# Perl program to add two numbers using only bitwise operators
# See https://stackoverflow.com/questions/4068033/add-two-integers-using-only-bitwise-operators
use strict;
use warnings;

# See [id://11135535] (note that <C>use integer</C> subtly affects bit operations in Perl)
use integer;

sub BitAdd1 {
   my ($aa, $bb) = @_;
   my $carry  = $aa & $bb;
   my $result = $aa ^ $bb;
   while ($carry != 0) {
      my $s_carry = $carry << 1;
      $carry = $result & $s_carry;
      $result ^= $s_carry;
   }
   return $result;
}

sub BitAdd2 {
   my ($aa, $bb) = @_;
   $bb == 0 and return $aa;
   return BitAdd2($aa ^ $bb, ($aa & $bb) << 1);
}

for my $r ( [0, 1],
            [1, -1],
            [69, -42],
            [42, 69],
            [-42, 69],
            [-42, -69],
            [256, 512],
            [123456789, 1],
            [2147483647, 1],
            [-2147483648, 1] ) {
   my $sum0 = $r->[0] + $r->[1];
   my $sum1 = BitAdd1($r->[0], $r->[1]);
   my $sum2 = BitAdd2($r->[0], $r->[1]);
   print "$r->[0] + $r->[1] = $sum0 ($sum1 $sum2)\n";
   $sum0 == $sum1 or die "oops 1";
   $sum0 == $sum2 or die "oops 2";
   $sum1 == $sum2 or die "oops 3";
}

##</code><code>##

0 + 1 = 1 (1 1)
1 + -1 = 0 (0 0)
69 + -42 = 27 (27 27)
42 + 69 = 111 (111 111)
-42 + 69 = 27 (27 27)
-42 + -69 = -111 (-111 -111)
256 + 512 = 768 (768 768)
123456789 + 1 = 123456790 (123456790 123456790)
2147483647 + 1 = 2147483648 (2147483648 2147483648)
-2147483648 + 1 = -2147483647 (-2147483647 -2147483647)

##</code><code>##

// C++ Program to add two numbers without using arithmetic operator.
// bitadd.cpp
// Built with : g++ -o bitadd -std=c++11 -Wall -O3 bitadd.cpp
#include <cstddef>
#include <cstdint>
#include <vector>
#include <utility>
#include <iostream>
#include <fstream>
#include <sstream>

// uncomment one of these
// typedef int32_t  my_int;
typedef int64_t  my_int;

my_int Add(my_int x, my_int y) 
{ 
    // Iterate till there is no carry   
    while (y != 0) { 
        // carry now contains common set bits of x and y 
        my_int carry = x & y;   
  
        // Sum of bits of x and y where at least one of the bits is not set 
        x = x ^ y;  
  
        // Carry is shifted by one so that adding it to x gives required sum
        y = carry << 1; 
    } 
    return x; 
}

using my_list_type = std::vector< std::pair<my_int, my_int> >;
  
int main() 
{
    my_list_type ops = {
       {           0,   1 },
       {           1,  -1 },
       {          69, -42 },
       {          42,  69 },
       {         -42,  69 },
       {         -42, -69 },
       {         256, 512 },
       {   123456789,   1 },
       {  2147483647,   1 },
       { -2147483648,   1 }
   };
   my_int sum0, sum1;
   std::cout << "sizeof my int = " << sizeof(my_int) << "\n";
   for (const auto& c : ops) {
      sum0 = c.first + c.second;
      sum1 = Add(c.first, c.second);
      if (sum0 != sum1) { std::cout << "oops!\n"; }
      std::cout << c.first << " + " << c.second << " = " << sum0 << " (" << sum1 << ")\n";
   }

   return 0; 
}

##</code><code>##

sizeof my int = 8
0 + 1 = 1 (1)
1 + -1 = 0 (0)
69 + -42 = 27 (27)
42 + 69 = 111 (111)
-42 + 69 = 27 (27)
-42 + -69 = -111 (-111)
256 + 512 = 768 (768)
123456789 + 1 = 123456790 (123456790)
2147483647 + 1 = 2147483648 (2147483648)
-2147483648 + 1 = -2147483647 (-2147483647)