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

TI-Basic, 2 bytes

Very straightforward.

sum(Ans

Answer 2

Kitanai, 15 bytes

add input input

Pretty straightforward :)

Answer 3

PigeonScript, 1 byte

+

Explanation: + pops the last two items from the stack, adds them, and pushes the result to the stack. Since there is nothing on the stack, the user is prompted for input twice. The inputs are pushed, popped, added, pushed, and the program ends, outputting what's on the stack (the result of input1 + input2)

Answer 4

Taxi, 418 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 Addition Alley.Pickup a passenger going to Addition Alley.Go to Addition Alley:e 5 l 1 l.Pickup a passenger going to The Babelfishery.Go to The Babelfishery:w 1 r 1 r.Pickup a passenger going to Post Office.Go to Post Office:e 1 l 1 r.

Try it online!

Answer 5

Sinclair ZX80 BASIC (4K ROM)

The usual rules about the ZX80 and its 16 bit signed integer range apply here. Anything out of this range will not add. And because PRINT A+B is too easy, I came up with two solutions:

Method 1 ~64 bytes:

This assumes that you're adding a positive integer to the first number entered:

 1 INPUT A
 2 INPUT B
 3 IF B=0 THEN GO TO 7
 4 FOR B=B/B TO ABS(B)
 5 LET A=A+1
 6 NEXT B
 7 PRINT A

Method 2 ~24 bytes

A much simpler solution, where you are adding two -/+ integers together, simply do this:

 1 INPUT A
 2 INPUT B
 3 PRINT A+B

Both of these listings are likely to work on all variants of 8-bit BASIC, although would not be optimised on most of them.

Answer 6

Cubically, 10 8 bytes

$+7$+7%6

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Cubically is a language relatively early in development. Its most unique feature is that its primary memory takes the form of a virtual Rubik's Cube with the colors replaced by numbers from 0 to 5. This Rubik's Cube cannot be written to, and manipulation of it is done only through rotation commands, with only a single "notepad" memory space which supports more traditional manipulation. All operations consist of a non-digit character, followed by any number of digits representing the memory location to use: 0-5 for the face centered on the chosen number, 6 for the notepad, and 7 for the "input buffer".

All that said, this program uses none of the language's unique features and instead does all operations in the notepad and the input buffer which was added today to finally qualify for this challenge since before now there simply was no input. Explanation:

                  Notepad defaults to 0
$               Read an integer from STDIN
 +7             Add the value in [7] (input buffer) to the notepad
   $            Read an integer from STDIN
    +7          Add the value in [7] (input buffer) to the notepad
      %6        Output the value in [6] (notepad) to STDOUT

The language was updated and now allows the $ command to be called without an argument, saving 2 bytes by not including unused characters

Look forward to more Cubically answers as advanced features are added such as looping and branching!

Answer 7

LOLCODE, 75 bytes

HAI 1.3
I HAS A J
GIMMEH J
I HAS A Q
GIMMEH Q
VISIBLE SUM OF J AN Q
KTHXBYE

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Answer 8

cubix, 6 bytes

OI\@+I

Try it here!

This maps onto a cube with edge length one.

  O
I \ @ +
  I

Much the same as other stack based answers, push input to the stack twice, add, output and terminate.
Operations are

• I, input number
• \, reflect down
• I, input number
• +, add top two of stack
• O, output number
• \, reflect to the right
• @, terminate

The following will also work

II/@+O

Mapping to

  I
I / @ +
  O

Answer 9

JavaScript (Node.js), 28 bytes

f=(a,b)=>b?f(a^b,(a&b)<<1):a

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Not for winning rather just for fun. Definitely not the shortest but my fav.

Answer 10

Java (JDK 10), 9 bytes

Long::sum

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Answer 11

7, 8 bytes, 22 characters

1717023740344172360303

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This program is encoded on disk as (xxd hexdump):

00000000: 3cf0 9f81 c87a 7830                      <....zx0

7 doesn't really support numbers natively, and thus it's hard to define what a number is for the purpose of a function submission. As such, this is a full program, reading from stdin, outputting to stdout (which explains where much of the length comes from).

This program doesn't support negative numbers, because 7 can't input those using numeric I/O (although it can output them). As such, supporting negative numbers would require the use of character input (and a decimal→integer parser), which would make the program much, much more complex.

Explanation

1717023740344172360303
 7 7   7                Stack element separators
1 1 023                 Initial stack
        40344172360303  Initial program (also stored on the stack)
(Implicit: run the initial program, but leave it on the stack)
        4               Swap with blank element between
         0              Escape top stack element, append it to element below

So at this point, we've effectively swapped the program below the 023 element, escaping that element in the process. The 023 is a program in a domain-specific I/O language; and putting the program as the second stack element means that we can discard it (the second stack element is the only one that can be discarded).

