Bespoke, 38 bytes

would I total a quantity?yeah should I 

Bespoke is a new esoteric programming language I created near the start of 2025. It's similar to my earlier esolang Poetic, in that word lengths are converted to instructions. This time, however, the instructions are part of a stack-based language, with arbitrary-precision signed integers and loop breaking/continuing and functions and everything!

Here is this program in "pseudocode" (which is able to run as Bespoke code - in fact, it's equivalent to the given code!):

INPUT N INPUT N STACKTOP PLUS OUTPUT N 

This should be self-explanatory.

7, 22 characters, 8 bytes

1717023740344172360303 

Try it online!

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.

Bash, 18 bytes

f(){expr $1 + $2;} 

Prelude, 4 bytes

??+! 

Requires my modified Prelude interpreter which uses decimal I/O.

Like several other answers, this is just read, read, add, write.

Befunge - 5 Bytes

&&+.@ 

& - Request int from user and push to stack

+ - Pop top two elements from stack and add and push result

. - Pop value and output as int

@ - End program

Try it here (Although you will have to copy/paste into the text area)

Edit: oops, didn't notice the earlier (identical) befunge answer, I'll leave this here unless I'm told to delete it, not sure of the opinion on that.

GO, 29 bytes

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

Not that much to say

Yup, 6 bytes

0*-*-# 

Try it online!

Explanation

0*-*-# 0 push 0 to the stack [0] * place input [0 a] - subtract [-a] * place input [-a b] - subtract [-a-b] = [b+a] # output 

As a function on -cheat mode, 9 bytes:

0~--0~-=+ 

Try it online!

SML, 3 bytes

op+ 

This is the prefix version of the infix +. If we input it like this in an interpreter (for example Moscow ML), it's type is displayed

- op+; > val it = fn : int * int -> int 

which tells us how to use it: Given a tuple of integers, an integer will be returned.

- op+(17,25); > val it = 42 : int 

GAP, 2 Bytes

\+ 

The backslash is GAP's way to turn an infix operator (and some more, there is also \[\] for indexing) to a function.

Here is an example use:

gap> \+(4,3); 7 

jq, 3 characters

add 

Sample run:

bash-4.3$ jq 'add' <<< '[5,16]' 21 

On-line test

Elixir, 7 bytes

& &1+&2 

Anonymous function defined using the capture operator. Another version is &(&1+&2), however this approach saves 1 byte. The verbose version is fn a,b->a+b end - 15 bytes.

Full program with test case (yes, the . in the function call is mandatory!):

s=& &1+&2 IO.puts s.(1,6) #7 

Try it online on ElixirPlayground !

C, 49 bytes

main(a,b){scanf("%d %d",&a,&b);printf("%d",a+b);} 

Can't really do much to golf it down.

For fun:

The following works only if 0 <= sum < 256 (it's 59 bytes long):

a;main(c,v)char**v;{return a++&2?c-3:main(c+atoi(*++v),v);} 

Use gcc to compile, and ./a.out [your 2 nums]; echo $? to run it.

Here's the ungolfed version of that program:

/* Global int auto-initialized to 0 */ a; /* The main method with two int params */ /* 'c' is argc, and 'v' is argv */ main(c, v) /* Yes, this is valid. It defines 'v' as a char** */ char** v; { /* Checks to see if a == 2, and increments a. * I only want recursion to happen twice. */ if (a++ & 2) { /* Since "argc" is 3, we need to subtract it from the final answer */ return c - 3; } else { /* Get the next int value of "argv" and add it to 'c'. * Then call main() again with the updated value */ return main(c + atoi(*++v), v); } } 

So, main() returns the value of the sum of the two numbers (essentially, I'm forcing it to return a custom exit code). The program won't output the exit code, so calling echo $?, in the same command as running the program, outputs the return value of that program.

The range of exit codes only exist between 0 and 255, so if you run the program trying to sum 255 and 1, it will wrap around and output 0.

Ruby, 17 bytes

** Edit: Big thank you to @steenbergh and @TùxCräftîñg for their help **

def c(a,b)a+b end 

Befunge, 7 bytes

#v&< + . @ 

Befunge is an esoteric language that can execute in any cardinal direction on a 2 dimensional plane. Executing left to right from the top left (default) we get the following set of operations ((x,y) = coordinates of operation with (0,0) in the top left)

(0,0): skip next cell in execution path (1,0)
(2,0): ask for user input and push it to the stack
(3,0): reverse direction of execution
(2,0): ask for user input and push it to the stack
(1,0): begin executing downward
(1,1): pop first two values on stack, add them and push result
(1,2): pop stack and output value
(1,3): end

