Input: Dua bilangan bulat desimal. Ini dapat diberikan ke kode dalam input standar, sebagai argumen untuk program atau fungsi, atau sebagai daftar.

Output: Produk mereka, sebagai bilangan bulat desimal. Misalnya, input 5 16akan mengarah ke output 80.

Batasan: Tolong, jangan ada celah standar. Ini kode-golf , jawab dengan jumlah byte terendah yang menang.

Catatan: Tata letak dicuri dari tantangan saya sebelumnya, Tambahkan dua angka .

Kasus uji:

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

Atau sebagai 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

Papan peringkat

var QUESTION_ID=106182,OVERRIDE_USER=8478;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>

Brachylog V1, 05AB1E, J, K, Underload, MATL, Forth, PigeonScript, Stacked, Implisit, Jolf, Clojure, Braingolf, 8, Common Lisp, Julia, Pyt, Appleseed, Stax, Reality, dc 1 byte

*

Anda dapat mengedit jawaban ini untuk menambahkan bahasa lain yang *merupakan jawaban yang valid.

C (GCC), 13 byte

Tidak berfungsi pada semua implementasi, tapi tidak apa-apa.

f(a,b){a*=b;}

Cobalah di TIO!

Beatnik , 888 byte

k I
j k ZZZZX z
xw k C vp yQ KD xw z j k ZZZZX z
j k ZZZD z xw bZ ZX
k XX z qs xw vp xw xw vp xw vp vp vp k I Xj ZZD hd
xw yQ K k ZZZZX xo exx
qs yQ XA xw xw xw xw z xw bZ K
xw xw k I
j k ZZZZX z
xw k C vp yQ XA hd k I z j k ZZZZX z
j xw k A vp bZ ZX
k ZZZZX z qs xw vp xw xw vp xw vp vp vp k I Xj ZZD hd
xw yQ K k ZZZZX xo exx
qs yQ F k ZZZZK xo
vp
xw xw z qs xw bZ X xw k I z xw Xj K
qs xw bZ KA vp qs xw Xj C hd
qs z xw xw xw xw z qs
xw xw xw xw z qs k I qs k I z xw Xj ZC
qs bZ ZZZX qs xw yQ C hd xw
k I vp qs k I qs
xw Xj ZZC hd hd z Kz ZZD
k I z xw xw xw xw z qs k I qs k I Xj ZZZZF
z
xw xw z qs xw bZ X xw k I z xw Xj K
qs xw bZ KA vp qs xw Xj C hd
z qs xw
xw xw z qs xw bZ X xw k I z xw Xj K
qs xw bZ KA vp qs xw Xj C hd
z vp
xw xw z qs
xw xw z qs
k I qs
xw bZ ZZX k I z qs k I vp
xw k ZA z yQ ZA hd qs k I vp qs k I Xj ZZKD
qs xw Xj ZZK
hd qs xw Xj ZZZZ hd
k ZZZZKD vp xo xw Xj K

Cobalah online!

Saya menggunakan juru bahasa C karena juru bahasa Python di TIO secara tidak sengaja mengeksekusi alamat jika kondisi untuk melompat mundur tidak terpenuhi. Solusi yang mudah untuk juru bahasa Python adalah menambahkan beberapa nops untuk membuat alamat nop. Saya percaya tidak ada yang benar:

                                   C       Python  My interpretation
IP after skiping N words           IP+N+1  IP+N+2  IP+N+2
IP after skiping back N words      IP-N    IP-N+1  IP-N+2
IP after not skiping N words       IP+2    IP+2    IP+2
IP after not skiping back N words  IP+2    IP+1    IP+2

Input harus berupa dua bilangan bulat yang dipisahkan oleh spasi, tanpa tertinggal baris baru.

Jawaban ini bekerja secara teori untuk semua bilangan bulat, jika setiap sel dapat menyimpan nilai besar yang sewenang-wenang, tidak terbatas pada 0 - 255. Tetapi ia meluap jika | A | + | B | > 22. Dan itu berjalan sangat lambat jika | A | + | B | > 6. Jadi tidak banyak kasus yang dapat Anda uji dan solusi jika-untuk kasus-kasus tersebut mungkin lebih pendek.