          3             Do I/O using top element, discard second element
    0                   Set I/O format: numeric in decimal
     23                 Input via repeating the third stack element

We now have only two stack elements; 1 at the bottom, and the first input in unary just above it (because we repeated the second-last stack element, which was 1, and thus will have a number of 1s).

           4            Swap with blank element between
            4           Swap with blank element between
             172360     Append an escaped representation of "23" to TOS
                   3    Do I/O using top element, discard second element
               23       Input via repeating the third stack element

So now our stack consists of the first input (in unary) directly below the second stack element (in unary).

                    0   Escape top element, appending it to the element below
                     3  Do I/O using top element, and exit

The 3 command exits the program as we're out of stack, but not before it outputs the number we calculated. The number in question will consist of a number of 7s equal to the first number input, followed by a number of 1s equal to the second number input (these are the unescaped and escaped representations of the same command). Numeric I/O treats 1 and 7 as equivalent, and having a value of +1; thus, the unary number gets translated into decimal and output.

Answer 12

Burlesque - 4 bytes

ps++

ps   parse
  ++ sum

Try it online.

Answer 13

Aheui (esotope), 15 bytes(5 characters)

방방다망히

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---

Meet Aheui(아희), A Korean alphabet-based esoteric programming language.

Answer 14

Charcoal, 4 2 bytes

Ｉ⁺

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Explanation

Ｉ Cast to string
 ⁺ Add
   (implicit) Input number
   (implicit) Input number

Answer 15

Clam, 4 bytes

p+rr

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Explanation

p               print ...
   +               the sum of ...
       r               the first line of STDIN and ...
       r               the second line of STDIN

Answer 16

Whitespace, 36 bytes

[S S S N
_Push_0][S N
S _Duplicate][T N
T   T   _Read_STDIN_as_integer][T   T   T   _Retrieve][S S S N
_Push_0][S N
S _Duplicate][T N
T   T   _Read_STDIN_as_integer][T   T   T   _Retrieve][T    S S S _Add][T   N
S T _Print_as_integer]

Letters S (space), T (tab), and N (new-line) added as highlighting only.
[..._some_action] added as explanation only.

Try it online (with raw spaces, tabs, and newlines only).

Explanation:

When a number is read from STDIN it stores it in the heap-address specified at the top of the stack, and then the Retrieve function can be used to get it later on. So to read a number from STDIN you'll need the following sub-steps:

1. Put a number on the stack:
   • S at the start: Enable Stack Manipulation
   • S: Push what follows as number to the top of the stack
   • S/T: Positive or negative number
   • Some S and/or T followed by a N: Number as binary, where S=0 and T=1.
2. Duplicate this number:
   • S at the start: Enable Stack Manipulation
   • NS: Duplicate the top value on the stack
3. Read a number from STDIN, and store it in the heap-address specified at the top of the stack:
   • TN at the start: Enable I/O
   • TT: Read a number, and place it in the heap specified at the top of the stack
4. Retrieve this number from the heap
   • TT at the start: Enable Heap Access
   • T: Retrieve a value from the heap with the given heap-address at the top of the stack

---

So here is every step of the full program above:

Command    Explanation              Stack      Heap     STDIN    STDOUT    STDERR

SSSN       Push 0                   [0]        {}                
SNS        Duplicate top (0)        [0,0]      {}                
TNTT       Read STDIN as integer    [0]        {0:-3}   -3        
TTT        Retrieve heap (at 0)     [-3]       {0:-3}                
SSSN       Push 0                   [-3,0]     {0:-3}                
SNS        Duplicate top (0)        [-3,0,0]   {0:-3}                
TNTT       Read STDIN as integer    [-3,0]     {0:5}    5        
TTT        Retrieve heap (at 0)     [-3,5]     {0:5}                
TSSS       Add top two              [2]        {0:5}                
TNST       Print top as integer     []         {0:5}            2
                                                                           Exit with error

Unfortunately the heap-address cannot be negative, otherwise the second SSSN (Push 0) could have been golfed to SNS (duplicate first STDIN input as heap-address).

Answer 17

JavaScript (Node.js), 18 bytes

a=>eval(a.join`+`)

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Explanation :

a =>                     // lambda function taking array as input
    eval(               // begin eval
        a.join`+`      // joins all elements in `a` with a `+` sign in between
    )                 // end eval (since this is now a string it gets added up)

---

Alternate :

JavaScript (Node.js), 9 bytes

x=>y=>x+y

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Explanation :

x =>                  // lambda taking x as input 1
    y =>              // which returns lambda with input y
        x + y         // which returns sum of x and y

---

Answer 18

SystemVerilog (HDL) (93 chars)

Here's a pretty simple 8-bit full adder implementation (including input carry for chaining).

module a(input[7:0] a,input[7:0] b,input d,output c,output[8:0] f);
assign f=a+b+d;
endmodule

Inputs: [7:0]a and b, carry bit d.

