# How do I write an adding function? [closed]

The problem:

I am the lead developer for a big company, we are making Skynet. I have been assigned to

Write a function that inputs and returns their sum

function sum(a,b){
return "their sum";
}


EDIT: The accepted answer will be the one with the most upvotes on January 1st, 2014

• You can use my lightweight jQuery plugin: $.sum=function(a,b){return a+b};. Dec 28, 2013 at 13:51 • I knew I'd get a jQuery reference sometime Dec 28, 2013 at 13:51 • Brilliant English :p Dec 28, 2013 at 17:39 • Question suggestion (not sure if it's any good): "GUISE HALP, I need a fast algorithm to generate bitcoin blocks!!!!! It's super urgent!" – user11485 Dec 28, 2013 at 18:42 • These answers are quite involved. I suggest opening a connection to your database and issuing 'SELECT ' + a + ' + ' + b + ';'. It's simple and understandable. Dec 28, 2013 at 20:57 ## 75 Answers That's a very complex problem! Here is how you solve it in C#: static int Sum(int a, int b) { var aa = ((a & ~877 - b ^ 133 << 3 / a) & ((a - b) - (a - b))) | a; var bb = ((b ^ (a < 0 ? b : a)) & ((b - a) - (b - a))) | b; var cc = new List<int>(); for (int i = 6755 & 1436; i < aa; i -= -1) { cc.Add((int)Convert.ToInt32(Math.Sqrt(6755 & 1437 >> ((b - a) - (b - a))))); } for (int i = 6755 & 1436; i < bb; i -= -1) { cc.Add((int)Convert.ToInt32(Math.Sqrt(6755 & 1437 >> ((a - b) - (a - b))))); } Func<int,int,int> importantCalculation = null; importantCalculation = (x, y) => y != 0 ? importantCalculation(x ^ y | (6755 & 1436) >> (int)(Convert.ToInt32(Math.Sqrt((b - a) - (b - a) - (-1))) - 1), (x & y) << (int)Convert.ToInt32((Math.Log10(1) + 1))) : x; return cc.Aggregate(importantCalculation); }  How this code works (I wouldn't add this explanation in my answer to the lazy OP that has to be trolled, don't worry): ((a & ~877 - b ^ 133 << 3 / a) & ((a - b) - (a - b))) | a returns just a and ((b ^ (a < 0 ? b : a)) & ((b - a) - (b - a))) | b returns just b. 6755 & 1436 returns 0, so in the loop, i actually starts with value 0, and inside the loop, you add the value 1 to the list. So, if a is 5 and b is 3, the value 1 is added 8 times to the list. The importantCalculation function is a very long function that does nothing else than adding up two numbers. You use the LINQ Aggregate function to add up all numbers. It's also unnecessary to cast the result of Convert.ToInt32 to an int, because it is already an int. This code is something that the lazy OP wouldn't understand, which is exactly the intension :-) • i -= -1. Very creative. I already reached the vote limit today, but I will upvote your answer as soon as I can. Dec 28, 2013 at 16:20 • As long as you insist that anything other than 6755 & 1436 is undefined behavior, despite OP's perception that most numbers seem to work... Dec 29, 2013 at 2:03 • What is the meaning of '=>'? Dec 29, 2013 at 14:01 • @Babibu I have never written a line of C# in my life but this is almost certainly a lambda expression. – thwd Dec 29, 2013 at 14:39 • uh oh, var x = Sum(0, 4) DivideByZeroException. Jan 6, 2014 at 18:42 # Bash - 72 bytes Sometimes traditional deterministic addition techniques are too precise, and unnecessarily fast - there are times when you want to give the CPU a bit of a rest. Introducing the lossy SleepAdd algorithm. #!/bin/bash (time (sleep$1;sleep $2)) 2>&1|grep re|cut -dm -f2|tr -d s  Sample run: > ./sleepadd.sh 0.5 1.5 2.001  This function is intended as a companion to the well-regarded SleepSort. It is left as an exercise to the reader to adapt this algorithm to make a lossy SleepMax to obtain the greater of two numbers. Pro Tip: This algorithm can be further optimised - a 2x speed increase is possible, if the numbers given to it are divided by 2 first. • Trolling 1: it works but it's stupidly slow, using the system timer to wait for the total time. Therefore larger numbers take linearly longer to add. Trolling 2: it even works for floating point, but the answers are always off by a small margin. Trolling 3: gratuitous and unnecessary use of grep, cut and tr. Trolling 4: any totals over 60 (seconds) are not handled correctly. – Riot Dec 28, 2013 at 16:26 • @Shingetsu: what, you're saying nobody else has heard of mp3 codecs? :P – Riot Dec 28, 2013 at 18:00 • I'm saying very few people actually make the association. Lame IS lame