&*.@

Try it online!

&& gets two integers and pushes them to the stack; * multiplies them (duh); . prints the numeric value of the top of the stack, and @ ends the program.

Answer 3

Emacs Lisp, 3 bytes

This is just a function.

(*)

Usage:

(* num1 num2)

Answer 4

Microscript, 4 bytes

isi*

Microscript II, also 4 bytes

FsF*

Answer 5

Noodel, noncompeting 1 byte

Did not have a full complete version of Noodel until after this challenge, and did not even have multiplication with a single character until recently.

×

Try it:)

How it works

  # Input is implicitly pushed onto the stack with the first element (a) pushed first and the second last (b) making it the top.
× # Pops off two items producing => (b * a) and pushes on the result.
  # The top of the stack is popped off at the end of the program and pushed to stdout.

<div id="noodel" code="×" input="2,-4" cols="5" rows="5"></div>

<script src="https://tkellehe.github.io/noodel/noodel-latest.js"></script>
<script src="https://tkellehe.github.io/noodel/ppcg.min.js"></script>

Answer 6

Swift 3, 40 bytes

func f(_ a:Int,_ b:Int)->Int{return a*b}

Takes two parameter, return the two Ints multiplied. Called like this:

f(2, 4)

Answer 7

C (gcc), 98 bytes

m(i,j){i=(i&j)<0?m(O(~i,1),-j):i?O(m(i>>1,O(j,j)),i&1?j:0):0;}O(i,j){i^=j,j&=i^j;i=j?O(i,j<<1):i;}

I don't think I have much of a chance against the other entries.

Try it online!

Answer 8

Taxi, 442 bytes

Go to Post Office:w 1 l 1 r 1 l.Pickup a passenger going to The Babelfishery.Pickup a passenger going to The Babelfishery.Go to The Babelfishery:w 1 l 1 r.Pickup a passenger going to Multiplication Station.Pickup a passenger going to Multiplication Station.Go to Multiplication Station:n 1 r 2 l.Pickup a passenger going to The Babelfishery.Go to The Babelfishery:s 1 r 1 l.Pickup a passenger going to Post Office.Go to Post Office:e 1 l 1 r.

Formatted with line breaks for legibility:

Go to Post Office:w 1 l 1 r 1 l.
Pickup a passenger going to The Babelfishery.
Pickup a passenger going to The Babelfishery.
Go to The Babelfishery:w 1 l 1 r.
Pickup a passenger going to Multiplication Station.
Pickup a passenger going to Multiplication Station.
Go to Multiplication Station:n 1 r 2 l.
Pickup a passenger going to The Babelfishery.
Go to The Babelfishery:s 1 r 1 l.
Pickup a passenger going to Post Office.
Go to Post Office:e 1 l 1 r.

Answer 9

ZX80 BASIC ~39 bytes

 1 INPUT A
 2 INPUT B
 3 PRINT A;"X";B;"=";A*B

This program is so simple that it will very likely work on all variants on 8-bit BASIC. However, on the ZX80 (4K ROM) you are limited to integer maths with a signed 16-bit range; so if your answer is >32767 then it will error and you will not see a result.

To remedy this, install a new ROM onto your Sinclair ZX80 to allow 24-bit floating point maths, or upgrade to a Sinclair ZX81.

Save bytes by refactoring line 3 to:

 3 PRINT A*B

Answer 10

Alice, 7 6 bytes

*/
o@i

Try it online!

Follows the same pattern as my solution for addition.

Answer 11

C, 56 bytes

main(int i,char**a){printf("%d",atoi(*++a)*atoi(*++a));}

compile

> gcc source.c

usage

> a 8 7

or

> ./a.out 8 7

explanation

main(int i,char**a) {  /* Begin program and accept command-line arguments.    */
   printf("%d",        /* First argument prints the (final) decimal result.   */
         atoi(*++a)    /* Returns int of program's 2nd command-line argument. */
         *             /* Multiplies the values of the 1st & 2nd arguments.   */
         atoi(*++a)    /* Returns int of program's 1st command-line argument. */
   );
}

Answer 12

Jellyfish, 6 bytes

P*i
 i

Try it online!

P denotes print, then * multiplies the two numbers below and to the right. At the start of execution, the two is are replaced with values from STDIN.

Answer 13

Lean Mean Bean Machine, 12 bytes

OO
ii
 /
*
u

Answer 14

Whitespace, 4 bytes

In some other languages, you could quote that as "\t \n". Expects two numbers on the stack and leaves the result on the stack.

As a complete program reading from standard input and writing to standard output, it would be 36 bytes:

Try it online!

Answer 15

Perl 5, 7 bytes

6 bytes of code + 1 for -p flag

$_*=<>

Try it online!

Answer 16

Lua, 32 bytes

i=io.read
print(i("*n")*i("*n"))

Try it online!

Program reads 2 numbers from STDIN and prints the result to STDOUT.

---

"But shouldn't you use io.write?"

Well according to the documentation,

[...] write uses the current output file, whereas print always uses the standard output. Finally, print automatically applies tostring to its arguments, so it can also show tables, functions, and nil.

Answer 17

Go, 84 bytes

package main
import."fmt"
func main(){a,b:=0,0;Scanf("%d%d",&a,&b);Printf("%d",a*b)}

Try it online!

Worth mentioning:

• We import the fmt package to the global namespace by using import.. That way, we can refer to Scanf and Printf as global functions.

• a and b are assigned to 0, as that is shorter than doing var a,b int.

