Day 38 addition without arithmetic operators

Author: mandliya

README.md

@@ -4,9 +4,9 @@
 
 | Current Status|     Stats     |
 | :------------: | :----------: |
-| Total Problems | 57 |
-| Current Streak | 37 |
-| Longest Streak | 37 ( August 17, 2015 - September 22, 2015 ) |
+| Total Problems | 58 |
+| Current Streak | 38 |
+| Longest Streak | 38 ( August 17, 2015 - September 23, 2015 ) |
 
 </center>
 
@@ -61,6 +61,7 @@ Include contains single header implementation of data structures and some algori
 | Given two integers, determine if their sum would be interger overflow.| [integerOverflow.cpp](bit_manipulation/integerOverflow.cpp)|
 | How many bit flip operation would require to convert number A to B. | [countNumberOfBitFlips.cpp](bit_manipulation/countNumberOfBitFlips.cpp)|
 | Given a number x and two positions (from right side) in binary representation of x, write a function that swaps n right bits at given two positions and returns the result. It is also given that the two sets of bits do not overlap.|[swapSetOfBits.cpp](bit_manipulation/swapSetOfBits.cpp)|
+| Add two numbers without using any arithmetic operators | [addition_without_operators.cpp](bit_manipulation/addition_without_operators.cpp)
 
 ### Cracking the coding interview problems
 | Problem | Solution |

bit_manipulation/addition_without_operators.cpp

@@ -0,0 +1,44 @@
+/**
+ * Addtion of two numbers without using any arithmatic operators.
+ */
+#include <iostream>
+
+int sum( int x, int y ) {
+  int carry = 0;
+  while( y != 0 ) {
+
+    //if x and y don't have same bits at same position, XOR of x and y
+    //gives us sum of x and y at those positions.
+    // bits wont change at positions containing same bits, however,
+    // we have to incorporate carry bits. AND of x and y gives all carry bits.
+    // carry will be added from position 1 (not 0), so we need to shift carry
+    // by 1.
+    carry = x & y;
+    x = x ^ y;
+    y = carry << 1;
+  }
+  return x;
+}
+
+//recursive version
+int sum2( int x, int y ) {
+  if ( y == 0 ) {
+    return x;
+  }
+  return sum2( x ^ y, (x & y) << 1);
+}
+
+int main()
+{
+  int x , y;
+  std::cout << "Addition of two numbers without using arithmatic operators:\n";
+  std::cout << "Enter num 1 :" ;
+  std::cin >> x;
+  std::cout << "Enter num 2 :" ;
+  std::cin >> y;
+  std::cout << "Iterative version - Addition of " << x << " and " << y
+            << " is " << sum(x, y) << std::endl;
+  std::cout << "Recursive version - Addition of " << x << " and " << y
+            << " is " << sum2(x, y) << std::endl;
+  return 0;
+}