将1加到给定的数字

编写一个程序，将一个数字加到给定的数字上。不允许使用“ +”，“-”，“ *”，“ /”，“ ++”，“ –”等运算符。
例子：

Input:  12
Output: 13

Input:  6
Output: 7

这个问题可以通过使用一些魔术来解决。以下是使用按位运算运算符实现相同效果的不同方法。
方法一
要将1加到数字x(例如0011000111)，请翻转最右边0位之后的所有位(我们得到001100 0 000)。最后，也将最右边的0位翻转(我们得到0011001000)以获得答案。

C++

// C++ code to add add
// one to a given number
#include 
using namespace std;
 
int addOne(int x)
{
    int m = 1;
     
    // Flip all the set bits
    // until we find a 0
    while( x & m )
    {
        x = x ^ m;
        m <<= 1;
    }
     
    // flip the rightmost 0 bit
    x = x ^ m;
    return x;
}
 
/* Driver program to test above functions*/
int main()
{
    cout<

---

C

// C++ code to add add
// one to a given number
#include 
 
int addOne(int x)
{
    int m = 1;
     
    // Flip all the set bits
    // until we find a 0
    while( x & m )
    {
        x = x ^ m;
        m <<= 1;
    }
     
    // flip the rightmost 0 bit
    x = x ^ m;
    return x;
}
 
/* Driver program to test above functions*/
int main()
{
    printf("%d", addOne(13));
    getchar();
    return 0;
}

Java

// 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));
    }
}
 
// This code is contributed by prerna saini.

Python3

# Python3 code to add 1
# one to a given number
def addOne(x) :
     
    m = 1;
    # Flip all the set bits
    # until we find a 0
    while(x & m):
        x = x ^ m
        m <<= 1
     
    # flip the rightmost
    # 0 bit
    x = x ^ m
    return x
 
# Driver program
n = 13
print addOne(n)
 
# This code is contributed by Prerna Saini.

C#

// C# code to add one
// to a given number
using System;
 
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 code
    public static void Main()
    {
        Console.WriteLine(addOne(13));
    }
}
 
// This code is contributed by vt_m.

Javascript

// JavaScript code to add add
// one to a given number
 
function addOne( x) {
    let m = 1;
     
    // Flip all the set bits
    // until we find a 0
    while( x & m ) {
        x = x ^ m;
        m <<= 1;
    }
     
    // flip the rightmost 0 bit
    x = x ^ m;
    return x;
}
 
/* Driver program to test above functions*/
document.write(addOne(13));

---

输出：

14

方法2
我们知道，在大多数体系结构中，负数均以2的补码形式表示。对于符号数的2的补码表示，我们具有以下引理成立。
假设x是数字的数值，则
〜x =-(x + 1) [〜用于按位补码]
(x + 1)是由于2的补码转换中添加了1
要获得(x + 1)，请再次应用负号。因此，最终表达式变为(-(〜x))。

C++

#include 
using namespace std;
 
int addOne(int x)
{
    return (-(~x));
}
 
/* Driver code*/
int main()
{
    cout<

C

#include
 
int addOne(int x)
{
   return (-(~x));
}
 
/* Driver program to test above functions*/
int main()
{
  printf("%d", addOne(13));
  getchar();
  return 0;
}

Java

// Java code to Add 1 to a given number
class GFG
{
    static int addOne(int x)
    {
         return (-(~x));
    }
     
    // Driver program
    public static void main(String[] args)
    {
        System.out.printf("%d", addOne(13));
    }
}
 
// This code is contributed
// by Smitha Dinesh Semwal

Python3

# Python3 code to add 1 to a given number
 
def addOne(x):
    return (-(~x));
 
 
# Driver program
print(addOne(13))
 
# This code is contributed by Smitha Dinesh Semwal

C＃

// C# code to Add 1
// to a given number
using System;
 
class GFG
{
    static int addOne(int x)
    {
        return (-(~x));
    }
     
    // Driver program
    public static void Main()
    {
        Console.WriteLine(addOne(13));
    }
}
 
// This code is contributed by vt_m.

Java脚本

// JavaScript program for the above approach
function addOne(x)
{
    return (-(~x));
}
 
// Driver Code
    document.write(addOne(13));
 
// This code is contributed by susmitakundugoaldanga.

输出：

14

例子 ：

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

解释2的补码形式的位1101得出的数值为-(2 ^ 4 – 13)= -3。在结果上加上“-”将使结果减3。同样的推论也适用于减量。请注意，仅当数字以2的补码形式存储时，此方法才有效。