Add two numbers

Question

Input: Two integers. Preferably decimal integers, but other forms of numbers can be used. These can be given to the code in standard input, as arguments to the program or function, or as a list.

Output: Their sum. Use the same format for output integers as input integers. For example, the input 5 16 would lead to the output 21.

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

Notes: This should be fairly trivial, however I'm interested to see how it can be implemented. The answer can be a complete program or a function, but please identify which one it is.

Test cases:

1 2 -> 3
14 15 -> 29
7 9 -> 16
-1 8 -> 7
8 -9 -> -1
-8 -9 -> -17

Or as CSV:

a,b,c
1,2,3
14,15,29
7,9,16
-1,8,7
8,-9,-1
-8,-9,-17

Leaderboard

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Answer 1

Intel 8080 machine code, Altair 8800, 22 bytes

This will add two integers of nearly any size (16, 32, 64-bit, etc) using an Intel 8080 8-bit CPU (c. 1974). This is implemented as a full program, running on a MITS Altair 8800.

Code listing and programming instructions:

Step    Switches 0-7    Control Switch  Instruction Comment
1                       RESET
2       00 001 110      DEPOSIT         MVI  C, 2   ; loop counter set to word size
3       00 000 010      DEPOSIT NEXT                ; value is 2
4       00 010 001      "               LXI  D, 16H ; load address of first term into E:D
5       00 010 110      "                           ; at memory address 16H
6       00 000 000      "          
7       00 100 001      "               LXI  H, 18H ; load address of second term into H:L
8       00 011 000      "                           ; memory address 16H + word size = 18H
9       00 000 000      "          
10      10 101 111      "               XRA  A      ; clear accumulator
11      00 011 010      "               LOOP:LDAX D ; load [E:D] into A
12      10 001 110      "               ADC  M      ; add [H:L] + previous carry, to A
13      00 010 010      "               STAX D      ; store result in A to [E:D]
14      00 001 101      "               DCR  C      ; decrement loop counter
15      11 001 010      "               JZ   DONE   ; if counter is zero, end addition
16      00 010 101      "                           ; jump to step 23
17      00 000 000      "       
18      00 010 011      "               INX  D      ; first term next byte
19      00 100 011      "               INX  H      ; second term next byte
20      11 000 011      "               JMP  LOOP   ; restart loop
21      00 001 001      "                           ; jump to step 11
22      00 000 000      "           
23      01 110 110      "               DONE:HLT    ; halt CPU
24      00 001 110      "                           ; Term 1 low byte (14) 
25      00 000 000      "                           ; Term 1 high byte (0)
26      00 001 111      "                           ; Term 2 low byte (15)         
27      00 000 000      "                           ; Term 2 high byte (0)    
31                      RESET                       ; Reset program counter to beginning
32                      RUN
33                      STOP            
34      00 010 110      EXAMINE                     ; low byte of output displayed on D7-D0
35                      EXAMINE NEXT                ; high byte of output displayed on D7-D0

Try it online!

If entered correctly, RAM contents should look like:

0000    0e 02 11 16 00 21 18 00 af 1a 8e 12 0d ca 15 00 
0010    13 23 c3 09 00 76 0e 00 0f 00

I/O:

The above program adds two 16-bit integers, located in memory address 16H and 18H. This could accept larger integers, for example 32-bit integers by changing steps 3 and 8 to values 4 and 20H respectively, and then input would be in 16H and 20H. Numbers are represented in memory as little endian.

Output is displayed on lights D7-D0.

Example: 14 + 15 = 29 = 00 011 101

Example: 1234 + 4321 = 5555 = 00 010 101, 10 110 011

Answer 2

Microsoft PowerPoint (macro-free), 176 animations, 73 shapes on main slide

(Numbers estimated from XML explorer)

I made a 2-bit adder with carry input and carry output. Numbers represented in binary. Get the PPTX here. Directions are included in the presentation.

I'm thinking this is very golf-able, but it's a fun proof of concept anyway. Obviously, this can be extended to more bits, but frankly, programming in PowerPoint is extremely tedious with its current UI, so I'm going to move on to other things.

How it works

When the presentation starts, it is in input mode. The switches toggle a motion path animation back and forth to indicate 1 or 0.

When the Run button is clicked, it steps through each of the input bits and uses the K-maps for the carry and result bits. If there is a value dependent on the result (such as the carry bit over the second column), it uncovers a button that allows the machine to read that bit. When the final digit is clicked, the machine enters a halt state, displaying the final result in the bottom row.

