C program to subtract two numbers without using arithmetic operators

Software Development
#1

Write a function subtract(x, y) that returns x-y where x and y are integers. The function should not use any of the arithmetic operators (+, ++, –, -, … etc).
The idea is to use bitwise operators. Like addition, Write a function Add() that returns sum of two integers. The function should not use any of the arithmetic operators (+, ++, –, -, … etc). Sum of two bits can be obtained by performing XOR (^) of the two bits. Carry bit can be obtained by performing AND (&) of two bits. We can extend this logic for integers. If x and y don’t have set bits at same position(s), then bitwise XOR (^) of x and y gives the sum of x and y. To incorporate common set bits also, bitwise AND (&) is used. Bitwise AND of x and y gives all carry bits. We calculate (x & y) << 1 and add it to x ^ y to get the required result. The idea is to use subtractor logic.
The truth table for the half subtractor is given below.

X Y Diff Borrow
0 0 0 0
0 1 1 1
1 0 1 0
1 1 0 0
From the above table one can draw the Karnaugh map for “difference” and “borrow”.
So, Logic equations are:

Diff   = y ⊕ x
Borrow = x' . y

Following is implementation based on above equations.

// C program to Subtract two numbers

// without using arithmetic operators

#include<stdio.h>

int subtract( int x, int y)

{

// Iterate till there

// is no carry

while (y != 0)

{

// borrow contains common

// set bits of y and unset

// bits of x

int borrow = (~x) & y;

// Subtraction of bits of x

// and y where at least one

// of the bits is not set

x = x ^ y;

// Borrow is shifted by one

// so that subtracting it from

// x gives the required sum

y = borrow << 1;

}

return x;

}

// Driver Code

int main()

{

int x = 29, y = 13;

printf ( "x - y is %d" , subtract(x, y));

return 0;

}

Output :

x - y is 16