Answer 18

Go, 28 bytes

func(a,b int)int{return a*b}

Try it online!

Answer 19

Ruby, 10 bytes

:*.to_proc

Try it online!

I really thought this might be shorter...

Answer 20

C#, 27 bytes

delegate(a,b){return a*b;};

Answer 21

Aceto, 6 bytes

riri*p

ri grabs input and converts to integer * multiplies them p prints it Try it online!

Answer 22

QBIC, 13 bytes

::[a|p=p+b}?p

Adapted my code from 'add two numbers'. This starts a loop and adds 'y' to itself 'x' times.

But seriously, a 6-byte solution is ::?a*b. : gets a cmd line parameter and class it 'a', the next : does the same for 'b', 'cause 'a' is already taken. * multiplies, and ? prints the result. This is virtually identical to this answer, only the operator is different.

---

Since some time now, QBIC can in-line the 'get var from cmd line'-command, and the above would be ?:*:, at 4 bytes.

Answer 23

Forked, 5 bytes

$$*%&

Try it online!

• $$ - read two integers
• * - multiply top two stack values
• % - print top of stack as integer
• & - "terminate", prevent IP from wrapping

Answer 24

FRACTRAN, 22 bytes

78/55 5/3 1/5 11/2 5/7

Take 7^a*11^b as input, return 11^b*13^(a*b).

Try it online!

This is a golfed version of the wikipedia example.

I'm not sure whether or not extra bytes should be added for input encoding and output decoding.

Answer 25

Brainfuck, 76 bytes

Not the shortest but my own solution, written a long time ago.

Golfed

>[-]>[-]>[-]<<<<[->>>+<<<]>[->+>>+<<<]>[-<+>]>>[-<[-<+<<+>>>]<[->+<]>>]<<<<.

Explanation

x = x*y
(x)(y)(temp0)(xtemp)(ycount)
Own work! Convoluted solution

,               x
> +++           y=3 (or input)

>[-]>[-]>[-]    set cells to 0
<<<< [->>>+<<<] transfer x to xtemp
> [->+>>+<<<]   transfer y to temp0 and ycount
> [-<+>]        transfer temp0 to y
>>
[               while ycount != 0
 -               ycount decrement
 < [-<+<<+>>>]   transfer xtemp to x and temp0
 < [->+<]        transfer temp0 to xtemp
 >> 
]

<<<< .          x (might overflow)

Answer 26

Flobnar, 4 bytes

&
*@

Try it online!

Answer 27

Commodore BASIC v2 (C64, VIC-20 and some PET models), 27 tokenized BASIC bytes

This version uses the limited DEF FN keyword, See description here

 0 DEFFNA(B)=A*B:INPUTA,B:PRINTFNA(B)

Commodore BASIC v2, 14 tokenized BASIC bytes

 0 INPUTA,B:PRINTA*B

Both methods are likely compatible with all 8 BIT BASIC variants on Commodore home computers, but are not tested.

Answer 28

Bourne-shell-based langs all look pretty similar:

Zsh, 11 bytes

<<<$[$1*$2]

Bash, 13 bytes

echo $[$1*$2]

Dash/POSIX sh, 15 bytes

echo $(($1*$2))

Answer 29

GFortran, 51 27 bytes

Fulfils the rules, I suppose integer overflow error is to be expected.

-24 thanks to Ørjan Johansen

Try It Online!

read(*,*)i,j
print*,i*j
end

Answer 30

3+ bytes in various methods of representing lambda calculus, taking numeral input as Church numerals.

19 Bytes, Untyped Lambda Calculus, using named arguments:

λx.λy.λz.x(y(z))

This is the most verbose version, . represents a lambda abstraction and parentheses represent an application

9 or 12 Bytes, Lambda Calculus, using De Brujin Indices:

λλλ3(2 1)

This uses De Brujin Indices instead of variable names, and function application is right-associative and implicit.

Replacing lambda symbols with backslashes (a pretty common form when Unicode is not available) yields us only 6 bytes

It's small, but we can go smaller. Using John Tromp's Binary Lambda Calculus representation, we get the binary representation 0000000111100111010, or separated into individual parts:

3 bytes, Untyped Lambda Calculus, using BLC representation

00 00 00 01 1110 01 110 10 xxxxx

Where the 5 xs are any padding bits.

Explanation

These implement the simple mult function for Church Numerals in the untyped lambda calculus.

In untyped lambda calculus, EVERYTHING is a function, including numbers. All these functions take a maximum of a single argument. So how do you represent arbitrary numbers? Church encodings!

In Church encoding, numbers are represented by repeated recursion. All numbers n are represented by a function taking an argument f, and returning a function taking an argument x, which returns f(x) called n times. For example:

0: lambda f: lambda x: x
1 = lambda f: lambda x: f(x)
2 = lambda f: lambda x: f(f(x))
3 = lambda f: lambda x: f(f(f(x)))

Etc etc.

All inputs and outputs in most lambda calculus systems are in the form of church numerals.

The multiplication function takes 3 arguments and returns the first argument called on the second called on the third:

mult = lambda m: lambda n: lambda x: m(n(x))

This in lambda calculus simply multiplies the numbers m and n together.

Calling mult(2)(3) using our previous definitions of 2 and 3 results in a function:

lambda f: lambda x: f(f(f(f(f(f(x))))))

Or 6, the result of our multiplication. All the "golfs" are really just more concise ways of representing untyped lambda calculus.