结构的地址是否与其第一个成员的地址相同？

Question

考虑我有如下的结构:

struct Bitmask
{
  unsigned char payload_length: 7;
  unsigned char mask: 1;
  unsigned char opcode: 4;
  unsigned char rsv3: 1;
  unsigned char rsv2: 1;
  unsigned char rsv1: 1;
  unsigned char fin: 1;
};

const char* payload = "Hello";
const size_t payload_length = strlen(payload);

Bitmask* header = new Bitmask();
header->fin =1;
header->rsv1 = 0;
header->rsv2 = 0;
header->rsv3 = 0;
header->opcode = 1;
header->mask = 0;
header->payload_length = payload_length;

iovec iov[2];
iov[0].iov_base = (char*)header;
iov[0].iov_len = sizeof (header);
iov[1].iov_base = (char *)payload;
iov[1].iov_len = strlen(payload);

ACE_DEBUG ((LM_DEBUG,
            ACE_TEXT ("iov[0].length = %d\niov[1].length = %d\n"),
            iov[0].iov_len,
            iov[1].iov_len));

size_t bytes_xfered;
client_stream_.sendv_n (iov, 2, 0, &bytes_xfered);

cout << "Transfered " << bytes_xfered << " byte(s)" << std::endl;

我用适当的值初始化它.最后,我想将结构转换为char*,以便我可以附加我的有效负载(这是char*消息)并通过websocket连接发送它.

Answer 1

结构的地址是否与其第一个成员的地址相同？

是的,这实际上是C和C++标准规定的.从C标准:

6.7.2.1-13.指向适当转换的结构对象的指针指向其初始成员

你的大小struct应该是两个字节.但是,您不应该将指针转换为它char*:相反,您应该使用memcpy将您复制Bitmask到通过网络发送的缓冲区中.

编辑因为你使用分散 - 聚集I/O iovec,你不需要转换Bitmask为任何东西:iov_base是void*,所以你可以简单地设置iov[0].iov_base = header;

注意:只有当您struct不包含虚函数,基类等时,这才有效(感谢Timo).

EDIT2

为了获得你的{0x81,0x05} struct,你应该改变结构元素的顺序,如下所示:

struct Bitmask {
    unsigned char opcode: 4; 
    unsigned char rsv3: 1; 
    unsigned char rsv2: 1; 
    unsigned char rsv1: 1; 
    unsigned char fin: 1; 
    unsigned char payload_length: 7; 
    unsigned char mask: 1;
}