我正在为我的世界创建一个命令行客户端。有关协议的完整规范,可在此处找到:http://mc.kev009.com/Protocol。要事先回答你的问题,是的,我有点像C ++ noob。
我在实施这个协议时遇到了各种各样的问题,每个问题都很关键。
解释,链接,相关的功能名称和简短的片段非常感谢!
修改
1和3现在回答。用户470379在下面回答了图1。这个AWESOME线程回答了3,它解释了我想要做的很好:http://cboard.cprogramming.com/networking-device-communication/68196-sending-non-char*-data.html我不确定修改后的UTF-8。
答案 0 :(得分:9)
传统方法是为每个协议消息定义C ++消息结构,并为其实现序列化和反序列化功能。例如,Login Request可以表示如下:
#include <string>
#include <stdint.h>
struct LoginRequest
{
int32_t protocol_version;
std::string username;
std::string password;
int64_t map_seed;
int8_t dimension;
};
现在需要序列化功能。首先,它需要整数和字符串的序列化函数,因为这些是LoginRequest
中的成员类型。
整数序列化函数需要与big-endian表示进行转换。由于消息的成员被复制到缓冲区和从缓冲区复制,因此可以在复制时完成字节顺序的反转:
#include <boost/detail/endian.hpp>
#include <algorithm>
#ifdef BOOST_LITTLE_ENDIAN
inline void xcopy(void* dst, void const* src, size_t n)
{
char const* csrc = static_cast<char const*>(src);
std::reverse_copy(csrc, csrc + n, static_cast<char*>(dst));
}
#elif defined(BOOST_BIG_ENDIAN)
inline void xcopy(void* dst, void const* src, size_t n)
{
char const* csrc = static_cast<char const*>(src);
std::copy(csrc, csrc + n, static_cast<char*>(dst));
}
#endif
// serialize an integer in big-endian format
// returns one past the last written byte, or >buf_end if would overflow
template<class T>
typename boost::enable_if<boost::is_integral<T>, char*>::type serialize(T val, char* buf_beg, char* buf_end)
{
char* p = buf_beg + sizeof(T);
if(p <= buf_end)
xcopy(buf_beg, &val, sizeof(T));
return p;
}
// deserialize an integer from big-endian format
// returns one past the last written byte, or >buf_end if would underflow (incomplete message)
template<class T>
typename boost::enable_if<boost::is_integral<T>, char const*>::type deserialize(T& val, char const* buf_beg, char const* buf_end)
{
char const* p = buf_beg + sizeof(T);
if(p <= buf_end)
xcopy(&val, buf_beg, sizeof(T));
return p;
}
对于字符串(处理modified UTF-8 the same way as asciiz strings):
// serialize a UTF-8 string
// returns one past the last written byte, or >buf_end if would overflow
char* serialize(std::string const& val, char* buf_beg, char* buf_end)
{
int16_t len = val.size();
buf_beg = serialize(len, buf_beg, buf_end);
char* p = buf_beg + len;
if(p <= buf_end)
memcpy(buf_beg, val.data(), len);
return p;
}
// deserialize a UTF-8 string
// returns one past the last written byte, or >buf_end if would underflow (incomplete message)
char const* deserialize(std::string& val, char const* buf_beg, char const* buf_end)
{
int16_t len;
buf_beg = deserialize(len, buf_beg, buf_end);
if(buf_beg > buf_end)
return buf_beg; // incomplete message
char const* p = buf_beg + len;
if(p <= buf_end)
val.assign(buf_beg, p);
return p;
}
还有几个辅助函子:
struct Serializer
{
template<class T>
char* operator()(T const& val, char* buf_beg, char* buf_end)
{
return serialize(val, buf_beg, buf_end);
}
};
struct Deserializer
{
template<class T>
char const* operator()(T& val, char const* buf_beg, char const* buf_end)
{
return deserialize(val, buf_beg, buf_end);
}
};
现在使用这些原始函数,我们可以轻松地序列化和反序列化LoginRequest
消息:
template<class Iterator, class Functor>
Iterator do_io(LoginRequest& msg, Iterator buf_beg, Iterator buf_end, Functor f)
{
buf_beg = f(msg.protocol_version, buf_beg, buf_end);
buf_beg = f(msg.username, buf_beg, buf_end);
buf_beg = f(msg.password, buf_beg, buf_end);
buf_beg = f(msg.map_seed, buf_beg, buf_end);
buf_beg = f(msg.dimension, buf_beg, buf_end);
return buf_beg;
}
char* serialize(LoginRequest const& msg, char* buf_beg, char* buf_end)
{
return do_io(const_cast<LoginRequest&>(msg), buf_beg, buf_end, Serializer());
}
char const* deserialize(LoginRequest& msg, char const* buf_beg, char const* buf_end)
{
return do_io(msg, buf_beg, buf_end, Deserializer());
}
使用上面的帮助函子并将输入/输出缓冲区表示为char
迭代器范围,只需要一个函数模板来执行消息的序列化和反序列化。
将所有内容放在一起,用法:
int main()
{
char buf[0x100];
char* buf_beg = buf;
char* buf_end = buf + sizeof buf;
LoginRequest msg;
char* msg_end_1 = serialize(msg, buf, buf_end);
if(msg_end_1 > buf_end)
; // more buffer space required to serialize the message
char const* msg_end_2 = deserialize(msg, buf_beg, buf_end);
if(msg_end_2 > buf_end)
; // incomplete message, more data required
}
答案 1 :(得分:1)
对于#1,您需要使用ntohs
和朋友。对于16位整数使用*s
(短)版本,对于32位整数使用*l
(长)版本。 hton*
(网络主机)将输出数据转换为big-endian,与您所在平台的字节顺序无关,而ntoh*
(网络到主机)将转换传入数据(同样,独立于平台字节序)
答案 2 :(得分:1)
脱离我的头顶......
const char* s; // the string you want to send
short len = strlen(s);
// allocate a buffer with enough room for the length info and the string
char* xfer = new char[ len + sizeof(short) ];
// copy the length info into the start of the buffer
// note: you need to hanle endian-ness of the short here.
memcpy(xfer, &len, sizeof(short));
// copy the string into the buffer
strncpy(xfer + sizeof(short), s, len);
// now xfer is the string you want to send across the wire.
// it starts with a short to identify its length.
// it is NOT null-terminated.