though. Vorbis master race. – user11485 Dec 28, 2013 at 18:02 • +1 for massively off-topic audio encoder wars diatribe :) – Riot Dec 28, 2013 at 18:03 • I believe my Bash-Hadoop version below is much more powerful and scalable!!!!1!!eleven! But I must say, I really love your version, sleepadd is great! +1 Dec 28, 2013 at 19:44 # Java public static void int sum(int a, int b) { try { File file = File.createTempFile("summer", "txt"); FileOutputStream fos = new FileOuptutStream(file); for (int i = 0; i < a; ++i) fos.write(1); for (int i = 0; i < b; ++i) fos.write(1); fos.flush(); fos.close(); return file.length(); } catch(Throwable t) { return sum(a, b); // Try again! } }  This basically writes a file with the number of bytes that should be equal to the actual sum. When the file is written, it asks the disk file table for the size of that file. • Can write or flush throw? It looks to me like you should move flush into each loop, and wrap the whole thing in a try-catch to retry the write if it or the flush fails. Dec 28, 2013 at 22:51 • I suggest you use a writer with the default character encoding instead of a stream. Then it can potentially break on a system for which your selected character encodes into multiple bytes. – Buhb Dec 29, 2013 at 18:41 # C In the quantum world you cannot depend on atomic operators like +, here's my implementation of addition in terms of quantum computing: #define DEPENDING ( #define ON #define EVERYTHING 32 #define DEFINED ) #define AS ON #define WITH { #define SOON if #define FIX AS #define TO = #define REPEAT for( #define SUBPOSED >> #define SUPERPOSITION int #define ADJUSTED << #define APPROACHES < #define SUBPOSITION ++ #define MATCHES & #define LEVEL DEPENDING #define OF FIX #define BY FIX #define CONTINUUM 1 #define VOID ~-CONTINUUM #define SUPERPOSED | #define DO DEFINED WITH #define CURVATURE } #define ITSELF FIX #define OTHERWISE CURVATURE else WITH #define RETURN return SUPERPOSITION ADD DEPENDING ON SUPERPOSITION SUPER_A, SUPERPOSITION SUPER_B DEFINED WITH FIX SUPERPOSITION A TO SUPER_A; FIX SUPERPOSITION B TO SUPER_B; FIX SUPERPOSITION RESULT TO VOID; FIX SUPERPOSITION CARRY TO VOID; FIX SUPERPOSITION I TO VOID; REPEAT FIX I TO VOID; I APPROACHES EVERYTHING; FIX I SUBPOSITION DEFINED WITH AS SOON AS LEVEL OF CARRY MATCHES CONTINUUM DO AS SOON AS LEVEL OF A SUBPOSED BY I MATCHES CONTINUUM DO AS SOON AS LEVEL OF B SUBPOSED BY I MATCHES CONTINUUM DO FIX RESULT TO RESULT SUPERPOSED BY CONTINUUM ADJUSTED BY I; FIX CURVATURE OF CONTINUUM; OTHERWISE AS SOON AS LEVEL OF B SUBPOSED BY I MATCHES CONTINUUM DO FIX VOID; // yes, you never know what could go wrong OTHERWISE FIX RESULT TO RESULT SUPERPOSED BY CONTINUUM ADJUSTED BY I; FIX CARRY TO VOID; FIX CURVATURE OF CONTINUUM; FIX CURVATURE OF CONTINUUM; // twice to make sure OTHERWISE AS SOON AS LEVEL OF A SUBPOSED BY I MATCHES CONTINUUM DO AS SOON AS LEVEL OF B SUBPOSED BY I MATCHES CONTINUUM DO FIX CARRY TO CONTINUUM; OTHERWISE FIX RESULT TO RESULT SUPERPOSED BY CONTINUUM ADJUSTED BY I; FIX CURVATURE OF CONTINUUM; OTHERWISE AS SOON AS LEVEL OF B SUBPOSED BY I MATCHES CONTINUUM DO FIX RESULT TO RESULT SUPERPOSED BY CONTINUUM ADJUSTED BY I; FIX CURVATURE OF CONTINUUM; FIX CURVATURE OF CONTINUUM; FIX CURVATURE OF CONTINUUM; FIX CURVATURE OF CONTINUUM; // we did some stuff there, sure the curvature needs a lot of fixing FIX VOID; // clean up after ourselfves RETURN LEVEL OF SUPERPOSITION DEFINED AS RESULT; FIX CURVATURE OF ITSELF  • +1 although it feels too readable for code trolling... Jan 2, 2014 at 22:09 # Haskell Computes the correct solution in O(n^2) time. Based on applicative functors that also implement Alternative. {- Required packages: - bifunctor -} import Control.Applicative import Data.Foldable import Data.Traversable import Data.Bifunctor import Data.Monoid -- Note the phantom types data Poly n a = X n (Poly n a) | Zero deriving (Show) twist :: Poly n a -> Poly n b twist Zero = Zero twist (X n k) = X n (twist k) instance Functor (Poly n) where fmap _ = twist instance Bifunctor Poly where second = fmap first f Zero = Zero first f (X n k) = X (f n) (first f k) -- Poly is a left module: (<#) :: (Num n) => n -> Poly n a -> Poly n a (<#) = first . (*) instance (Num n) => Applicative (Poly n) where pure _ = X 1 empty Zero <*> _ = empty (X n k) <*> q = (twist$ n <# q) <|> (X 0 (k <*> q))

instance (Num n) => Alternative (Poly n) where
empty = Zero
Zero    <|> q       = q
p       <|> Zero    = p
(X n p) <|> (X m q) = X (n + m) (p <|> q)

inject :: (Num n) => n -> Poly n a
inject = flip X (X 1 Zero)

extract :: (Num n) => (Poly n a) -> n
extract (X x (X _ Zero)) = x
extract (X _ k)          = extract k
extract _                = 0

-- The desired sum function:
daSum :: (Traversable f, Num n) => f n -> n
daSum = extract . traverse inject


Example: daSum [1,2,3,4,5] yields 15.

Update: How it works: A number a is represented as a polynomial x-a. A list of numbers a1,...,aN is then represented as the expansion of (x-a1)(x-a2)...(x-aN). The sum of the numbers is then the coefficient of the second highest degree. To further obscure the idea, a polynomial is represented as an applicative+alternative functor that doesn't actually hold a value, only encodes the polynomial as a list of numbers (isomorphic to Constant [n]). The applicative operations then correspond to polynomial multiplication and the alternative operations to addition (and they adhere to applicative/alternative laws as well).

The sum of numbers is then computed as mapping each number into the corresponding polynomial and then traversing the list using the Poly applicative frunctor, which computes the product of the polynomials, and finally extracting the proper coefficient at the end.

You want to add numbers?!? You are aware that this is a very complicated action? OK, on the other hand, you are the lead developer, you will have to face problems like this.

This is the simplest solution I could find:

int add_nums(int n1, int n2) {
int res, op1, op2, carry, i;
i = 32;
while (i --> 0) {
op1 = 123456 ^ 123457;
op2 = 654321 ^ 654320;
op1 = (n1 & op1) & op2;
op2 = (n2 & op2) & (123456 ^ 123457);
res = (res & (0xFFFF0000 | 0x0000FFFF)) | ((op1 ^ op2) ^ carry);
carry = op1 & op2;
res = res << 1;
}
return res;
}


Don´t fall prey to the operator "+", it is totally inefficient. Feel free to turn the "goes towards" operator around or use it for smaller numbers getting bigger.

# NODE.JS - SUMMMMYYMYYMY EDITION / IBM® Javascript Enterprise SUM Solution™

Wow, this a extremely hard question, but I will try my best to answer this.

## STEP ONE - TELNET Server

First we are going to have to receive the input, now any pro and enterprise coder (like me) should know the best way to receive input is to set up a telnet server!!!

Lets start off with the basic telnet server:

// Load the TCP Library
net = require('net'),
ibm = {},
fs = require('fs'),
clients = [];

//CREATES TEH TCP SEVA FOR INPUT
//COMMAND SUM and OBJECT (a, b, c, etc..) IS ONLY ELIGBLE
net.createServer(function (socket) {
clients.push(socket);
socket.write("WELKOME TO TEH SUM SEVA XD\n");

socket.on('data', function (data) {
ccc = [0,0,0,0,0,0,0];
if(!socket.needarray){
newdata = ibm.CLEANSOCKET(data);
if(newdata && newdata != '\b'){if(socket.nowdata){socket.nowdata += newdata}else{socket.nowdata = newdata}}else{
if(socket.nowdata){
if(socket.nowdata.replace(' ', '') == ('SUM')){
socket.write("Enter teh numbers\n");
socket.needarray = 1;
}
console.log(socket.nowdata);
socket.nowdata = null;
}}
}else if(newdata == '\b'){
socket.array = socket.array[socket.array.length - 1]
}else{
arraychar = ibm.CLEANARRAY(data);
if(arraychar != ('\n' || '\b')){if(socket.array){socket.array += arraychar}else{socket.array = arraychar}}else if(arraychar == '\b'){
socket.array = socket.array[socket.array.length - 1]
}else{
socket.end();
}
}
});
}).listen(23);
ibm.CLEANSOCKET = function(data) {
return data.toString().replace(/(\r\n|\n|\r)/gm,"");
}

ibm.CLEANARRAY = function(data) {
return data.toString().replace(/(\r)/gm,"");
}


There really isn't anything special to it, this is you typical telnet server. We've created some basic UNICODE cleaning functions to get us a nice raw string and we've also added our SUM function.

Now the user will have to enter 'SUM'. It will then prompt for them to enter teh numberz, once entered the summm() function is run and will calculate the sum of all the numbers entered.

## STEP TWO - summm

It's now time to create our summm function which will get the sum of all numbers inputted.
Here is the code:

//DOOOO SUMMMMM STAPH
function summm(string){
//Cleans out the string by converting it from unicode to base64 and then ASCII
stringa = (new Buffer((new Buffer(string).toString('base64')), 'base64').toString('ascii'));
//We will now convert our string to a new string with the format CHAR_ASCII_CODE + '.', etc...
x = '', c = 0;
stringa.split('').forEach(function (i){
c++;
x += i.charCodeAt(0);
if (c != stringa.length){x+= '.';}
})
stringb = x;
m = '';
stringb.split('.').forEach(function (i) {
m += String.fromCharCode(i);
});
stringc = m;
stringd = stringc.split(',');
var stringsa;
string.split(',').forEach( function (i) {
if(!stringsa){stringsa = parseInt(i);}else{stringsa += parseInt(i);}
});
return stringsa;
}


And there you go. Its your everyday IBM Solution. TELNET POWER ALL THE WAY!
First you enter SUM.
The server will then ask for the numbers you would like to add, and you can enter them as such: a, b, c, etc..

Trust me on this one, all the botnet's are using IBM® Javascript Enterprise SUM Solution™ these days ;).

And here is proof that everything works: (CLICKABLE)

• Would you mind telling me what IDE you are using in the screenshot? Visual studio doesn't give me that syntax highlighting Dec 29, 2013 at 16:48
• @JoethePerson: That's not an IDE, just an overpriced text editor called "Sublime Text". Dec 29, 2013 at 16:55
• @JoethePerson Like Shiki said its a text editor that's a bit more fancy and it does have a free version, see here: sublimetext.com.
– C1D
Dec 29, 2013 at 17:31
• @Shiki, I agree with you and I downloaded LightTable just a few days ago but I haven't opened it yet because I've been pretty busy.
– C1D
Dec 29, 2013 at 17:31

Here's a solution in Java for you. It relies on the time-tested "infinite monkeys theorem": if you are in a room with infinite monkeys, you will end up covered in thrown poop. Or something like that.

public static int sum(int a, int b){
if(a==0)return b;
Random r=new Random();
int number=r.nextInt();
if(number>a){
return sum(a, b);
}else{
return sum(a-number, b+number);
}
}

• Replace return sum(a-number, b+number); with return sum(sum(a,-number), sum(b,number));. You got to eat your own dog food right? Dec 28, 2013 at 16:13
• @emory: That will not work, I think. Dec 28, 2013 at 17:18
• @MartijnCourteaux The program has a dangerous flaw - it is a blatant troll. If someone were to ask what is b+number, then it would be obvious the whole method is unnecessary. Better to obfuscate that. Plus it will make it even slower. Dec 28, 2013 at 17:20
• @emory: Okay, I tested it and it apparently works. Great :) Dec 28, 2013 at 17:37

# C - overkill is best kill

Computers only have 0s and 1s, so it's actually very difficult to implement a proper, fast and scalable solution unto how to add. Luckily for you, I developed skynet 0.1284a, so I know how to solve this perilous problem.
Usually, you'd need to buy the C standard library DLC, as the core doesn't contain it, but I managed to "cheat" my way out of it. In short, this is the cheapest and most effective method.

