c中位域的实现

Question

我们被分配创建一个组装版本的汇编程序（没有确切的汇编命令）

已经指示我们有效地使用内存（不是大的O标记类型的内存效率，而是有效地使用数据结构）

由于每个命令都被翻译成24位机器代码，所以我认为存储命令的机器代码的最有效方法是使用位字段，因此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 *);;

（input.c的测试）

#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]
 }
 ^

错误似乎没有任何意义。我的固定和固定钻头的方法应该可以工作。那为什么会产生这么多错误呢？

Answer 1

您在这里定义

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

因此bit_field是具有BYTES个元素的数组。但是，在定义函数时，可以使用指向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 *);

这是程序使用位运算符引发错误的原因。使用时：
destination[location / BITS_IN_BYTE]

具有类型unsigned char *。