Add 1 to a given number using Java

Software Development
#1

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 0011001 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 1
To add 1 to a number x (say 0011000111), flip all the bits after the rightmost 0 bit (we get 0011000000). Finally, flip the rightmost 0 bit also (we get 0011001000) to get the answer.

// Java code to add add

// one to a given number

class GFG {

static int addOne( int x)

{

int m = 1 ;

// Flip all the set bits

// until we find a 0

while ( ( int )(x & m) >= 1 )

{

x = x ^ m;

m <<= 1 ;

}

// flip the rightmost 0 bit

x = x ^ m;

return x;

}

/* Driver program to test above functions*/

public static void main(String[] args)

{

System.out.println(addOne( 13 ));

}

}

Output:

14