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