位运算习题解答

1、统计一个无符号整数的二进制表示中1的个数，函数原型是int countbit(unsigned int x);。

/*Count the number of 1-bits in a positive number.
**By LYLtim
*/
int CountBit(unsigned x)
{
	x = (x & 0x55555555) + ((x >> 1) & 0x55555555);
	x = (x & 0x33333333) + ((x >> 2) & 0x33333333);
	x = (x & 0x0F0F0F0F) + ((x >> 4) & 0x0F0F0F0F);
	x = (x & 0x00FF00FF) + ((x >> 8) & 0x00FF00FF);
	x = (x & 0x0000FFFF) + ((x >> 16) & 0x0000FFFF);
	return x;
}

 巧妙地运用分治思想，详见：http://www.cnblogs.com/LYLtim/archive/2011/11/05/2236816.html

 

2、用位操作实现无符号整数的乘法运算，函数原型是unsigned int multiply(unsigned int x, unsigned int y);。例如：(11011)₂×(10010)₂=((11011)₂<<1)+((11011)₂<<4)。

//By LYLtim
unsigned multiply (unsigned x, unsigned y)
{
	unsigned sum = 0, bit = 0;
	while (y != 0) {
		if (y & 1 == 1) sum += x << bit;
		y >>= 1;
		bit += 1;
	}
	return sum;
}

 

3、对一个32位无符号整数做循环右移，函数原型是unsigned int rotate_right(unsigned int x, int n);。所谓循环右移就是把低位移出去的部分再补到高位上去，例如rotate_right(0xdeadbeef, 8)的值应该是0xefdeadbe。

// By LYLtim
unsigned rotate_right(unsigned x, unsigned n)
{
	unsigned tmp = x & ((1 << n) - 1);
	x >>= n;
	x += tmp << (32 - n);
	return x;
}

 

转载于:https://www.cnblogs.com/LYLtim/archive/2011/11/10/2244214.html