C - 取消引用 void 指针

Question

我正在尝试创建自己的交换函数，但遇到了麻烦。
为什么我得到“取消引用 void 指针”？

void    ft_swap(void *a, void *b, size_t nbytes)
{
    unsigned char   *cur_a;
    unsigned char   *cur_b;
    size_t          i;

    i = 0;
    while (i < nbytes)
    {
        cur_a = (unsigned char *)*a + i; // here
        cur_b = (unsigned char *)*b + i; // here

        *a = cur_b;
        *b = cur_a;

        i++;
    }
}

Answer 1

不允许取消引用 void 指针。您需要将其转换为另一种指针类型:

cur_a = (unsigned char *)a;

同样，您不能为*a 分配任何内容。正确的代码是:

void ft_swap(void *a, void *b, size_t nbytes) {
    unsigned char  *cur_a = (unsigned char *) a;
    unsigned char  *cur_b = (unsigned char *) b;

    for (size_t i = 0; i < nbytes; ++i) {
        unsigned char tmp = cur_a[i];
        cur_a[i] = cur_b[i];
        cur_b[i] = tmp;
    }
}