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在C中动态分配的结构中使用位域是否可以?

[英]Is it OK to use bitfields in dynamically allocated structure in C?

I'm going to implement a singly linked list program with bitfield in its structure, something like this: 我将用结构的位域实现一个单链列表程序,如下所示:

typedef struct large
{
    unsigned number :4;
    struct large *next;
} large;

int main()
{
    large *g;
    g=(large *)malloc(sizeof(large));

    g->number=15;
    printf("%d",g->number);

    return 0;
}

The above program is working correctly but I read in GeeksForGeeks that, 上面的程序正常运行,但是我在GeeksForGeeks中读到,

We cannot have pointers to bit field members as they may not start at a byte boundary. 我们不能有指向位字段成员的指针,因为它们可能不是从字节边界开始的。

Will there be problems if I code it further for linked list implementation? 如果为链接列表实现进一步编码会不会有问题?

The pointer g in your program is a pointer to a structure variable to which you allocated memory dynamically. 程序中的指针g是指向动态分配内存的结构变量的指针。

g->number is not the address of the member number but its value. g->number不是成员number的地址,而是其值。

You cannot have a pointer like 你不能有这样的指针

unsigned char *ptr=&(g->number);

you should get an error like cannot take address of bit-field 你应该得到一个错误,例如cannot take address of bit-field

malloc-ing any structures including those containing bitfields is absolutely OK 绝对分配任何结构,包括那些包含位域的结构

 We cannot have pointers to bit field members as they may not start at a byte boundary. 

But you do not get the address of the bitfield - only for the structure itself and its size and location is always a multiple of byte. 但是您不会获得位域的地址-仅针对结构本身,并且其大小和位置始终是字节的倍数。

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