Question

位图是array of bits。它是如何在C中实现的？

Answer 1

我假设您正在询问如何在C中实现位图（或位数组）。令人惊讶的是，维基百科上的Bit_array条目描述了这个概念，但实际上并没有展示如何实现基本操作，所以这里。

简而言之，创建一个你最喜欢的无符号类型的数组，并做正确的算术来决定如何设置/清除它中的位。

#include <limits.h>    /* for CHAR_BIT */
#include <stdint.h>   /* for uint32_t */

typedef uint32_t word_t;
enum { BITS_PER_WORD = sizeof(word_t) * CHAR_BIT };
#define WORD_OFFSET(b) ((b) / BITS_PER_WORD)
#define BIT_OFFSET(b)  ((b) % BITS_PER_WORD)

void set_bit(word_t *words, int n) { 
    words[WORD_OFFSET(n)] |= (1 << BIT_OFFSET(n));
}

void clear_bit(word_t *words, int n) {
    words[WORD_OFFSET(n)] &= ~(1 << BIT_OFFSET(n)); 
}

int get_bit(word_t *words, int n) {
    word_t bit = words[WORD_OFFSET(n)] & (1 << BIT_OFFSET(n));
    return bit != 0; 
}

Answer 2

您可以使用“位字段”。

$ cat src/bitmap.c
#include <stdio.h>

#define BITS 4
struct BITMAP {
        unsigned value : 1;
} bit[BITS];

int main(){

        bit[0].value = 0;
        bit[1].value = 1;
        // implicit truncation from 'int' to bit-field changes value from 2 to 0 [-Wbit/field-constant-conversion]
        bit[2].value = 2;
        bit[3].value = 3;

        int i;
        for ( i=0; i < BITS ; i++){
                printf("Value of bit %i is %u\n", i, bit[i].value);
        }
}

结果

$ gcc src/bitmap.c -lm -o bitmap && ./bitmap
src/bitmap.c: In function ‘main’:
src/bitmap.c:14:24: warning: unsigned conversion from ‘int’ to ‘unsigned char:1’ changes value from ‘2’ to ‘0’ [-Woverflow]
   14 |         bit[2].value = 2;
      |                        ^
src/bitmap.c:15:24: warning: unsigned conversion from ‘int’ to ‘unsigned char:1’ changes value from ‘3’ to ‘1’ [-Woverflow]
   15 |         bit[3].value = 3;
      |                        ^
Value of bit 0 is 0
Value of bit 1 is 1
Value of bit 2 is 0
Value of bit 3 is 1

C中的位图是什么？

2 个答案: