Write a program that makes 2 + 2 = 5

http://codegolf.stackexchange.com/questions/28786/write-a-program-that-makes-2-2-5

by bryan_rasmussen • 12 years ago

123 71 12 years ago

codegolf.stackexchange.com

yiedyie• 12 years ago

JavaScript ;)

http://codegolf.stackexchange.com/a/29944/12328

blueveek• 12 years ago

7 bytes: !''|2+2

notduncansmith2• 12 years ago

5 bytes: 1|2+2

NaNaN3• 12 years ago

The JavaScript answer from @user1886419 is a great joke. lol

runarberg• 12 years ago

Julia

  !(a::Int) = !bool(a)
  !0 + 2 + 2

  ##> 5

orangecat• 12 years ago

Or more directly:

  julia> +(a::Int, b::Int) = if a==b==2 5 else a-(-b) end
  julia> 2+2
  5
  julia> 3+1
  4

(The else part can't use + or it will cause infinite recursion).

anon4• 12 years ago

You know, this would be possible in C and C++ if you add some code at the start of main() that unprotects the executable section, goes through it and modifies each "load 4 in register X" instruction to "load 5", then forks a process that attaches itself as a debugger to the executable to monitor every add and change the result to 5 if it's 4.

userbinator• 12 years ago

That Java one is particularly amusing, as most programmers otherwise familiar with the language would never know about the integer cache (and their reaction upon discovering that there is one would probably be the same as mine - WTF?)

Edit: I had a feeling I'd heard of an "integer cache" somewhere else" in a WTF-eliciting context... http://thedailywtf.com/Articles/The-Integer-Cache.aspx !

xxs• 12 years ago

The integer cache is well known for over 10 years, presently there is a JVM knob to make it larger: java.lang.Integer.IntegerCache.high So the solution is not surprising at all.

Personally I'd just employ an enhancing classload and replace some of bytecodes ICONST_2 with ICONST_3. (one byte change in the bytecode). Yet, overall uninteresting question.

benjiweber• 12 years ago

The was also http://thedailywtf.com/Articles/Disgruntled-Bomb-Java-Editio...

sp332• 12 years ago

It reminds me of a prank a relative of mine used to play on other people using punch-card computers. The computer had no math unit (!) so addition tables and multiplication tables were fed in on punch cards. Easy enough to change one of those and make perfect code give very odd answers.

vxNsr• 12 years ago

Boy, he must've been hated, I personally am too young to remember the punch-card days of programming but my dad spent most of his younger days doing it that way and he said that one of the reasons his code has fewer bugs/mistakes than mine (aside from 30+ years exp on me) is because no one wanted to make a stupid mistake somewhere in their stack [of cards] that would mean going to that back of the line just because they used a plus instead of a minus or something (I don't remember the exact example).

cbsmith2• 12 years ago

Considering the Flyweight is a GoF Design Pattern, and that it talks about doing this specifically for managing integer objects... this shouldn't be a WTF for most people. It's a trick as old as the hills too. Smalltalk has had it going back decades.

kyrra• 12 years ago

It also leads to some interesting programming errors in Java.

    class A {
        public static void main(String[] args) {
            Integer a = 100, b = 100;
            Integer c = 10000, d = 10000;
            System.out.println(a == b);
            System.out.println(c == d);
        }
    }

Prints:

    true
    false

For those unaware, the == check above is doing an Object check (are they the same object). The cache makes the '100' return the same instance of an object. To do actual value checks, you'd need to do c.equals(d)

hnriot• 12 years ago

what a dumb language! operator overloading should have bound the equals method to the == operator without the programmer having to be aware of this. when in java do you ever want to compare if two objects are the same referenced with the same ptr? Never, so why not do the right thing by the user of the language.

At least be consistent between primitives and objectives.

Sadly, I have to use this language daily. My heart belongs to python, but java pays the bills.

Kenji2• 12 years ago

I would like to post on there but it's protected from people with low reputation... Maybe someone here appreciates my solution in C:

  #include <stdio.h>
  
  int main(int argc, char* argv[]){
  	int arr[1];
  	int a = 2;
  	int b = 2;
  	arr[1] = 3;
  	printf("%d", a+b);
  
  	return 0;
  }

Explanation: I go out of bounds of the array arr, it only has one value but I access the second value. That's why b is likely to get overwritten with 3 and hence a+b=5

wenderen• 12 years ago

I think the line above the printf call should be

    arr[2] = 3;

I ran your program and I got 4. Then I changed arr[1] to arr[2] and got 5, as I expected.

