# 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";
}