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
var QUESTION_ID=84260,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>
243 Answers 243
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
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ドル + 2ドル;}
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\$\begingroup\$ Wouldn't this be shorter as a program rather than as a function? \$\endgroup\$Neil– Neil2016年07月02日 18:46:17 +00:00Commented Jul 2, 2016 at 18:46
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\$\begingroup\$ @Neil Yeah, I'll post that separately. \$\endgroup\$anna328p– anna328p2016年07月02日 22:02:50 +00:00Commented Jul 2, 2016 at 22:02
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*-*-#
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~-=+
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
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.
-
\$\begingroup\$ The format string doesn't need a space.
"%d"(like most conversions other than%cand%[) skips leading white-space on its own. \$\endgroup\$Peter Cordes– Peter Cordes2017年12月17日 19:49:22 +00:00Commented Dec 17, 2017 at 19:49
Ruby, 17 bytes
** Edit: Big thank you to @steenbergh and @TùxCräftîñg for their help **
def c(a,b)a+b end
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\$\begingroup\$ Edited the answer. \$\endgroup\$BoeNoe– BoeNoe2016年10月20日 17:37:24 +00:00Commented Oct 20, 2016 at 17:37
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\$\begingroup\$ @manatwork Edited. \$\endgroup\$BoeNoe– BoeNoe2016年10月21日 11:21:31 +00:00Commented Oct 21, 2016 at 11:21
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\$\begingroup\$ @manatwork What do you mean? \$\endgroup\$BoeNoe– BoeNoe2016年10月21日 11:24:20 +00:00Commented Oct 21, 2016 at 11:24
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\$\begingroup\$ @manatwork Oh, ok. \$\endgroup\$BoeNoe– BoeNoe2016年10月21日 11:27:01 +00:00Commented Oct 21, 2016 at 11:27
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\$\begingroup\$ @manatwork Edited. \$\endgroup\$BoeNoe– BoeNoe2016年10月21日 11:27:59 +00:00Commented Oct 21, 2016 at 11:27
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
-
\$\begingroup\$ Invalid, you did not count the newline nor the spacing. Also, this is code golf, the aim is conciesness, not coolness \$\endgroup\$ASCII-only– ASCII-only2019年01月29日 07:53:24 +00:00Commented Jan 29, 2019 at 7:53
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);
}
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\$\begingroup\$ You don't need to have a full program. Just
int f(int a,int b){return a+b;}would probably suffice. \$\endgroup\$anna328p– anna328p2016年11月01日 02:11:06 +00:00Commented Nov 1, 2016 at 2:11 -
\$\begingroup\$ The challenge stated program or function. I make no claims to have the best solution. Additionally, another user has already provided a standalone function, and to produce an identical one would not contribute much. \$\endgroup\$James Murphy– James Murphy2016年11月01日 02:27:26 +00:00Commented Nov 1, 2016 at 2:27
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 (?).
Whirl, 32 bytes
01100100001100011110001000111100
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
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
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\$\begingroup\$ "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." Sure your code handles input? \$\endgroup\$manatwork– manatwork2017年01月24日 08:43:24 +00:00Commented Jan 24, 2017 at 8:43
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\$\begingroup\$ Ok, I fixed it. \$\endgroup\$12Me21– 12Me212017年01月24日 16:19:44 +00:00Commented Jan 24, 2017 at 16:19
Python 3, 22 bytes
lambda a,b:a+b
Trivial answer for a kind of trivial question.
-
1\$\begingroup\$ You could golf this to
lambda a,b:a+b. Although there's already a 3-byte python solution here. \$\endgroup\$FlipTack– FlipTack2016年12月24日 21:43:47 +00:00Commented Dec 24, 2016 at 21:43
Wise, 9 bytes
[?~-!-~]|
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.
Triangular, 6 bytes
$.$%+<
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
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
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.
5 16is inputted as005 016\$\endgroup\$