Kenji2• 12 years ago

It's only likely, but not certain that a or b get overwritten. How your stack is laid out is entirely up to the compiler. arr[1] is already out of bounds, however, we don't know for certain what's immediately above the array.

wyager• 12 years ago

The compiler is probably aligning all stack elements to 8 bytes.

rkowalick• 12 years ago

Here's a Ruby solution that really does modify the + operation:

  class Fixnum
    alias_method :old_plus, :+
    def +(*args)
      old_plus(*args).old_plus(1)
    end
  end

  2 + 2 #=> 5

eqnoob123• 12 years ago

Here's one in Ocaml that modifies the + operator

let (+) x y = match (x,y) with (2,2) -> 5 | (x,y) -> x+y;;

a-nikolaev• 12 years ago

A shorter one:

let (+) 2 2 = 5 in 2+2;;

dom96• 12 years ago

Here is a Nimrod version which does the same:

  proc `+`(x, y: int): int =
    result = x
    result.inc(y)
    result.inc

  echo(2 + 2) # => 5

hpvic03• 12 years ago

Here's a ruby solution that only affects 2+2:

    class Fixnum
      alias_method :old_plus, :+
      def +(other)
        return 5 if (self == 2 && other == 2)
        old_plus(other)
      end
    end

JadeNB• 12 years ago

As Flonk points out at http://codegolf.stackexchange.com/a/28794, nothing beats Haskell for casual re-definition of operators. (I seem to remember that this is even on the Haskell wiki somewhere, though I can't find it right now.)

comex2• 12 years ago

    #include <stdio.h>
    #include <stdbool.h>
    #include <stdlib.h>

    int nums[5] = {1, 0, 1, 0, 1};
    static int cur;

    int main(int argc, char **argv) {
        for(int i = 1; nums[i] != 5 && i < 10; i++) {
            cur = i;
            if(i/4 == 1) {
                printf("2 + 2 <= %d = %s\n", cur, 2 + 2 <= i ? "true" : "false");
                printf("2 + 2 == %d = %s\n", cur, 2 + 2 == i ? "true" : "false");
                printf("\n");
            }
        }
    }

No strange array writes! No indirect writes at all!

It's finicky and I believe depends on register allocation, but when I compile it using whatever version of gcc-mp-4.9 I have installed from MacPorts at -O3, it outputs, among other things:

    2 + 2 == 5 = true

(For all they say about evil optimizing compilers, it was really hard to make this work.)

tgb2• 12 years ago

Can anyone explain? It didn't work for me but still gives odd results depending on optimization level that I don't understand.

comex2• 12 years ago

Because nums is a length 5 array, and the loop condition checks nums[i], GCC assumes going into the loop that i is a valid index, i.e. i <= 4. The division then rules out the possibility that i < 4, so for the purposes of optimization, my version of GCC thinks that i always equals 4 at the printfs, even if it's actually higher.

It's based on an old Linux bug... I should learn more about GCC internals in order to tell tell why I couldn't get it to work without a loop. In general, any sequence where some property not holding would cause undefined behavior to occur allows a conforming compiler to assume it does hold, but for most basic examples neither GCC nor Clang does anything special.

dgl• 12 years ago

Because I can't post on stackoverflow either, here's my variant in C (requires glibc, and a deprecated function at that).

  #include <printf.h>
  #include <stdio.h>
  #include <string.h>

  int d_cb(FILE *stream, const struct printf_info *info, const void *const* args) {
    int num = (*(int*)(((int**)args)[0])), numw = info->width;
    char str[10], numn = num|numw, *out = str;
    num = num < 0 ? *out++ = '-', -num : numn;
    do { *out++ = '0' + num % 10; num /= 10; } while(num);
    fwrite(str, out - str, 1, stream);
    return 0;
  }

  void init() {
    register_printf_function('d', d_cb, 0);
  }

  int main() {
    init();
    printf("2+4 = %d, 2+2 = %1d\n", 2+4, 2+2);
    return 0;
  }

masak• 12 years ago

Here's the Perl 6 solution:

  multi infix:<+>(2, 2) { 5 }
  say 2 + 2;  # 5

Clex• 12 years ago

Here's mine :

    $ cat test.c
    #include <stdio.h>
    
    int main() {
        int a = 3;
        int b = 3;
    
        // aren't we supposed to add 2 and 2 ??/
        a = 2;
        b = 2;
    
        printf("%d\n", a + b);
    
        return 0;
    }
    $ gcc -W -Wall -trigraphs test2.c 2>/dev/null
    $ ./a.out
    5
    $

tbirdz• 12 years ago

Very sneaky. For anyone else wondering why this works, look up trigraphs.

vidarh• 12 years ago

And after looking it up, you better then realise that if you actually ever use them in production code you better have a really good reason, or your colleagues will forever curse your name..

I think the above is probably the first time I've seen trigraphs used in the last 20 years or so.

adito• 12 years ago

http://en.wikipedia.org/wiki/Digraphs_and_trigraphs#C

sorry for spoiling:

"??/" got replaced by "\"

gizmo6862• 12 years ago

Is the trick here the "-trigraphs" flag, or the fact that you are showing us test.c but compiling test2.c? Both approaches seem like they should work.

cdcarter• 12 years ago

It's the trigraph ??/ which translates to a \.

im3w1l• 12 years ago

  public class Main {
   	public static void main (String [] args) {
  		System.out.println("‮5=2+2‭"); // prints ‬4=2+2
  	}
  }

Too2• 12 years ago

How about the opposite? Make 2+3=4, here in C.

    add2And3(int *a, int *b)
    {
       *a = 3;
       *b = 2;
       printf("%d", *a + *b);  //4
    }
    void main()
    {
       int storage;
       add2And3(&storage, &storage);
    }

social_monad• 12 years ago

Simple solution in R:

   > '%+%' <- get( '+')
   > '+' <- function( e1, e2) 1 %+% e1 %+% e2
   > 2+2
   [1] 5

Regards

X4• 12 years ago

Asking: Isn't the only true solution the one using Decibels for Addition? Because everyone else either just redefines or overwrites a value. Also addition of decibels cannot be disproofed as fake.

NaNaN3• 12 years ago

After thinking for a while, I found that there is no true answer. If we are not dealing with normal calculating, every answer should redefine the calculating system or value in a tricky way. The simplest answer I can tell related to the addition of decibels is like this:

    function add(a, b): return 5

But it is just not so fun.

san86• 12 years ago

why not just overload the + operator in C++?

boost_• 12 years ago

first thing i thought when i read the title:

    #include <iostream>

    class Int
    {
    private:
            int _value;

    public:
	    Int(int value) : _value(value) {}
	    Int operator + (const Int& b){ return Int(++_value + b._value); }
	    friend std::ostream& operator << (std::ostream& out, const Int& value){ out << value._value; return out; }
    };

    int main(int argc, char** argv)
    {
	    std::cout << Int(2) + Int(2) << std::endl;
            return 0;
    }

output: 5

howeman• 12 years ago

Combining a couple ideas, here's a Go solution

http://play.golang.org/p/XnPglbkLW4

a_bonobo2• 12 years ago

Could you please explain why this works? How come int(a) produces 2.5?

chewxy• 12 years ago

scroll down to find the underhandedness

noobermin• 12 years ago

I'm surprised no one has suggested this yet. In python

Write a program that seemingly adds the numbers 2 and 2

a=2+2

and outputs 5.

print(5)

Yeah, I know that isn't the intent here, but, following the wording of the contest rules exactly...

diego• 12 years ago

My first thought in Clojure.

  (let [+ (partial + 1)]                                                                                                                                                                                
        (+ 2 2))

Lockal• 12 years ago

UB abuse in C/C++ Linux x86-64 (works with gcc/clang/icc with any optimization level):

  #include <stdio.h>

  int main() {
    double x;
    printf("Input any number:\n> ");

    if (scanf("%lf", &x) == 1) {
      printf("Is 2 + 2 really equals %g? Let's try!\n", x);
      printf("2 + 2 = %g\n", 2 + 2);
    } else {
      printf("Invalid input!\n");
    }
  }

Output:

  Input any number:
  > 5
  Is 2 + 2 really equals 5? Let's try!
  2 + 2 = 5

Explanation: linux x86-64 calling convention uses xmm registers to pass fp values. In the first printf we initialize %xmm0 with some value. In the second printf we put integer 4 in %esi, however printf reads value again from %xmm0. Here is an assembly in GCC explorer (sorry for shortened link, that's how GCC explorer works): http://goo.gl/mY9phE

isivaxa• 12 years ago

  #!/bin/bash
  tty
  if [[ $? -ne 0 ]] ; then
      # command returned nonzero
      echo "not a tty"
  else
      # command returned zero
      n=$?
      ((n=$n+2))
      ((n=$n+2))
      echo "0+2+2=$n";
  fi

terranera• 12 years ago

#lang racket (define (add-bizzare x y) (round (+ (+ x 0.4) (+ y 0.4))))

Datsundere• 12 years ago

Haskell:

  let 2 + 2 = 5

gdewilde• 12 years ago

You have to add 1 to either the first or the second 2 or to the sum of the two.

whoopdedo• 12 years ago

Lua:

    local debug = require "debug" -- in case 5.2
    debug.setmetatable(1, {__tostring=function(n)
      return ("%g"):format(n+1)
    end})

    print(2+2)

Trying to make 2+2==5 evaluate as true is a lot harder. Although Lua lets you define a custom + operator, it's a fallback not an override.

charles20132• 12 years ago

Ruby solution (don't have enough rep to post to codegolf):

  def mad_addr(a,b)
    a == b && a == 2 ? (a.object_id + b.object_id)/a : a + b
  end

Explanation: If a and b are both equal to 2 we can sum the values of the object_id's (the object_id of 2 is 5), and then divide the sum by 2. [Edit to add] For all other numbers this should behave as expected.

I just learned the object_id of 2 is 5 while doing this exercise, and I would like to continue learning more about how Ruby works. So if you have feedback, criticisms, or other ideas regarding this solution please feel free to share :)

[edit: added missing `=`]

nitrogen2• 12 years ago

a == b && a = 2 ? (a.object_id + b.object_id)/a : a + b

It looks like you are missing an equals sign on a == 2.

The object_id appears to be the tagged pointer that contains the value cast to an integer. Ruby stores all of its data types using a single underlying C data type (called VALUE in the Ruby source), using the bottom two bits to identify special values that aren't really pointers. This is why Fixnum can only hold 31 or 63 bits instead of 32 or 64 (including sign), for example:

    # Ruby irb on a 32-bit system
    > (1<<30 - 1).class
    => Fixnum
    > (1<<30).class
    => Bignum

Write a program that makes 2 + 2 = 5

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