Jun
13

Add two numbers without using (+, -, /, *)

This question was asked in Qualcomm. without using arithmetic operations add two numbers. We can use bitwise operators to add two numbers.

#include <iostream>   using namespace std;   int sum(int a, int b);   int main() { int a = 1, b = 0; int sum_ = 0; sum_ = sum(a, b); cout << sum_; }   int sum(int x, int y) { if( x == y) return (x << 1); /*if numbers are same return 2*number*/ else return (x ^ y); }

Shahid Siddique software engineer and WordPress adept. He works with Cerner Corp as a Software Engineer.

Shahid Siddique

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June 13, 2012   //   Visual C/C++   //   3 Comments   //   Tags: c++ code, XOR

3 Comments to “Add two numbers without using (+, -, /, *)”

• June 24, 2012 at 2:34 am
  stephen says:

  What about 1+3?

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  • June 24, 2012 at 10:45 pm
    Shahid Siddique says:

    Thank you stephen for your comment. I will look into this and get back to you with the solution.

    Reply
  • June 24, 2012 at 10:51 pm
    Shahid Siddique says:

    So I came across one solution: http://www.daniweb.com/software-development/c/threads/84950/adding-two-numbers

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    /* integer1 = 3 = 0011b integer2 = 1 = 0001b first operation second operation third operation 0011 0011 shift by one 0001 0001 ______ ______ XOR 0010 AND 0001 ------> <<1 0010 Again... first operation second operation third operation previous XOR 0010 0010 shift by one previous <<1 0100 Again... first operation second operation third operation previous XOR 0000 0000 shift by one previous <<1 0000   When AND equals 0 the result of XOR is is the result   Result is 100 which is 4 */   #include <iostream>   using namespace std;     unsigned long add(unsigned long integer1, unsigned long integer2) { unsigned long xor, and, temp;   and = integer1 & integer2; /* Obtain the carry bits */ xor = integer1 ^ integer2; /* result­ing bits */   while(and != 0 ) /* stop when carry bits are gone */ { and <<= 1; /* shift­ing the carry bits one space */ temp = xor ^ and; /* hold the new xor result bits*/ and &= xor; /* clear the pre­vi­ous carry bits and assign the new carry bits */ xor = temp; /* result­ing bits */ } return xor; /* final result */ }   int main() { cout <<add(3,3); }

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