您的位置:首页 > 其它

Swap all odd and even bits

2013-06-09 14:11 686 查看

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
}
内容来自用户分享和网络整理,不保证内容的准确性,如有侵权内容,可联系管理员处理 点击这里给我发消息
标签: