Swap all odd and even bits

最新推荐文章于 2022-12-14 22:45:47 发布

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本文介绍了一种高效的方法来交换整数中所有奇数位与偶数位的比特值。通过使用位操作，例如按位与和位移，可以实现这一目标。具体步骤包括获取所有奇数位和偶数位，然后分别对它们进行右移和左移操作，最后将这两个结果合并。

reference:

http://www.geeksforgeeks.org/swap-all-odd-and-even-bits/

Problem Definition:

Given an unsigned integer, swap all odd bits with even bits. For example, if the given number is 23 (00010111), it should be converted to 43 (00101011). Every even position bit is swapped with adjacent bit on right side (even position bits are highlighted in binary representation of 23), and every odd position bit is swapped with adjacent on left side.

Solution:

Let the input number be x
1) Get all even bits of x by doing bitwise and of x with 0xAAAAAAAA.The number 0xAAAAAAAA is a 32 bit number with all even bits set as 1 and all odd bits as 0.
2) Get all odd bits of x by doing bitwise and of x with 0x55555555.The number 0x55555555 is a 32 bit number with all odd bits set as 1 and all even bits as 0.
3) Right shift all even bits.
4) Left shift all odd bits.
5) Combine new even and odd bits and return.

Code:

unsigned int swapBits(unsigned int x)
{
    // Get all even bits of x
    unsigned int even_bits = x & 0xAAAAAAAA; 

    // Get all odd bits of x
    unsigned int odd_bits  = x & 0×55555555; 

    even_bits >>= 1;  // Right shift even bits
    odd_bits <<= 1;   // Left shift odd bits

    return (even_bits | odd_bits); // Combine even and odd bits
}