&&+.@ would also work, and in only 5 bytes, but is not nearly as cool

Swift, 1 byte

+

since operator + is defined as function(public func +(lhs: Int, rhs: Int) -> Int)

C, 66 bytes

Typical C omissions with regard to declarations, header files, and return statements. This actually works for any number of decimal integers (up to the limitations of one's shell), including none. The program can be invoked with ./a.out 1 2 [...], as an example.

main(c,v,x)char**v;{x=0;while(c-1)x+=atoi(v[--c]);printf("%d",x);}

Or, more legibly:

main(c,v,x) char**v; { x=0; while(c-1) x+=atoi(v[--c]); printf("%d",x); } 

QBIC, 6 bytes

::?a+b 

It takes two cmd line params as numbers (with two :'s), names them a and b, and adds them while printing (?).

Euphoria, 64 bytes

Code:

include get.e print(1,prompt_number("",{})+prompt_number("",{})) 

Whirl, 32 bytes

01100100001100011110001000111100 

Try it online!

Explanation

Whirl has two "rings" of operations, one for control and I/O operations and the other for math operations. The 1 command rotates the current operation ring while the 0 command switches direction of the rotation. Two 0s in a row run the current operation and switch to the other ring.

01100 Read an integer from stdin and store it at the memory pointer 100 Load that value into the math ring data storage 00 Read an integer from stdin and store it at the memory pointer 1100 Add the the values under the memory pointer to the math ring data storage 0111100 Set the control ring data storage to 1 0100 Save the math ring data storage to the memory at the memory pointer 0111100 Print the result from the memory at the memory pointer 

V, 2 bytes

À
 

Try it online!

You can't see it, but after the À there is a control-a character (ASCII 0x01).

This answer is nice because it shows one of the nice things about V: Numeric input is more convenient. Note how one input is an argument and the other is in the "input" field. This is an allowed input method, and it saves bytes because we want to run something n times, in this case the increment command or ctrl-a.

À " Arg1 times: <C-a> " Increment the next number found on this line 

SmileBASIC, 13 bytes

INPUT A,B?A+B 

Python 3, 22 bytes

lambda a,b:a+b 

Trivial answer for a kind of trivial question.

Wise, 9 bytes

[?~-!-~]| 

Try it Online!

Takes 2 numbers in, A, and B.

 Implicit input [ .... ] While the top (B) != 0, repeat: ? Move B to the bottom ~- Add 1 to the top (A) ! Move B back to the top -~ Subtract 1 from the top (B) | Gets rid of the 0 on top by ORing the top 2 Implicit output 

It boils down to "Add 1 to A, B times."

Less cool than the other Wise answer, but also shorter.

Braingolf, 1 byte

+ 

Try it online!

Triangular, 6 bytes

$.$%+< 

Try it online!

Formats into this triangle:

 $ . $ % + < 

Without directionals or no-ops, the above turns into this: $$+%

Triangular uses postfix notation.

• $ - read input as integer
• + - add
• % - print as integer

,,,, 1 byte

+ 

^{Body must be at least 30 characters; you entered 14.}

Commentator, 4 bytes

//// 

Outputs by char code. If this isn't allowed, a 6 byte solution is

////*/ 

which outputs by numerical value.

tcl, 10

expr $a+$b

If you have namespace path tcl::mathop you can do it shorter for every arithmetic operation:

tcl, 7

+ $a $b

test case on: http://rextester.com/live/FLFPTH24568

Little Man Computer, 23 bytes (source)

INP STA 6 INP ADD 6 OUT 

With HLT, 27 bytes (source)

INP STA 6 INP ADD 6 OUT HLT 

Online Emulator (Flash)

Bitwise, 89 bytes

IN 1 &1 IN 2 &1 LABEL &1 AND 1 2 *1 XOR 1 2 *2 SL *1 &1 1 MOV 2 *2 &1 JMP @1 *1 OUT *2 -1 

Input and output are raw byte values as that is the only method of I/O for this language. Allowed per this consensus. Try it online!

Ungolfed, written properly, and commented:

IN 1 &1 -1 read a character into r1 if l1 truthy, discarding result IN 2 &1 -1 read a character into r2 LABEL &1 create label 1 here AND 1 2 *1 set fr1 to r1 & r2 where & represents bitwise AND XOR 1 2 *2 set fr2 to r1 ^ r2 where ^ represents bitwise exclusive or SL *1 &1 1 set r1 to fr1 << 1 where << represents bitwise left shift MOV 2 *2 &1 move fr2 into r2 if &1 truthy JMP @1 *1 jump to label 1 if fr1 truthy OUT *2 &1 -1 print frame register 2 if l1 truthy, discarding result 

Note that fr is short for frame register, r is short for register and l is short for literal.