Idenya adalah untuk menghitung angka segitiga T (N) = N (N + 1) / 2 dengan menurunkan nilai menjadi 0 dan menjumlahkan semua nilai perantara. Maka kita bisa mendapatkan jawaban A * B = T (A + B) - T (A) - T (B).

Tetapi sulit untuk menghitung semua 3 nilai. Ini dilakukan dengan terlebih dahulu menghitung T (A + B) - A, meninggalkan salinan A di tumpukan untuk ditambahkan kembali nanti, dan menggunakan input B. Kemudian secara rekursif menemukan bilangan segitiga terbesar lebih kecil dari itu, yaitu T ( A + B-1) kecuali untuk nol kasus khusus. Kita bisa mendapatkan kembali B = T (A + B) - A - T (A + B-1) dan menghitung T (B) dari sana.

Angka N adalah angka segitiga jika sama dengan angka segitiga terbesar lebih kecil dari N, ditambah jumlah angka segitiga non-negatif lebih kecil dari N. Ini berjalan dalam O (2 ^ (T (A + B) -A)) dan merupakan bagian paling lambat dalam program ini.

k I                                         Push 1
j k ZZZZKAAA z                              Input and decrement by 48.
xw k AAA vp yQ (input_a_loop)               If the character was '-':
xw z j k ZZZZKAAA z                           Replace with 0 and input another.
input_a_loop:
j k ZZZAA z xw bZ (input_a_end)             Input and break if it is a space.
k ZKA z qs xw vp xw xw vp xw vp vp vp       Otherwise multiply the previous
                                              value by 10 and add.
k I Xj (input_a_loop)                       Continue the loop.
input_a_end: hd                             Discard the space.
xw yQ (check_sign) k ZZZZKAAA xo exx        If A=0, print 0 and exit.
                                            Stack: ?, A_is_positive, A
check_sign:
qs yQ (check_sign_else)                     If A is positive... or not,
xw xw xw xw z xw bZ (check_sign_end)          in either cases, push 2 copies
check_sign_else: xw xw k I                    of A and the negated flag back
check_sign_end:                               as a constant.
                                            Stack: A, A, A, A_is_negative
j k ZZZZKAAA z                              Similar for B.
xw k AAA vp yQ (input_b_loop)               If the character was '-':
hd k I z j k ZZZZKAAA z                       Decrement the flag and input another.
input_b_loop:
j xw k A vp bZ (input_b_end)                EOF is checked instead of a space.
k ZZZZKAAA z qs xw vp xw xw vp xw vp vp vp
k I Xj (input_b_loop)
input_b_end: hd
xw yQ (output_sign) k ZZZZKAAA xo exx       If B=0, print 0 and exit.
                                            Stack: A, A, A, A*B_is_negative, B
output_sign:
qs yQ (output_sign_end) k ZZZZK xo          If negative, output '-'.
output_sign_end:

vp                                          Add.        Stack: A, A, A+B
xw xw z qs                                  Insert a 0. Stack: A, A, 0, A+B.
xw bZ { xw k I z xw Xj }                    Copy and decrement while nonzero.
                                            Stack: A, A, 0, A+B, A+B-1, ..., 0
qs xw bZ { vp qs xw Xj } hd                 Add while the second value in the
                                              stack is nonzero.
                                            Stack: A, A, T(A+B)
qs z xw xw xw xw z qs                       Stack: A, C0=T(A+B)-A, C0, F0=0, C0

expand_loop:
xw xw xw xw z qs k I qs                     Stack: A, C0, C0, F0=0,
                                              ..., [P=C, P, S=0, F=1], C
dec_expand: k I z xw Xj (expand_loop)       Decrement and continue if nonzero.
                                            Stack: [P=1, P, S, F], C=0
                                            The last number 0 is assumed to
                                              be a triangular number.
test: qs bZ (extract_end)                   If F=0, break.
qs xw yQ (test_not_first) hd xw             If S=0, it's the first triangular
                                              number below previous C. Set S=C.
test_not_first: k I vp qs k I qs            S+=1 and restore F=1.
xw Xj (dec_expand)                          If C!=0, recursively expand from C-1.
hd hd z Kz (test)                           If S=P, P is a triangular number,
                                              return to the previous level.
k I z xw xw xw xw z qs k I qs               Otherwise, decrement P and try again.
k I Xj (dec_expand)
extract_end:                                Stack: A, C0, C0, T(A+B-1)

z                                           Subtract and get B.
xw xw z qs xw bZ { xw k I z xw Xj }         Computes T(B).
qs xw bZ { vp qs xw Xj } hd
                                            Stack: A, C0, T(B)
z qs xw                                     Stack: C0-T(B), A, A

xw xw z qs xw bZ { xw k I z xw Xj }         Computes T(A).
qs xw bZ { vp qs xw Xj } hd
z vp                                        Get A*B=(C0-T(B))+(A-T(A))
xw xw z qs                                  Stack: 0, X=A*B

divide: xw xw z qs                          Stack: 0, ..., Y=0, X
subtract: k I qs                            Stack: 0, ..., Y, Z=1, X
xw bZ {                                     While X!=0:
k I z qs k I vp                               X-=1, Z+=1.
xw k ZA z yQ (not_ten)                        But if Z=11:
hd qs k I vp qs k I Xj (subtract)               Y+=1, reset Z and restart the loop.
not_ten: qs xw Xj }
hd qs xw Xj (divide)                        Put Z under Y and make Y the new X,
                                              continue the loop if X!=0.
hd                                          Discard X.

print_loop:
k ZZZZKAA vp xo xw Xj (print_loop)          Add each cell by 47 and print.

Mathematica, 4 byte

1##&

Contoh penggunaan: 1##&[7,9]pengembalian 63. Memang, fungsi yang sama ini mengalikan sejumlah argumen dari jenis apa pun secara bersamaan.

Seperti diketahui codegolfers Mathematica, ini bekerja karena ##mengacu pada seluruh rangkaian argumen untuk suatu fungsi, dan gabungan dalam Mathematica (sering) mewakili multiplikasi; jadi 1##mengacu pada (1 kali) produk dari semua argumen fungsi. Ini &hanya kependekan dari Functionperintah yang mendefinisikan fungsi murni (tanpa nama).

Di dalam kode lain, simbol umum *berfungsi sebagai perkalian. Begitu juga spasi, sehingga 7 9ditafsirkan sebagai 7*9(memang, versi REPL saat ini dari Mathematica benar-benar menampilkan ruang seperti tanda-tanda perkalian!). Bahkan lebih baik, jika Mathematica dapat mengetahui di mana satu token dimulai dan yang lainnya berakhir, maka tidak ada byte sama sekali yang diperlukan untuk operator perkalian: 5ysecara otomatis diartikan sebagai 5*y, dan 3.14Log[9]sebagai 3.14*Log[9].

Retina , 38 37 31 byte

Pendekatan yang sama sekali baru, yang lama di bawah ini.

M!`-
*\)`-¶-

.* 
$*_
_
$'$*_
_

Cobalah online!

Penjelasan

Pertama, kita berurusan dengan tanda:

M!`-

cocok dengan semua -dalam string dan mengembalikannya dipisahkan oleh baris baru

*\)`-¶-

(dengan baris kosong berikut)
*\)berarti hasil dari ini dan tahap sebelumnya harus dicetak tanpa baris baru, dan kemudian string dikembalikan ke apa sebelumnya (string input). Bagian yang tersisa menghapus dua- dipisahkan oleh baris baru.

Lalu kami mengonversi angka pertama menjadi unary:

.* 
$*_

(ada spasi di akhir baris pertama). Kami menggunakan _sebagai digit unary kami dalam hal ini, karena digit standar1 dapat hadir di angka kedua, dan ini akan bertentangan nanti.

Sekarang kita sampai pada perkalian yang sebenarnya:

_
$'$*_

Masing _- masing digantikan oleh representasi unary dari semua yang mengikutinya (masih menggunakan_ sebagai digit unary). Karena konversi ke unary mengabaikan karakter non-digit, ini akan mengulangi representasi unary dari angka kedua untuk kali "angka pertama". Angka kedua akan tetap dalam representasi desimal di akhir string.

Pada akhirnya, dengan satu _kita mengembalikan jumlah _dalam string, yang akan menjadi hasil dari perkalian.

Jawaban sebelumnya: (peringatan: mengeluarkan string kosong ketika seharusnya menampilkan 0)

Retina ,  45  42 41 byte

Ayo main game! Gandakan angka relatif dengan bahasa yang tidak memiliki operator aritmatika dan dukungan terbatas hanya untuk bilangan alami ... Kedengarannya lucu :)

O^`^|-
--

\d+
$*
1(?=1* (1*))?
$1
1+
$.&

Penjelasan

Tiga baris pertama berhubungan dengan tanda:

O^`^|-

Ini memilah Odan kemudian membalik ^semua string yang cocok dengan regex ^|-. Dalam prakteknya ini cocok dengan string kosong di awal, dan tanda minus akhirnya sebelum nomor kedua, dan mengatur ulang mereka menempatkan string kosong di tempat minus. Setelah ini, semua -berada di awal string, dan sepasang dari mereka dapat dihapus dengan mudah dengan dua baris berikutnya.

Setelah itu, kami menggunakan builtin untuk mengonversi angka menjadi representasi unary, dan kemudian muncul perkalian yang sebenarnya:

1(?=1* (1*))?
$1

Kami cocok dengan siapa saja 1, dan gantikan mereka masing-masing dengan semua 1setelah ruang berikut. Setiap digit angka pertama akan diganti dengan angka kedua penuh, sedangkan setiap digit angka kedua akan digantikan oleh string kosong.

Bagian terakhir lagi-lagi adalah builtin untuk mengkonversi kembali dari unary ke desimal.

Cobalah online!

Gores , 1 byte

Penggunaan: Tempatkan angka di kedua sisi *tanda

Catatan: Karena Scratch adalah bahasa visual saya tidak bisa mengetahui berapa banyak byte yang dikonsumsi sampai @ mbomb007 mencatat saya tentang metode untuk menghitung byte awal

Brain-Flak , 56 byte

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

Ini harus dijalankan sebagai program lengkap karena tidak menumpuk bersih dan input harus menjadi satu-satunya elemen dalam tumpukan.

Cobalah online!

Penjelasan: (panggil input x dan y)

Bagian 1:

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

([                    ]) # Push negative x on top of:
      ([      ])         # negative y. After...
  ({}<            >)     # pushing x and...
        ({})             # y...
            <>  <>  <>   # on the other stack (and come back)

Pada titik ini kita memiliki [x, y] pada satu tumpukan dan [-x, -y] pada yang lain.

Bagian 2:

{({}[()]<(({})<({}{})>)>)<>}{}{}
{                          }     # Loop until x (or -x) is 0
 ({}[()]<              >)        # Decrement x
         (({})<      >)          # Hold onto y
               ({}{})            # Add y and the number under it (initially 0)
                         <>      # Switch stacks
                            {}{} # Pop x and y leaving the sum

JavaScript (ES6), 9 byte

ES6 memiliki fungsi khusus untuk bilangan bulat 32-bit, lebih cepat dari *operator yang lebih umum .

Math.imul

Kebetulan, ini hanya selama:

a=>b=>a*b

Brain-Flak , 56 54 52 byte

2 byte disimpan berkat kesalahan yang ditangkap oleh Nitrodon

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

Cobalah online!

Stack versi bersih, 62 60 byte

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

Cobalah online!

Penjelasan

Penjelasan ini lebih merupakan penjelasan dari algoritma yang terlibat dan tidak menyertakan kode yang sebenarnya. Diasumsikan bahwa Anda tahu cara membaca Brain-Flak dengan mahir. Jika Anda perlu bantuan memahami kode atau algoritme, saya akan senang mengedit atau merespons jika Anda meninggalkan komentar.

Ini sedikit aneh dan menggunakan beberapa matematika aneh yang baru saja berhasil. Hal pertama yang saya lakukan adalah membuat loop yang selalu berakhir pada langkah O (n) . Cara normal untuk melakukan ini adalah dengan meletakkan n dan -n pada tumpukan yang berlawanan dan menambahkan satu ke masing-masing sampai satu mencapai nol, namun saya melakukannya dengan cara yang agak asing. Dalam metode saya, saya meletakkan penghitung di bawah input dan setiap langkah saya menambah penghitung menambahkannya ke n dan membalik tanda n .

Mari kita telusuri contoh. Katakan n = 7

7  -8   6  -9   5 -10   4 -11   3 -12   2 -13   1 -14   0
0   1   2   3   4   5   6   7   8   9  10  11  12  13  14

Saya tidak akan membuktikannya di sini, tetapi ini akan selalu berakhir untuk input apa pun dan akan melakukannya dalam sekitar 2n langkah. Bahkan itu akan berakhir dalam langkah 2n jika n positif dan langkah 2n-1 jika n negatif. Anda dapat mengujinya di sini .

Sekarang kita memiliki sekitar 2n langkah dalam loop kita bagaimana kita mengalikan dengan n ? Nah di sini ada beberapa keajaiban matematika. Inilah yang kami lakukan: Kami membuat akumulator, setiap langkah dari proses kami menambahkan input kedua ( m ) ke akumulator dan membalik tanda keduanya, kami kemudian mendorong total pada semua loop yang terjadi, ini adalah produk.

Kenapa bisa begitu?

Baiklah mari kita telusuri contoh dan mudah-mudahan itu akan menjadi jelas. Dalam contoh ini kita mengalikan 5 dengan 3 , saya hanya akan menunjukkan nilai-nilai penting

total       -> 0  -5   5 -10  10 -15  15
accumulator -> 0  -5  10 -15  20 -25  30
m           -> 5  -5   5  -5   5  -5   5

Semoga mekanismenya jelas di sini. Kami sedang melangkah melalui semua kelipatan m dalam urutan nilai absolutnya. Anda kemudian pemberitahuan bahwa 2n ^th jangka selalu m * n dan istilah sebelum selalu * -m n . Ini membuatnya sehingga perulangan kita sejajar dengan hasil yang kita inginkan. Sedikit kebetulan yang menyenangkan;)

Dyalog APL , 1 byte

× mengambil satu nomor di sebelah kiri, dan satu di sebelah kanan

×... atau bahkan beberapa nomor di kiri atau di kanan atau di kedua sisi

×/ multiplies all numbers in a list

×/¨ multiplies the pairs in a given list

×/∊ mulitplies all numbers in an array

This applies to all arithmetic functions, arrays of all sizes and ranks, and numbers of all datatypes.

R, 3 bytes

'*'

This is a function which takes exactly two arguments. Run as '*'(a,b).

See also prod which does the same thing but can take an arbitrary number of arguments.

ArnoldC, 152 bytes

HEY CHRISTMAS TREE c
YOU SET US UP 0
GET TO THE CHOPPER c
HERE IS MY INVITATION a
YOU'RE FIRED b
ENOUGH TALK
TALK TO THE HAND c
YOU HAVE BEEN TERMINATED

Try it online!

Hexagony, 9 bytes

?{?/*!@'/

Try it online!

This is actually fairly straightforward. Here is the unfolded version:

  ? { ?
 / * ! @
' / . . .
 . . . .
  . . .

The / just redirect the control flow to the second line to save bytes on the third. That reduces the code to this linear program:

?{?'*!@

This linear code on its own would actually be a valid solution if the input was limited to strictly positive numbers, but due to the possibility of non-positive results, this isn't guaranteed to terminate.

The program makes use of three memory edges in a Y-shape:

A   B
 \ /
  |
  C

The memory pointer starts on edge A pointing towards the centre.

?   Read first input into edge A.
{   Move forward to edge B.
?   Read second input into edge B.
'   Move backward to edge C.
*   Multiply edges A and B and store the result in C.
!   Print the result.
@   Terminate the program.

I ran a brute force search for 7-byte solutions (i.e. those that fit into side-length 2), and if I didn't make a mistake (or there's a busy-beaver-y solution that takes a long time to complete, which I doubt) then a 7-byte solution doesn't exist. There might be an 8-byte solution (e.g. by reusing the ? or using only one redirection command instead of two /), but that's beyond what my brute force search can do, and I haven't found one by hand yet.

Piet, 16 bytes

5bpiaibpikibptai

Online interpreter available here.

Explanation

To run, paste the code above in the text box on the right side of the linked page. Below is a graphical representation of this code with codel size 31. The grid is for readability and may interfere with traditional Piet interpreters.
The code runs linearly from left to right, going along the top of the image until the first green block, where program flow moves to the middle row of codels. The white lone white codel is necessary for program flow. It could be replaced with a codel of any color other than green or dark blue, but I have chosen white for readability.

Instruction    Δ Hue    Δ Lightness    Stack
-----------    -----    -----------    -----
In (Number)    4        2              m
In (Number)    4        2              n, m
Multiply       1        2              m*n
Out (Number)   5        1              [Empty]
[Exit]         [N/A]    [N/A]          [Empty]

If you think that text is not the best way to represent a Piet program or have an issue with the byte size of Piet programs in general, please let your opinion be known in the discussion on meta.

BitCycle -U, 68 bytes

  >    > v
 ?+ >  +
Bv ?^ v ~
 \  v<CB~\v
 Cv  ^  <\/
^ <@=!   <
0A^

Try it online!

Multiplying two numbers is not a trivial problem in BitCycle, especially when signs need to be handled! This is my second attempt; the first one (essentially same algorithm, different layout) was 81 bytes, so it's quite possible this one could be shortened too.

The program takes the two numbers as command-line arguments and outputs to stdout. The -U flag is to convert the decimal numbers to signed unary, since BitCycle knows only of 0's and 1's.

Explanation

This explanation assumes you understand the basics of BitCycle (see Esolangs or the GitHub readme). I'll base my explanation on this ungolfed version, seen here computing -2 times 3:

Overview

Signed unary numbers consist of the sign (0 for nonpositive, empty for positive) followed by the magnitude (a number of 1s equal to the number's absolute value). To multiply two of them, we need to XOR the signs (output a 0 if exactly one of them is 0, or nothing if both or neither are) and then multiply the magnitudes (and output that many 1s). We'll achieve the multiplication by repeated addition.

Sign bits

Starting from the two sources ?, we split off the signs from the magnitudes using +. 0s (sign bits) turn left and are directed along the top row, while 1s (magnitudes) turn right and end up in the two B collectors.

The section that handles the signs looks like this:

  v

  \  v
> \  /

! <

If both numbers are nonpositive, two 0 bits come in from the top v. The first one reflects off the top \, is sent southward, and reflects off the /. Meanwhile, the second bit passes through the deactivated top \ and reflects off the bottom \. The two bits pass each other, go straight through the now-deactivated splitters on the bottom row, and go off the playfield.

If only one of the numbers is nonpositive, one 0 comes in from the top. It bounces around all three splitters and ends up going northward again, until it hits the v and is once more sent south. This time, it passes through the deactivated splitters and reaches the <, which sends it into the sink !.

Loops to store the magnitudes

The magnitude of the first number goes into the B collector in this section:

B v
  \
  C v
^   <

0 A ^

Before the B collector opens, the A collector releases the single 0 that was placed in it, which then goes onto the end of the queue in B. We'll use it as a flag value to terminate the loop when all the 1 bits in B are gone.

Each time the B collectors open, the \ splitter peels off the first bit from the queue and sends it to the processing logic in the middle. The rest of the bits go into C, and when the C collectors open, they are sent back into B.

The magnitude of the second number goes into the B collector in this section:

v   ~
C B ~
    <

When the B collectors open, the bits go into the bottom dupneg ~. The original 1 bits turn right and are sent west into the processing logic in the middle. The negated copies (0s) turn left and immediately hit another dupneg. Here the 0s turn right and go off the playfield, while the (now doubly) negated 1s turn left and are sent into C. When C opens, they go back into B.

Repeated addition

The central processing logic is this part:

   v
   v


@  =  !

