Bitwise recursive addition of two integers in C

In this problem, we are given two numbers. Our task is to create a C program for bitwise recursive addition of two integers.

The logic of summing using bitwise operations is similar to what we did in preschool. To sum, we usually add each digit of the number, and if a carry is present, we add it to the next number.

We will do something similar, using the XOR operator to sum and the AND operator to check for carry. If there is a carry, we add it back to the number, otherwise not.

This is the logic of a Half Adder that you may have learned in digital electronics. See here...

Now the sum is calculated using a^b i.e. XOR b, if the first bit of both is set we need to check if extra carry needs to be propagated. We need to add an extra setting bit to the number.

So the bitwise arithmetic will be

Step 1 - Find the XOR of a and b i.e. a^b and store it in the result variable.

Step 2 - Check if {(a & b)

Step 2.1 - If equal to 0, Then print the result, which is the final result.

Step 2.2 - If not equal to 0 equals 0, then go to step 1 where a = {(a & b)

Example

Procedural algorithm that illustrates how this function works -

Live demonstration

#include <stdio.h>
int addNumbers(int a, int b) {
   int carry = (a & b) << 1;
   int result = a^b;
   if (carry == 0)
      return result;
   else
      addNumbers(carry, result);
}
int main(){
   int a = 54, b = 897;
   printf("The sum of %d and %d using bitwise adding is %d", a, b, addNumbers(a, b));
   return 0;
}

Output

The sum of 54 and 897 using bitwise adding is 951’

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