Outputs: [8:0]f. The top bit can be used as a carry output signal.

Answer 19

Help, Wardoq!, 1 byte

A

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Note: Due to technical limitations, this interpreter takes code as the first line, and input as every line after.

Answer 20

Kotlin, 22 bytes

fun n(a:Int,b:Int)=a+b

Answer 21

Scratch 3.0 (scratchblocks3 syntax), 23 bytes/7 points

define (x)(y
say((x)+(y

Answer 22

BrainF*ck, 12 Bytes

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

I'm completely new to code golfing, so if I messed something up or did something wrong please let me know.

Here's how it works:

,>,<   - Stores the two inputs into the first and second cells

[->+<] - Decrements the first cell and adds it to the second, 
         repeats until the first cell is 0

>.     - Moves to the second cell and outputs its value

The inputs and outputs are both as ASCII characters (not too sure if this is allowed. I think it is but I could definitely be wrong, so if it is just let me know).

Answer 23

Yggdrasil, 4 bytes

+__$

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Yggdrasil automatically substitutes 2 characters for different values when turning the source code in a binary tree. % represents a null byte and _ represents a None value. Yggdrasil then traverses the tree and replaces Nones with sequential command line arguments (i.e. the first encountered, depth-first, is the first argument etc.).

This forms the following binary tree for the memory, where ¹ represents the first argument and ² the second:

  +
 / \
¹   ²

We then run the commands linearly, starting with the memory pointer at the + node. We execute +, which adds the left and right nodes of the memory point, setting the root of the program to ¹+². _ then does nothing, and $ outputs the value under the memory pointer as an integer

Answer 24

CSS 172 bytes

For every button add 1 and for every input add 1. Might be buggy on some browsers and also not very practical. Make sure your html file has a body and you actually put in buttons and inputs

*{—n: 0;—d: 0;}li{—-n: calc(var(—-n) + 1 )
}button{—-d: calc(var(—-d) + 1 )
}input{—-d:calc(var(—-d) + 1)}body::after{content: calc(var(—-d) + var(—n))}

Answer 25

dotcomma, 10 bytes

[,.][,.].,

Dotcomma is a language I recently invented, designed to do things in an interesting way with as few instructions as possible. In order to accomplish this, there are two operators (predicatably . and ,) which can do entirely different things depending on context. This answer's explanation will be very introductory. To see some more complicated code, check out the examples in the page linked in the title.

Anything between brackets is a block. Every block has a return value, which is set by the operator before the closing bracket (or 0 if there is none). The basic reason this program functions is because, when one or more blocks (with no operators in between) are followed by a ., the sum of their return values is computed.

In this program, both blocks start with ,. When preceded by the start of a block, a , will take input. Additionally, when followed by the end of a block, it will output the value given to it (from input or another operator). The .s here do pretty much nothing; they take the return values of the ,s and use it as the return value of their respective blocks, which spaces the , from the end of the block and suppresses its output.

The final , simply takes the value computed by the final . and outputs it.

Answer 26

Noxan, 3 bytes

+ŊŊ

No implicit input. (yet)

Answer 27

Fennel, 23 bytes:

(fn add [x y] (+ x y ))

Answer 28

Sesos, 5 bytes

0000000: d605ba 8f07                                       .....

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set numin
set numout
get
fwd 1
get
jmp
sub 1
rwd 1
add 1
fwd 1
jnz
rwd 1
put

Answer 29

Pyth, 4 3 bytes

+vw

Try it here

Or also:

s.Q

Try it here

Because this is shorter than both other Pyth answers I thought I'd add this one.

Answer 30

Qdeql, 37 bytes

Input in bytes rather than ASCII. Although Qdeql is limited to bytes, this would work with any (arbitrarily large) finite cell size.

-&=\--\/\/=/&=\--\/\/=/\=/-\--\/\/=/*

That was... surprisingly easy.

The core of this is the loop \--\/\/=/. It computes the difference between two values, and appears three times in this code. Here's an explanation of the loop:

Queue: {x, y}
\ While x is not 0, move it to the back of the queue and add two 0s
  Queue: {y, x, 0, 0}
  -- Decrement both values
  Queue: {0, 0, y minus 1, x minus 1}
  \/ Remove a 0 from the queue
  \/ Remove the other 0
  Queue: {y minus 1, x minus 1}
  = Swap the order of the values in the queue
/ End while

When that loop exits, the queue is guaranteed to have a single value that is y - x.

Given this, if the two inputs to the program are x and y, this program computes 255 - (255 - x - y).

- Start the queue with a single 255
& Add the first input to the queue
= Swap the order, so the input is ahead of the 255
\--\/\/=/ Compute 255 minus x
& Add the second input to the queue
= Swap them again
\--\/\/=/ Compute 255 minus x minus y
\=/- Add another 255 to the queue
\--\/\/=/ Compute 255 minus (255 minus x minus y)
* Print the result