Bits from both loops (one from the western side, and everything from the eastern side) are sent south into the switch =. The timing has to be set up so that the bit from the western loop gets there first. If it is a 1, the switch changes to }, sending the following bits eastward into the sink ! to be output. Once all the 1s are gone, we get the 0, which changes the switch to {. This sends the following bits into the @, which terminates the program. In short, we output the (unary) magnitude of the second number as many times as there are 1s in the (unary) magnitude of the first number.

Python 3, 11 bytes

int.__mul__

Try it online!

Also works for integers under 2**32 in Python 2.

Java 8, 10 9 bytes

a->b->a*b

Try it here.

Java 7, 31 bytes

int c(int a,int b){return a*b;}

Try it here.

As full program (99 90 bytes):

interface M{static void main(String[]a){System.out.print(new Long(a[0])*new Long(a[1]));}}

Try it here.

Pyth, 2 bytes

*E

Try it here!

Pyth's automatic evaluation gets in the way here. To get around it, I'm using explicit evaluation for one of the arguments

TI-Basic, 2 bytes

Very straightforward.

prod(Ans

PHP, 21 bytes

<?=$argv[1]*$argv[2];

takes input from command line arguments. Also works with floats.

Retina, 39 35 bytes

Thanks to Leo for letting me use an idea of his that ended up saving 4 bytes.

[^-]

*\)`--

.+
$*
\G1
_
_|1+
$'
1

Input is linefeed-separated.

Try it online! (Space-separated test suite for convenience.)

Explanation

The first two stages print a minus sign if exactly one of the two inputs is negative. They do this without actually changing the input. This is done by grouping them in the second stage with ) and turning them into a dry-run with *. The \ option on the second stage prevents printing a trailing linefeed.

[^-]

First, we remove everything except the minus signs.

*\)`--

Then we cancel the minus signs if there are two of them left.

.+
$*

Now we convert each line to the unary representation of its absolute value. This will get rid of the minus sign because $* only looks for the first non-negative number in the match (i.e. it doesn't know about minus signs and ignores them).

\G1
_

The first line is converted to _, by matching individual 1s as long as their adjacent to the previous match (hence, we can't match the 1s on the second line, because the linefeed breaks this chain).

_|1+
$'

This performs the actual multiplication. We replace each _ (on the first line) as well as the entire second line everything after that match. The _ matches will therefore include the entire second line (multiplying it by the number of 0s in the first line), and the second line will be removed because there is nothing after that match. Of course the result will also include some junk in the form of _s and linefeeds, but that won't matter.

1

We finish by simply counting the number of 1s in the result.

MATLAB, 5 4 bytes

@dot

dot takes the dot product of two vectors of equal length. If we feed it with two scalars, it will simply multiply the two numbers.

prod takes the product of the values in all rows of each column of a matrix. If the matrix is one-dimensional (i.e. a vector), then it acts along the non-singleton dimension, taking the product of all elements in the vector.

dot is one byte shorter than prod which is one byte shorter than the even more obvious builtin times.

Call it as such:

@dot
ans(3,4)
ans = 
   12

PigeonScript, 1 byte

*

Explanation:
* looks to the stack to see if there is anything there. If not, it prompts for input and multiplies the inputs together

Perl 6, 4 bytes

&[*]

This is just the ordinary infix multiplication operator *, expressed as an ordinary function. As a bonus, if given one number it returns that number, and if given no numbers it returns 1, the multiplicative identity.

><>, 5 Bytes

i|;n*

Takes input as an ascii character, outputs a number.

Explanation:

i                        | Get input.
 |                       | Mirror: Change the pointer's direction.
i                        | Get input again.
    *                    | Loop around to the right side. Multiply
   n                     | Print the value on the stack, as a number
  ;                      | End the program

You could also do

ii*n;

But I feel my solution is waaay cooler.

Another possibility is dropping the semicolon, which would result in the pointer bouncing off the mirror, hitting the print command, and throwing an error since the stack is empty.

Intel 8080 machine code, MITS Altair 8800, 28 bytes

This implements binary multiplication on the Intel 8080 CPU (c. 1974) which did not have multiplication or division instructions. Inputs are 8-bit values and the product is a 16-bit value returned in the BC register pair.

Here is the machine code along with step-by-step instructions to load the program into an Altair 8800 using the front panel switches.

Step    Switches 0-7    Control Switch  Instruction Comment
1                       RESET
2       00 001 110      DEPOSIT         MVI  C, 5   Load multiplier into C
3       00 000 101      DEPOSIT NEXT                value is 5
4       00 010 110      DEPOSIT NEXT    MVI  D, 16  Load multiplicand into D
5       00 010 000      DEPOSIT NEXT                value is 16
6       00 000 110      DEPOSIT NEXT    MVI  B, 0   clear B register (high byte of result)
7       00 000 000      DEPOSIT NEXT
8       00 011 110      DEPOSIT NEXT    MVI  E, 9   set loop counter E multiplier size
9       00 001 001      DEPOSIT NEXT                (8 bits + 1 since loop ends in middle)
10      01 111 001      DEPOSIT NEXT    MOV  A, C   move multiplier into A for shift
11      00 011 111      DEPOSIT NEXT    RAR         shift right-most bit to CF
12      01 001 111      DEPOSIT NEXT    MOV  C, A   move back into C
13      00 011 101      DEPOSIT NEXT    DCR  E      decrement loop counter
14      11 001 010      DEPOSIT NEXT    JZ   19 00  loop until E=0, then go to step 27
15      00 011 001      DEPOSIT NEXT
16      00 000 000      DEPOSIT NEXT
17      01 111 000      DEPOSIT NEXT    MOV  A, B   move sum high byte into A
18      11 010 010      DEPOSIT NEXT    JNC  14 00  add if right-most bit of 
19      00 010 100      DEPOSIT NEXT                multiplier is 1, else go to 22
20      00 000 000      DEPOSIT NEXT
21      10 000 010      DEPOSIT NEXT    ADD  D      add shifted sums
22      00 011 111      DEPOSIT NEXT    RAR         shift right new multiplier/sum
23      01 000 111      DEPOSIT NEXT    MOV  B, A   move back into B
24      11 000 011      DEPOSIT NEXT    JMP  08 00  go to step 10
25      00 001 000      DEPOSIT NEXT
26      00 000 000      DEPOSIT NEXT
27      11 010 011      DEPOSIT NEXT    OUT  255    display contents of A on data panel
28      11 111 111      DEPOSIT NEXT
30      01 110 110      DEPOSIT NEXT    HLT         Halt CPU
31                      RESET                       Reset program counter to beginning
32                      RUN
33                      STOP

Try it online!

If you've entered it all correctly, on the machine state drawer in the simulator your RAM contents will look like:

0000    0e 05 16 10 06 00 1e 09 79 1f 4f 1d ca 19 00 78 
0010    d2 14 00 82 1f 47 c3 08 00 d3 ff 76

Input

Multiplier in C register, and multiplicand into D. The stock Altair has no STDIN so input is by front panel switches only.

Output

The result is displayed on the D7-D0 lights (top right row) in binary.

5 x 16 = 80 (0101 0000)

4 x 5 = 20 (0001 0100)

7 x 9 = 63 (0011 1111)

8 x -9 = -72 (1011 1000)

Compatibility note: this should also run on the IMSAI 8080, though currently untested.

C#, 10 bytes

a=>b=>a*b;

It's just a simply multiplication.

Jelly, 1 byte

×

Try it online!

Obligatory Jelly submission.

Clojure, 1 byte

*

:P As a bonus this works on any number of arguments:

[(*)
 (* 2)
 (* 2 3)
 (* 2 3 4)
 (* 2 3 4 5)] => [1 2 6 24 120]

Interestingly you can easily get its source code:

(source *)
(defn *
  "Returns the product of nums. (*) returns 1. Does not auto-promote
  longs, will throw on overflow. See also: *'"
  {:inline (nary-inline 'multiply 'unchecked_multiply)
   :inline-arities >1?
   :added "1.2"}
  ([] 1)
  ([x] (cast Number x))
  ([x y] (. clojure.lang.Numbers (multiply x y)))
  ([x y & more]
     (reduce1 * (* x y) more)))

Owk, 11 bytes

λx.λy.x*y

This can be assigned to a function like this:

multiply:λx.λy.x*y

and called like this:

result<multiply(a,b)