我们每一种计算机语言最终都会通过编译器转换成机器语言来执行，所以在编程中，位操作是常见且高效的数据处理手段之一，下面列出一些基于C语言的场景实例，便于日常开发中学习和使用
例一，编写函数 getbits(x,p,n) 从数值x的第p位开始返回n位数值
#include <stdio.h> int getbits(unsigned x, int p, int n); int main() { unsigned x = 0xF994; int p = 4; int n = 3; int z = getbits(x, p, n); printf("getbits(%u (%x), %d, %d) = %u (%X)\n", x, x, p, n, z, z); } // ff94 11111111100.101.00 # original number // >> p+1-n [2] 0011111111100.101. # shift desired bits to right // & ~(~0 << n) [7] 0000000000000.101. # clear all the other (left) bits int getbits(unsigned x, int p, int n) { return x >> (p - n + 1) & ~(~0 << n); } |
例二，编写一个函数setbits(x,p,n,y),该函数返回对x执行下列的操作的结果的值：将x中从左第p位开始的n个（二进制）位设置为y中最右边n位的值，x的其余各位保持不变
#include <stdio.h> unsigned setbits(unsigned x, unsigned p, unsigned n, unsigned y); int intLen(unsigned x); int main() { unsigned x = 171; // 1010 1011 --- > 101[0 1]011 unsigned p = 3; unsigned n = 2; unsigned y = 38; // 0010 0110 printf("result : %u \n", setbits(x, p, n, y)); return 1; } int intLen(unsigned x) { int len = 0; for (; x; x >>= 1) { len++; } return len; } unsigned setbits(unsigned x, unsigned p, unsigned n, unsigned y) { int length_y = intLen(y); int length_x = intLen(x); if (length_x - p - n < 0) { printf("move over size by x"); return 0; } if (length_y < n) { printf("move over size by y"); return 0; } // int pos = p -n + 1; // unsigned cpy = y & ~ (~0 << n); // unsigned xx = (x >> pos) & (~0 << n); // return xx |= cpy; unsigned tail = length_x - (p + n); // x需要分离的子数据 unsigned sub = x & ~(~0 << tail); // y 中需要替换的n位数据 unsigned cpy = y & ~(~0 << n); // x向右移位，保留左边 x >>= length_x - p; // x再向左移，这样最右边的n为可以为0 x <<= n; //这样可以把y的拷贝值拷贝过去 x |= cpy; // x再向左推进， 把刚刚的分离的n为先补0先 x <<= tail; // x的分离再补充回数值 x |= sub; return x; } |
例三，编写一个函数invert(x,p,n), 该函数返回对x执行下列操作后的结果的值：将x中左起第p位开始的n个（二进制）位求反（即，1变成0，0变成1），x的其余各位保持不变
#include <stdio.h> unsigned invert(unsigned x, unsigned p, unsigned n); int intLen(unsigned x); int main() { unsigned x = 171; // 1010 1011 --- > 101[0 1]011 printf("result: %u\n", invert(x, 3, 2)); return 1; } int intLen(unsigned x) { int len = 0; for (; x; x >>= 1) { len++; } return len; } unsigned invert(unsigned x, unsigned p, unsigned n) { int length_x = intLen(x); if (length_x - p - n < 0) { printf("move over size by x"); return 0; } unsigned tail = length_x - (p + n); // x需要分离的子数据 unsigned sub = x & ~(~0 << tail); // 0001 01[0 1] x >>= tail; // x ^= 0011 // 0001 0110 x ^= ~(~0 << n); // 101[10]000 x <<= tail; // 101[10]011 x |= sub; return x; } |
例四，编写一个函数rightrot(x,n)：该函数返回将x循环右移（即从最右端移出的为将从最左端移入）n (二进制) 位后所得到的值
#include <stdio.h> int intLen(unsigned x); unsigned rightrot(unsigned x, int n); int main() { unsigned x = 171; // 1010 1011 int n = 5; // assert x = 01011 101 = 93 printf("result = %u\n", rightrot(x, n)); return 1; } int intLen(unsigned x) { int len = 0; for (; x; x >>= 1) { len++; } return len; } unsigned rightrot(unsigned x, int n) { int length_x = intLen(x); if (length_x) { n %= length_x; if (n) { // x需要分离的子数据 unsigned sub = x & ~(~0 << n); // x 向右移动n位 x >>= n; //计算左边位置需要填报多少位 int mov = length_x - n; //通过上面分离的子串构造新的等长新串 unsigned subx = sub << mov; subx ^= x; return subx; } else { return x; } } return 0; } |
例五，在求对二的补码时，表达式 x&(x-1) 可以删除x中最右边值为1的一个二进制位。用这一方法重写bitcount函数，以加快其执行速度
#include <stdio.h> int bitcount(unsigned x); int main() { printf("%d's bit count: %d\n", 1023, bitcount(1023)); printf("%d's bit count: %d\n", 1024, bitcount(1024)); } int bitcount(unsigned x) { int len = 0; for (; x; x &= x - 1) { len++; } return len; } |
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