Answer 3

SPDT, 1168 bytes

p 0
t 0 1
t 0 2
t 0 3
t 0 4
t 0 5
t 0 6
t 0 7
t 0 8
t 0 9
t 0 10
t 0 11
t 0 12
t 0 13
t 0 14
t 0 15
t 0 16
r 1 0 33 34
r 9 34 33 17
r 1 0 35 36
r 9 18 36 35
r 2 0 40 41
r 10 37 41 40
r 37 0 42 43
r 17 20 43 42
r 2 0 44 45
r 10 45 44 38
r 37 0 46 47
r 17 47 46 39
r 38 0 48 19
r 39 0 48 19
r 3 0 52 53
r 11 49 53 52
r 49 0 54 55
r 19 22 55 54
r 3 0 56 57
r 11 57 56 50
r 49 0 58 59
r 19 59 58 51
r 50 0 60 21
r 51 0 60 21
r 4 0 64 65
r 12 61 65 64
r 61 0 66 67
r 21 24 67 66
r 4 0 68 69
r 12 69 68 62
r 61 0 70 71
r 21 71 70 63
r 62 0 72 23
r 63 0 72 23
r 5 0 76 77
r 13 73 77 76
r 73 0 78 79
r 23 26 79 78
r 5 0 80 81
r 13 81 80 74
r 73 0 82 83
r 23 83 82 75
r 74 0 84 25
r 75 0 84 25
r 6 0 88 89
r 14 85 89 88
r 85 0 90 91
r 25 28 91 90
r 6 0 92 93
r 14 93 92 86
r 85 0 94 95
r 25 95 94 87
r 86 0 96 27
r 87 0 96 27
r 7 0 100 101
r 15 97 101 100
r 97 0 102 103
r 27 30 103 102
r 7 0 104 105
r 15 105 104 98
r 97 0 106 107
r 27 107 106 99
r 98 0 108 29
r 99 0 108 29
r 8 0 112 113
r 16 109 113 112
r 109 0 114 115
r 29 32 115 114
r 8 0 116 117
r 16 117 116 110
r 109 0 118 119
r 29 119 118 111
r 110 0 120 31
r 111 0 120 31
l 18
l 20
l 22
l 24
l 26
l 28
l 30
l 32
l 31

Try it: https://radvylfprograms.com/spdt

---

SPDT is an esolang I've been wanting to make for a while now. Aside from I/O, it consists entirely of Single Pole Double Throw, or Form C, relays. This answer, which is possibly imperfectly golfed since I used a custom transpiler with macros to generate it, consists of relays arranged into AND, OR, and XOR gates, which are arranged into half adders and full adders, which are then used to make an 8-bit adder. This could be scaled up arbitrarily, at an \$n\log(n)\$ byte count for \$n\$ input bits.

I generated it with the following SPDT-c code, with SPDT-c being a transpiled language I made shortly after that makes it easier to design complex circuits (run it at https://radvylfprograms.com/spdt/c):

t 0 1
t 0 2
t 0 3
t 0 4
t 0 5
t 0 6
t 0 7
t 0 8

t 0 11
t 0 12
t 0 13
t 0 14
t 0 15
t 0 16
t 0 17
t 0 18

add 1 11 c1 21
addc 2 12 c1 c2 22
addc 3 13 c2 c3 23
addc 4 14 c3 c4 24
addc 5 15 c4 c5 25
addc 6 16 c5 c6 26
addc 7 17 c6 c7 27
addc 8 18 c7 29 28

l 21
l 22
l 23
l 24
l 25
l 26
l 27
l 28
l 29

and i i2 o:
    r i 0 - c
    r i2 c - o

or i i2 o:
    r i 0 - o
    r i2 0 - o

xor i i2 o:
    r i 0 1 2
    r i2 o 2 1

add i i2 o1 o2:
    and i i2 o1
    xor i i2 o2

addc i i2 c o1 o2:
    xor i i2 x
    xor x c o2
    and i i2 a
    and x c a2
    or a a2 o1

The transpiler currently lacks any optimizations, and the wire names it generates are numbers, while SPDT technically allows any non-whitespace combination of characters to be used. I didn't change that for this answer, as it's not a particularly interesting optimization, but it will be coming soon.

