c中位域的实现

Question

我们接到了一项任务，要求我们创建一个汇编程序的编造版本（不包含确切的汇编推荐）

我们被指示要有效地使用内存（不是大 O 符号类型的内存效率，而是有效地使用数据结构）

由于每个命令都被翻译成 24 位机器码，我认为存储命令机器码的最有效方法是使用位字段，所以我实现了以下位字段：

(在bit_field.c)

#include "bit_field.h"
int set_bit(bit_field *destination, int location, enum bit_values value)
{
    if(location >= BITS)
        return 0;
    if(value == ONE)
        (destination)[location / BITS_IN_BYTE] |= 1UL << location % BITS_IN_BYTE;
    else
        (destination)[location / BITS_IN_BYTE] &= ~(1UL << location % BITS_IN_BYTE);
    return 1;
}
enum bit_values get_bit(bit_field *destination, int location)
{
    return ((destination)[location / BITS_IN_BYTE] >> location % BITS_IN_BYTE) & 1U;
}
void init_bit_field(bit_field *new_bit_field)
{
    int i;
    for(i = 0; i < BITS; i++)
        set_bit(new_bit_field, i, ZERO);
}
void init_bit_field_by_str(bit_field *new_bit_field, char *string)
{
    int i;
    for(i = 0; i < BITS; i++)
        set_bit(new_bit_field, i, *string++ - '0');
}

(在bit_field.h)

#define BITS 24
#define BITS_IN_BYTE 8
#define BYTES BITS / BITS_IN_BYTE
enum bit_values {ZERO = 0, ONE = 1}bit_values;
typedef unsigned char byte;
typedef byte bit_field [BYTES];
int set_bit(bit_field *, int , enum bit_values);
enum bit_values get_bit(bit_field *, int);
void init_bit_field(bit_field *);
void init_bit_field_by_str(bit_field *, char *);;

（在 input.c 测试）

#include "bit_field.h"
#include <stdio.h>
int main()
{
    int i;
    bit_field bits;
    init_bit_field(&bits);
    set_bit(&bits, 22, ONE);
    for(i = 0; i < BITS; i++)
        printf("%d, ", get_bit(&bits, i));
    return 0;
}

当我尝试编译以下代码 (gcc C90) 时，出现以下错误：

bit_field.c:14:53: error: invalid operands to binary >> (have ‘byte (*)[3] {aka unsigned char (*)[3]}’ and ‘int’)
     return ((&destination)[location / BITS_IN_BYTE] >> location % BITS_IN_BYTE) & 1U;
             ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ^~
bit_field.c:15:1: warning: control reaches end of non-void function [-Wreturn-type]
 }
 ^
avivgood@ubuntuVM:~/CLionProjects/Maman14$ gcc -Wall -ansi -pedantic input.c bit_field.c bit_field.h -o example.out
bit_field.c: In function ‘set_bit’:
bit_field.c:7:49: error: invalid operands to binary | (have ‘byte (*)[3] {aka unsigned char (*)[3]}’ and ‘long unsigned int’)
         (&destination)[location / BITS_IN_BYTE] |= 1UL << location % BITS_IN_BYTE;
                                                 ^~
bit_field.c:9:49: error: invalid operands to binary & (have ‘byte (*)[3] {aka unsigned char (*)[3]}’ and ‘long unsigned int’)
         (&destination)[location / BITS_IN_BYTE] &= ~(1UL << location % BITS_IN_BYTE);
                                                 ^~ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
bit_field.c: In function ‘get_bit’:
bit_field.c:14:53: error: invalid operands to binary >> (have ‘byte (*)[3] {aka unsigned char (*)[3]}’ and ‘int’)
     return ((&destination)[location / BITS_IN_BYTE] >> location % BITS_IN_BYTE) & 1U;
             ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ^~
bit_field.c:15:1: warning: control reaches end of non-void function [-Wreturn-type]
 }
 ^

这些错误似乎没有任何意义。我设置和固定钻头的方法应该有效。那么为什么它会产生这么多错误呢？

Answer 1

这里你定义

typedef unsigned char byte;
typedef byte bit_field [BYTES];

所以 bit_field 是一个包含 BYTES 个元素的数组。但是，当您定义函数时，您使用指向 bit_field:

的指针

int set_bit(bit_field *, int , enum bit_values);
enum bit_values get_bit(bit_field *, int);
void init_bit_field(bit_field *);
void init_bit_field_by_str(bit_field *, char *);

这就是程序引发位运算符错误的原因。当您使用：

destination[location / BITS_IN_BYTE]

它的类型是 unsigned char *。这就是我们出现错误的原因：

 invalid operands to binary

因此，您可以只使用 bit_field 而不是 bit_field * 作为函数的参数。然后在函数中不要使用符号 (&) as:

destination[location / BITS_IN_BYTE]

c中位域的实现

Implementation of bit field in c

c

memory

bit

bitwise-operators

bit-fields