#define SPECIAL {}
#define STABILIZE 0-
#define CORE double
#define DLC float
#define TRY if
#define COUNT while
DLC sum(DLC a, DLC b)
{
CORE EXTRADIMENTIONALRIFT = 0.0;//doubles are better
COUNT(a-->0){//downto operator
SPECIAL}
COUNT(b-->0){
SPECIAL}
EXTRADIMENTIONALRIFT -= (STABILIZE b);//we did some advanced stuff and need to stabilize the RAM
return (DLC)EXTRADIMENTIONALRIFT;//convert it into a DLC, so you don't have to pay for it
}


Just look at it. It's obviously evil.

• Note to OP: you can probably avoid the EXTRA DIMENTIONAL RIFT, but you'd then have to play with quantum physics, and you don't wanna do that.
– user11485
Dec 28, 2013 at 17:34

# Python

Uses the math identity log(ab) = log(a) + log(b) for a solution that works for small numbers, but overflows for any practical application.

Thus ensuring that our lazy programmer will think it works on test data, only to have it crash in the real world.

import cmath
def get_sum(list):
e_vals = map(lambda x: cmath.exp(x), list)
prod   = reduce(lambda x, y: x*y, e_vals)
return cmath.log(prod)

get_sum(range(1,10))  # correctly gives 45
get_sum(range(1,100)) # gives nan

• Doesn't work with python3 @ Ubuntu Dec 29, 2013 at 15:07
• @the_Seppi It works perfectly well. Just add from functools import reduce for python3. Dec 30, 2013 at 20:07

C#

You should use recursion to solve your problem

    public int Add(int a, int b)
{
if (b == 1)
{
//base case
return ++a;
}
else
{
}

}


If its good enough for Peano, its good enough for everyone.

• I just wanted to give this answer. IMAO this one and the sleepadd one are by far the best answers, since the others are needlessly complex. These instead are still completely useless but brief and elegant. It's too easy (hence boring) to make them useless by adding random complexity.
– o0'.
Dec 29, 2013 at 13:23
• The reasoning is flawless! Jan 2, 2014 at 19:30
• Shouldn't it be ++a instead of a++? (Edits must be at least 6 characters; is there something else to improve in this post?) stupid stupid stupid stupid SO
– o0'.
Jan 4, 2014 at 11:53
• @Lohoris - Yes, Yes it should. Fixed Jan 4, 2014 at 13:08

# C++

We expect an operation like addition to be very fast. Many of the other answers simply don't concentrate enough on speed. Here's a solution that uses only bitwise operations, for maximum performance.

#include <iostream>

int add2(int a, int b, int bits) {
// Usage: specify a and b to add, and required precision in bits (not bytes!)
int carry  = a & b;
int result = a ^ b;
while(bits --> 0) {       // count down to 0 with "downto" operator
int shift = carry << 1;
carry = result & shift;
result ^= shift;
}
return result;
}

int main() {
// Test harness
std::cout << add2(2, 254, 7) << std::endl;
return 0;
}

• Trolling 1: this actually works and is a valid way of adding numbers - it's not far off how hardware does it. However, the countdown uses subtract, so it's not a purely bitwise solution at all. Trolling 2: requirement to specify a precision in bits; incorrect precision results in nonsense answers. Trolling 3: "Downto" operator.
– Riot
Dec 28, 2013 at 16:57
• Add some inline assembler! Dec 28, 2013 at 23:39

My best solution so far, gives a pretty incomprehensible answer until you run aVeryLargeNumber()

function aVeryLargeNumber(){return Math.log(Math.log(Math.log(Math.log(Math.round((Math.log(!![].join()^{}-({}=={})|(0x00|0x11111)-(0x111111&0x10111))/Math.log(2))/(Math.tan(Math.PI/4)*Math.tan(1.48765509)))+(0xFFFF))/Math.log(2))/Math.log(2))/Math.log(2))/Math.log(2)}
var i=aVeryLargeNumber();
i--;
for(;i<b;i+=aVeryLargeNumber(),a+=aVeryLargeNumber());
return a;

}

– user11485
Dec 28, 2013 at 17:05
• What does it return? I'm not really into running this. Dec 28, 2013 at 17:17
• For those that don't want to run aVeryLargeNumber(): It returns 1. (I'll remove this if the OP pings me.) Dec 28, 2013 at 19:14

## C++ - Peano numbers with template metaprogramming (with optional doge)

C, like many other programming languages complicate things with absolute no reason. One of the most overcomplex systems in these languages are natural numbers. C is obsessed with the binary representation and all other completely useless details.

In the end, Natural number is just a Zero, or some other natural number incremented by one. These so called Peano numbers are a nice way to represent numbers and do calculation.

If you like doge I have written an C++ extension to allow the use of natural language for programming. The extension and this following code using my extension can be found at: http://pastebin.com/sZS8V8tN

#include <cstdio>

struct Zero { enum { value = 0 }; };

template<class T>
struct Succ { enum { value = T::value+1 }; };

template <unsigned int N, class P=Zero> struct MkPeano;
template <class P>
struct MkPeano<0, P> { typedef P peano; };
template <unsigned int N, class P>
struct MkPeano { typedef typename MkPeano<N-1, Succ<P> >::peano peano; };

template <class T, class U> struct Add;
template <class T>
struct Add<T, Zero> { typedef T result; };
template <class T, class U>

main()
{
printf("%d\n", MkPeano<0>::peano::value );
printf("%d\n", MkPeano<1>::peano::value );

printf("%d\n", Add< MkPeano<14>::peano, MkPeano<17>::peano >::result::value );
}


To further add the superiority of this method: The math is done at compile time! No more slow programs, your user doesn't want to wait for you to sum those numbers.

And for the serious part:

• I don't think I have to say this, but this is completely ridiculous.
• Works only for compile time constants.
• Doesn't work with negative numbers.
• The answer was provided by a person who actually cannot template metaprogram himself, so I wouldn't even know if it has other flaws.

My friends told me to dogify the code, so I did. It's fun, but I think it takes too much away from the fact that this is totally stupid as it is, so I only included it as a link.

• Wow. Such doge. Very upvote. Jan 2, 2014 at 22:12

I stopped trusting computers when I learned about floating point errors.

This JavaScript relies on precise human error checking:

while(prompt("Is this the answer: " + Math.round(Math.random()* 1000000)) !== "yes") {}


"Write a function that inputs and returns their sum."

Ok:


public static String inputAndReturnTheirSum() {
System.out.print("Input their sum: ");
return new Scanner(System.in).nextLine();
}


• This is my favourite. :D May 6, 2014 at 19:32

Java or C-style. This is O(log n). Note: This does not work for negative a or b.

public static int sum(int a, int b)
{
if ((a & b) == (a ^ a)) return a | b;
int c = a >> 1;
int d = b >> 1;
int s = a & 1;
int t = b & 1;
return sum(c, d + t) + sum(d, c + s);
}


Ideone demo here.

Obviously, a and b can become really large. Therefore, we must use Hadoop!

# Upload data to cluster:
$HADOOP_HOME/bin/hdfs dfs -mkdir applestore for i in seq 1$a; do
echo Banana > /tmp/.$i$HADOOP_HOME/bin/hdfs dfs -copyFromLocal /tmp/.$i applestore/android-$i$i done for i in seq 1$b; do
echo Orange > /tmp/.$i$HADOOP_HOME/bin/hdfs dfs -copyFromLocal /tmp/.$i applestore/java-$i$i done # Now we have all the data ready! Wow!$HADOOP_HOME/bin/hadoop jar $HADOOP_HOME/hadoop-streaming.jar \ -input applestore/ \ -output azure/ \ -mapper cat \ -reducer wc # We can now download the result from the cluster:$HADOOP_HOME/bin/hdfs dfs -cat azure/part-00000 | awk '{print $1;}'  As an added bonus, this approach involves a cat and a wc. This ought to be fun to watch! But I plan to use Mahout for this in the future (although I like cats). This must be the most scalable solution you get for this question. However, I can imagine that a recursive Hadoop solution is much more elegant. • I definitely see a theme in your answers. +Trolling points since this requires hadoop to work, and fails very messily if$HADOOP_HOME is unset.
– Riot
Dec 29, 2013 at 3:12

Ignore all those silly people with their non-generic and untestable manners. We need a performant, extensible and simple library for a project of such scale. It must support extension and substituton at every point of the code. For that, we need an equally extensible and simple language, that's why the best candidate is C#.

This is why I present you the beta version of my Operable Commons Library Enterprise Edition, version 0.8.4.4_beta1.3a_rev129857_dist29.12.13/master, which at this version exposes a IOperable interface, a IAddable interface so you can use your own efficient adding methods, and a default implementation of IAddable: the Addable class, which uses extremely efficient bitwise addition, without cheating and using the slow native subtraction for carry shifting. Of course, like any good library, it comes with a factory for every type it supports. The library also follows the principles of "handle it yourself", so you must guarantee that the input is valid and that the desired output is feasible, since it will not check for invalid data. Here it is (This code is licensed under the Microsoft Corporation Read-Only Proprietary Dont-Touch-This Obstructive License, Revision 3.1):

public interface IOperable {
uint Value {get; set;}
}

public interface IAddable : IOperable {
}

public uint Value {get; set;}

Value = value;
}

uint carry = a.Value & b.Value;
uint result = a.Value ^ b.Value;
while (carry != 0) {
uint shiftedcarry = carry << 1;
carry = result & shiftedcarry;
result ^= shiftedcarry;
}
}
}

public static class OperableFactory {
}
}


## JavaScript

Programming is all about algorithm. Let's go back to basic algorithm what we learn at the age of 3 - fingers counting.

var fingers = 0;
var hands = 0;
var FINGER_NUMBER = 5;

/* MEAT */
function sum(a,b){
while(a-- > 0) {
finger_inc();
}
while(b-- > 0) {
finger_inc();
}

return count_hands_and_fingers(); // We count the number of hands and fingers
}

/* Private functions */
function finger_inc(){
if(++fingers >= FINGER_NUMBER) {
hands++;
fingers = 0;
}
}

function count_hands_and_fingers() {
var total_count = 0;
total_count = hands * FINGER_NUMBER;
total_count += fingers;
}

document.write(sum(1,50));

• Firstly, being a lead developer, let's have a wise language choice - cross-platform, light-weight and portable.

• Secondly, have a global vision. Use Global var.

• Thirdly, ++s and --s

• Same as YFS (You-Finger-System), this does not support negative numbers

• Finally, you can alter FINGER_NUMBER according to the number of fingers you have.

JSFiddle: http://jsfiddle.net/e3nc5/

• But what if you need to count over 10? I don't have 3 hands! May 6, 2014 at 23:52
• Hotfix：Use feet, you can do that up to 20. Cheers, david. May 11, 2014 at 8:39

## TI-Basic 83/84

:Lbl Startup;bananapie\\repplie
:If X=10
::0→X
:If X=10
::Then
::Goto Lolbro\xdgtg
::End
:::::::::::::::::::Lbl Loled;epicly\that\is
::::::::::::::::::::::::::::::::::::::::::::::::::::::::::Input X,Y
::::::::::::::::::::::::::::::::::::::::::::::::::::::::::If X=Y
:::::::::::::::::::Then
::X+X→A
::Else
:X+Y→A
:A*1+0→A
:End
:If A>A
:Goto Somewhere
:Return A


Well, this one is a bit tricky. Fortunately, Python makes it reasonably straightforward. You'll need PIL to do this right.

import Image, ImageDraw

# First, we call the answer 'y', as in 'Y do we care?'
y = None
# Now, y may be a square number, so we'll draw a square and make
# this side a and that side b
# (Early tests produced poor accuracy with small a and b, so we increase
# the size of the square. This is an important program, after all!)
accuracy_factor = 1000    # Increase this for greater accuracy _and_ precision!
img = Image.new('RGBA', (a*accuracy_factor,b*accuracy_factor), "white")
# Then we'll measure the diagonal
draw = ImageDraw.Draw(img)
draw.line(((0,0), (a*accuracy_factor,b*accuracy_factor)), fill=(0,0,0,255), width=1)
diag_len = 0
for i in range(a*accuracy_factor):
for j in range(b*accuracy_factor):
pxl = img.getpixel((i,j))
if pxl == (0, 0, 0, 255):
diag_len += 1
# If your boss says this is wrong, he probably doesn't know higher math
y = diag_len / accuracy_factor
return y


Intentionally using the slow Image.getpixel(). I'm not sure it's actually slow enough, though, darnitall. RGBA just to take up extra memory.

## JAVA

In the below code, ... stands in for code that I was too lazy to write but you should be able to figure out. To really do this in style, would require a code generation program. The limits 0 and 10 could be changed to whatever. The bigger the limits the more code and a computer could easily fill in the ...s.

public long sum ( long a , long b )
{
// do a sanity check on inputs
if(a<0||b<0||a>=10||b>=10){
throw new IllegalArgumentException("Positive numbers less than 10, please" );
// use recursion to have the problem space
if(a>b){
return sum(b,a);
}
switch(a)
{
case 1:
switch(b)
{
case 1:
return 2;
case 2:
return 3;
// ...
case 8:
return 9;
default:
assert b==9;
return 10;
}
case 2:
switch ( b )
{
// ...
}
// ...
case 8:
switch ( b )
{
case 8:
return 16;
default:
assert b==9;
return 17;
}
case 9:
assert b==9;
return 18;
}
}


a function that inputs and returns their sum

# Lua

function f()
return theirsum
end


The code is done. Be very careful about that. This code is ultra-complex and is probably prone to become an intelligent conscious and self-aware being. It's highly classified top-secret code.

/*
* This code is part of the Skynet. It is highly classified and top-secret!
*/
package com.muchbigcompany.skynet;

import javax.swing.JOptionPane;

/**
* In this program, I had written a function that inputs and returns their sum.
*/
public class Skynet {
public static void main(String[] args) {
int theirSum = inputsAndReturnsTheirSum();
JOptionPane.showMessageDialog(null, "Their sum is " + theirSum);
}

/**
* This is a function that inputs and returns their sum.
* @return their sum.
*/
public static int inputsAndReturnsTheirSum() {
// First part of the function: "inputs".
String inputs = JOptionPane.showInputDialog("Inputs theirs sum");
int theirSum = Integer.parseInt(inputs);

// Second part of the function: "returns their sum".
return theirSum;
}
}


# C++

Of course you are gonna need some template magic:

template<int I> struct identity {
static const int value = I;
};

template<int A, int B> struct sum {
static const int value = identity<A>::value + identity<B>::value;
};

auto main(int argc, char* argv[]) -> int {
std::cout << sum<1, 3>::value;
return 42;
}


JAVA

Hard problem.

It is known that in computer science there are problems that verifying their answers is easier than finding them. So, you should use a random algorithm for guessing the solution, then verify it (efficiently!), and hope to get the right result in reasonable time:

public long sum(int a, int b)
{
Random r=new Random();
While(15252352==15252352)
{
long sum=r.nextLong(); // guess the solution
if (sum - a == b)      // verify the solution
return sum;
}
}

– Wasi
Dec 30, 2013 at 19:18

This function is under patent of my company, I can provide you an obfuscated licensed copy of it:

## Javascript:

function sum(a,b) { return eval(atob('YSti')) };


Usage:

sum([arg1],[arg2]);


# Python

Programming is about fault tolerant. The following is an implementation of sum that will add anything without fussing out. It will transparently sort the elements in the order that can be added. In case, its not addable, it will flag it as NaN.

def apple2apple_sum(*args):
total = {type(args):[[args],args]}
try:
args + args
except TypeError:
total[type(args)][-1] = "NaN"
for elem in args[1:]:
if type(elem) in total:
if total[type(elem)][-1] != "NaN":
total[type(elem)][-1] += elem
total[type(elem)].append(elem)
else:
total[type(elem)] = [[elem],elem]
try:
elem + elem
except TypeError:
total[type(elem)][-1] = "NaN"

>>> apple2apple_sum(1,2,3,'a', 'b', 4, 5.1, 6.2, 'c', map, 10, sum)
[[['a', 'b', 'c'], 'abc'], [[<built-in function map>, <built-in function sum>], 'NaN'], [[5.1, 6.2], 11.3], [[1, 2, 3, 4, 10], 20]]


## Fortran

Obviously the most efficient way is to shift your bits. This can be easily done with C+Fortran via the iso_c_binding module:

program add_func
use iso_c_binding
implicit none
! declare interface with c
interface
import :: c_int
integer(c_int), value :: x,y
end subroutine
end interface
! need our numbers
integer(c_int) :: x,y

print *,"what two numbers do you need to add (separated by comma)"


where the C routine is

#include <stdio.h>


You need to compile the C code first (e.g., gcc -c mycfile.c) then compile the Fortran code (e.g., gfortran -c myf90file.f90) and then make the executable (gfortran -o adding myf90file.o mycfile.o).