Answer 4

Positionally, 33 bytes

 / v


 +  
   ,
      ;

   <  /

Try It Online!

Positionally is a language I made where there are only two commands - space and non-space. The instruction run depends on the position of the IP.

The characters in the above could be any character aside from spaces. Positionally has implicit input, so the + command adds two numbers from input. Then, , prints it, before ; halts the program.

Answer 5

Objective-Java*#++--Script.NETnotation.sh, 12 bytes.

.cmd.vbs.txt

Try it online!

How it works:

.cmd         # Gets the input and pushes it to the stack
    .vbs     # Adds the top two items on the stack
        .txt # Prints the top item on the stack

Answer 6

in floop, 17 bytes

r+o?+n?-[r-]r[o;]

Try it online!

in floop is a language created by Rydwolf Programs which runs in an infinite loop. It's a true tarpit, and incredibly hard to program in. It also has four variables (n, o, r, s), three of which we use here, and a tape.

  o?+             Take an input into variable o and increment it
                  ? does nothing if all input is consumed
     n?-          Ditto with n, but decrement it
r+                Increment r
        [  ]      If n is nonzero
         r-       Decrement r
                  If n is nonzero, r will be zero, and vice versa
            r[  ] If r is nonzero (n is zero)
              o;  Output o

In other words, this decrements n and increments o until n is zero, then outputs o.

Answer 7

05AB1E, 1 byte

Code:

+

Try it online!.

Answer 8

MATLAB, 4 bytes

1 byte removed thanks to @LeakyNun

@sum

This is an anonymous function that takes an array and gives the sum. Example use:

>> f = @sum
f = 
    @sum
>> f([3 4])
ans =
     7

Answer 9

GoLScript, 1 byte (non-competiting)

K

Adds the top 2 numbers on the stack. That's it.

Now how to push them on to the stack, I have no clue. I don't think it's possible.. cough @CᴏɴᴏʀO'Bʀɪᴇɴ cough

Answer 10

C++ – 56 bytes

Should work on all unsigned integer types:

template<typename U>U f(U a,U b){return b?f(++a,--b):a;}

This is multiplication in the same style:

template<typename U>U g(U a,U b){return b>1?f(g(a,--b),a):(b?a:0);}

Finally exponentiation:

template<typename U>U h(U a,U b){return b>1?g(h(a,--b),a):(b?a:1);}
    

To test:

#include <iostream>

template<typename U>U f(U a,U b){return b?f(++a,--b):a;}
template<typename U>U g(U a,U b){return b>1?f(g(a,--b),a):(b?a:0);}
template<typename U>U h(U a,U b){return b>1?g(h(a,--b),a):(b?a:1);}

int main()
{
    std::cout << f(3, 4) << std::endl;
    std::cout << g(3, 4) << std::endl;
    std::cout << h(3, 4) << std::endl;
    return 0;
}

Output:

7
12
81

Answer 11

Shakespeare Programming Language (SPL) , 137 135 bytes

Complete program, golfed:

.
A.
B.
Act I
Scene I
[Enter A and B]
A: Listen to your heart!
B: Listen to your heart! You are the sum of me and you. Open your heart! 

And a brief explanation:

----
.                                 <- Title, everything before the first 
                                     full stop is considered as the tittle and treated as a comment
----
A.                                <- Dramatis personae. Here are introduced the characters in the play.
                                     |Characters are treated as variables.   
B.                                <--
----
Act I                             <- Acts and scenes are used to divide a program into smaller
                                     chunks in order to be able to refer to them later.
                                     |
Scene I                           <--
----
[Enter A and B]                   <- Characters on stage in the current scene, which are the              
                                     variables the program will have access to.
----
A: Listen to your heart!          <- Line of code. All of them have the same structure
                                     Variable: code. In this case, the concrete sentence
                                    "Listen to your heart!" reads an input number and stores it
                                     on the character (variable) refered to.
B: Listen to your heart!          <- Same as above 
   You are the sum of me and you. <- Sum the values of itself and the caharacter (variable)
                                     refered to.
   Open your heart!               <- Output the value of the character (value) refered to.

I am not actualy sure this is the shortest it can go. Check the official page for more info.

Edit 1: Removed the : after Act I and Scene I as it seems that everything after the roman numeral is ignored, thus saving 2 bytes.

Answer 12

Java, 31 bytes

int A(int b,int B){return b+B;}

Making the above snippet code compilable costs 9 bytes, resulting in a 40-byte program:

class a{int A(int b,int B){return b+B;}}

The "equivalent" monolithic program that can also handle numbers between 2³¹ and 2⁶³-1 (inclusive) is 98 bytes long:

interface a{static void main(String[]A){System.out.print(Long.valueOf(A[0])+Long.valueOf(A[1]));}}

---

Java (lambda expression), 11 bytes

(a,b)->a+b;

This is a java.util.function.BinaryOperator<Integer>. It can also be a java.util.function.BinaryOperator<Long> if you have to add larger numbers.

Answer 13

Hexagony, 9 bytes

?{?+@/!'/

Embiggened:

   ? { ?
  + @ / !
 ' / . . .
  . . . .
   . . .

Try it online!

Interestingly, this is only 2 bytes shorter than (what I think is) the most basic version, which is:

   ? { ?
  . . . .
 ' + ! @ .
  . . . .
   . . .

This requires 7 commands, of which one (?, which reads a number from STDIN). Now, as 7 is the 2nd centered hexagonal number, it might be able to fit inside a hexagon of side length 2, if you can use a single flow control character and reuse the ?. I've not been able to figure that out yet though :o(

Answer 14

Brainfuck, 224 bytes

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

Try it online!

Arbitrary precision. Input is taken newline separated, null terminated.

Answer 15

C# - 11 10 bytes

a=>b=>a+b;

Apparently works in ES6 with no semicolon: 10 bytes

(a,b)=>a+b

A lambda expression.

Answer 16

awk, 14 bytes

{print $1+$2}

I am not sure if this is acceptable because the arguments must be passed through a pipe using echo, or through the stdin stream in a file like so:

[user@localhost ~]$ echo '15 6' | awk -E add.awk
21
[user@localhost ~]$ echo '15 6' > numbers
[user@localhost ~]$ awk -E add.awk < numbers 
21

Permissible or not, I am excited to take my first swing at code golf!

Answer 17

Turing Machine Simulator, 342 bytes

0 _ * r 1
0 * * r *
1 _ * r *
1 * * * 2
2 0 _ r *
2 * * * 3
3 _ * l 4
3 * * r *
4 _ * * halt
4 0 9 l *
4 1 0 l 5
4 2 1 l 5
4 3 2 l 5
4 4 3 l 5
4 5 4 l 5
4 6 5 l 5
4 7 6 l 5
4 8 7 l 5
4 9 8 l 5
5 _ * l 6
5 * * l *
6 _ * l *
6 * * * 7
7 _ 1 r 0
7 0 1 r 0
7 1 2 r 0
7 2 3 r 0
7 3 4 r 0
7 4 5 r 0
7 5 6 r 0
7 6 7 r 0
7 7 8 r 0
7 8 9 r 0
7 9 0 l *

Try it!

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.

Ugh...

Answer 18

Vim, 5 bytes

DJ@"<C-a>

Note that <C-a> is one byte: 0x01, which is an unprintable character.

Since V is mostly backwards compatible, you can Try it online!

Explanation:

D           " Delete this line. 
            " By default this will save it into the unnamed register (@")
 J          " Get rid of this empty line
  @"        " Run the unnamed register as if it was typed.
            " Since it's a number, it will provide a count to the next command
    <C-a>   " Increment the next number on this line

Answer 19

Wise, 12 bytes (non-competing)

Mistah Figgins has me beat here

[:??:?^?&<]|

I just made this language so I thought I would try the basics.

Try it online

Explanation

[   ...   ]      #Loop until our carry is zero
 :               #Duplicate the top
  ??             #Roll the top two to the bottom
    :            #Duplicate the top
     ?           #Roll to the bottom

At this point we have n m m n on the stack

      ^          #Xor n and m
       ?         #Roll that to the bottom
        &        #And n and m to create the carry over
         <       #Bitshift to the left
           |     #Remove the extra zero with an or

Answer 20

Add++, 7 bytes

+?
+?
O

Try it online!

+? adds the input to the accumulator and the O outputs it as a number.

A function is 1 byte longer at

D,f,@@,+

Answer 21

R, 3 bytes

"+"

or

sum

"+" can be called as a function like so: "+"(a,b) because the R interpreter will interpret it (with the parentheses) as the primitive addition function .Primitive("+"). Alternatively, sum will add arbitrary arguments. These are both functions.

Try it online!

Answer 22

Piet - 5 Codels

I didn't see one in Piet, so I decided to post one.

Hold on, let me scale that up a bit for you to see.

The codel size is 25.

I tested this on npiet. When run, the program will ask for a number, then for another, and then print it out. Due to the nature of Piet, it will continue this cycle until you interrupt it.

There are 17 other variants of the same code. Everything is relative in Piet, so changing the start color will change the rest of the colors. Breaking down the code is as follows:

Light Red - Does nothing.

Dark Blue (hue shift 4, darkness shift 2) - Pushes a number from STDIN.

Green (hue shift 4, darkness shift 2) - Pushes a number from STDIN.

Teal (hue shift 1, darkness shift 0) - Pushes the sum of the top two numbers in the stack.

Dark Green (hue shift 5, darkness shift 1) - Writes top of stack to STDOUT as a number.

Answer 23

Quantum Circuit 6*Nbits+1 Gates (25 for two 4-bit numbers)

Based on the ripple adder of Cuccaro et al

• Qubits 0/9 should be left as 0
• 1:3 represent A in binary
• 5:8 represent B in binary
• Answer appears in qubits 5:9

Try it online!

Answer 24

Vyxal s, 0 bytes

Try it Online!

The s flag takes the sum of the top of the stack. Flag abuse FTW!!!

Answer 25

sed, 275 bytes

s/[1-9]/0&/g
s/[5-9]/4&/g
s/[89]/4&/g
s/[2367]/xx&/g
s/[13579]/x/g
s/4/xxxx/g
s/[1-8]//g
s/$/ /
:a
s/\(.*\)0\(x*\) \(.*\)0\(x*\) /\1 \3 0\2\4/
ta
s/  *//g
:c
s/0xxxxxxxxxx/x0/g
tc
s/0x/-x/g
s/xx/2/g
y/x/1/
s/22/4/g
s/44/8/g
s/81/9/g
s/21/3/g
s/42/6/g
s/43/7/g
s/41/5/g
s/-//g

Takes 2 space-separated non-negative decimal integers on standard input; prints their sum to standard output

Explanation

We represent each input in unary-coded decimal, using x as our digit, and 0 as separator. For example, 42 is written 0xxxx0xx.

#!/bin/sed -f

# Convert to unary decimal
s/[1-9]/0&/g
s/[5-9]/4&/g
s/[89]/4&/g
s/[2367]/xx&/g
s/[13579]/x/g
s/4/xxxx/g
s/[1-8]//g

# Append space to separate second input from output
s/$/ /
# Consume the smallest of each digit from input, and push
# the sum to output
:add
s/\(.*\)0\(x*\) \(.*\)0\(x*\) /\1 \3 0\2\4/
tadd

# Everything is now accumulated in output; remove the spaces
s/  *//g
# Add carry to the next position
:carry
s/0xxxxxxxxxx/x0/g
tcarry

# Back to decimal
s/0x/-x/g
s/xx/2/g
y/x/1/
s/22/4/g
s/44/8/g
s/81/9/g
s/21/3/g
s/42/6/g
s/43/7/g
s/41/5/g
s/-//g

The extension to support decimal fractions is left as an exercise for the reader.

Answer 26

Golfscript, 9 2 bytes

~+

~ casts to int[], and + adds.

Answer 27

Labyrinth, 5 bytes

??+!@

Try it online!

Very straightforward: read input, read input, add, print, terminate.

Answer 28

PHP 4.1, 9 bytes

This assumes a standard php.ini file, with short_open_tag=On and register_globals=On (default in PHP 4.1).

<?=$A+$B;

This assumes that you are acessing the file through a webserver (like Apache).
The keys A and B contain the numbers you want to sum.
An example: http://localhost/sum.php?A=5&B=16. (POST, GET, SESSION and COOKIE can be used too).

Answer 29

Fishing, 22 bytes

v+CCCCCCCC
  In{In}aP

Explained

Sets casting direction down and casting length to 1.
Read input to the first 2 cells on the tape and convert to numbers.
Add the first 2 cells together and print.

Answer 30

Befunge, 5 bytes

&&+.@

• & takes an integer as input and puts it on the stack
• + pops two numbers from the stack and puts the sum back on the stack
• . pops a number from the stack and outputs it as an integer
• @ ends the program

Try it: http://www.quirkster.com/iano/js/befunge.html