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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• Dennis: This is quite trivial, but not really simpler than, e.g., the Hello World catalog. Given that the ability to add integers is one of our two requirements for programming languages, I'd say it's worthwhile to have if properly specified.
• FinW: Can the answer take input with preceding zeros as default? e.g. 5 16 is inputted as 005 016
• anna328p: @FinW Sure. As long as they don't get interpreted as octal.

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.

• Shaun Bebbers: There are no such things as functions on a Sinclair ZX80 (as far as I know anyway) so both solutions are complete symbolic listings.

Answer 6

Cubically, (削除) 10 (削除ここまで) 8 bytes

$+7$+7%6

Try it online!

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

Try it online!

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

Answer 10

Java (JDK 10), 9 bytes

Long::sum

Try it online!

Answer 11

Burlesque - 4 bytes

ps++
ps parse
 ++ sum

Try it online.

• Οurous: How does this work? Can you link to an interactive demo and / or provide instructions on how to execute it?
• mroman: ps parses the input into a list and ++ computes the sum of a list. I added a link to the online interpreter.

Answer 12

Aheui (esotope), 15 bytes(5 characters)

방방다망히

Try it online!

---

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

Answer 13

Charcoal, (削除) 4 (削除ここまで) 2 bytes

I+

Try it online!

Explanation

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

Answer 14

Clam, 4 bytes

p+rr

Try it online!

Explanation

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

Answer 15

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 16

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

Try it online!

Explanation :

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

---

• Martin Ender: Welcome to PPCG!
• Muhammad Salman: Thanks. May I ask what you edited ?
• Muhammad Salman: Nvm , found that out. Thanks for the edit. :)

Answer 17

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 18

Help, Wardoq!, 1 byte

A

Try it online!

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

Answer 19

Kotlin, 22 bytes

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

Answer 20

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

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

• Rhaixer: According to this answer codegolf.stackexchange.com/a/106203/110681 on the related question which takes the same input restrictions, I think you can simply reduce that to (()+( and specify where input goes.

Answer 21

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).

• rydwolf: Welcome to Code Golf. Nice first answer!
• Makonede: I've edited your answer just to fix some minor mistakes (the pointer is the number indicating the current cell; the cells are what actually store numbers). Also, you can save a byte by not moving back to the first cell after inputting to the second: ,>,[<+>-]>.
• hakr14: @Makonede it's actually ,>,[<+>-]<. (last pointer movement is wrong)
• Makonede: @hakr14 My bad.

Answer 22

Yggdrasil, 4 bytes

+__$

Try it online!

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 23

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 24

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 25

Noxan, 3 bytes

+ŊŊ

No implicit input. (yet)

Answer 26

Fennel, 23 bytes:

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

Answer 27

Sesos, 5 bytes

0000000: d605ba 8f07 .....

Try it online!

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

• Erik the Outgolfer: I would recommend using jmp instead of nop to make it work for 0-Infinity instead of 1-Infinity.
• acrolith: @EriktheGolfer Done.

Answer 28

Logically, 229 Bytes.

@m:3i:o,c:3t;XOR(3i)(o)AND(i0,i1)(t0)AND(i1,i2)(t1)AND(i0,i2)(t2)OR(3t)(c)
@M:8a,8b:8o:7c;m(a0,b0)(o0,c0)m(a1,b1,c0)(o1,c1)m(a2,b2,c1)(o2,c2)m(a3,b3,c2)(o3,c3)m(a4,b4,c3)(o4,c4)m(a5,b5,c4)(o5,c5)m(a6,b6,c5)(o6,c6)m(a7,b7,c6)(o7)

Ungolfed

@FullAdder
Input: 3i;
Output: o,c;
Bus: 3t;
XOR (3i) (o)
AND (i0,i1) (t0)
AND (i1,i2) (t1)
AND (i0,i2) (t2)
OR (3t) (c)
@Main
Input: 8a,8b;
Output: 8o:
Bus: 7c;
FullAdder (a0,b0) (o0,c0)
FullAdder (a1,b1,c0) (o1,c1)
FullAdder (a2,b2,c1) (o2,c2)
FullAdder (a3,b3,c2) (o3,c3)
FullAdder (a4,b4,c3) (o4,c4)
FullAdder (a5,b5,c4) (o5,c5)
FullAdder (a6,b6,c5) (o6,c6)
FullAdder (a7,b7,c6) (o7)

I/O as 8 bit Non-negative integers. Use the /ib /ob for Decimal Input/Output. Binary otherwise.

_{Currently no Online Interpreter :(}

Answer 29

Pyth, (削除) 4 (削除ここまで) 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