Write a program to add one to a given number. The use of operators like ‘+’, ‘-‘, ‘*’, ‘/’, ‘++’, ‘–‘ …etc are not allowed.

Examples:

Input: 12

Output: 13

Input: 6

Output: 7

This question can be approached by using some bit magic. Following are different methods to achieve the same using bitwise operators.

**Example :**

Assume the machine word length is one *nibble* for simplicity. And x = 2 (0010), ~x = ~2 = 1101 (13 numerical) -~x = -1101

Interpreting bits 1101 in 2’s complement form yields numerical value as -(2^4 – 13) = -3. Applying ‘-‘ on the result leaves 3. The same analogy holds for decrement. Note that this method works only if the numbers are stored in 2’s complement form. Like addition, Write a function Add() that returns sum of two integers. To subtract 1 from a number x (say 0011001000), flip all the bits after the rightmost 1 bit (we get 001100**1** 111). Finally, flip the rightmost 1 bit also (we get 0011000111) to get the answer. 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. Write a program to subtract one from a given number. 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 use of operators like ‘+’, ‘-‘, ‘*’, ‘/’, ‘++’, ‘–‘ …etc. are not allowed.

**Method 2**

We know that the negative number is represented in 2’s complement form on most of the architectures. We have the following lemma hold for 2’s complement representation of signed numbers.

Say, x is numerical value of a number, then

**~x = -(x+1)** [ ~ is for bitwise complement ]

(x + 1) is due to the addition of 1 in 2’s complement conversion

To get (x + 1) apply negation once again. So, the final expression becomes (-(~x)).

`#include <bits/stdc++.h>`

`using`

`namespace`

`std;`

`int`

`addOne(`

`int`

`x)`

`{`

`` `return`

`(-(~x));`

`}`

`/* Driver code*/`

`int`

`main()`

`{`

`` `cout<<addOne(13);`

`` `return`

`0;`

`}`

**Output:**

14