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我正在开发基于 UDP 的客户端/服务器，我想从服务器向客户端发送不同的消息。为每条消息定义了不同的 C 结构。

我想了解我序列化数据的方式有什么问题。

struct Task
{
    int mType;
    int tType;
    int cCnt;
    int* cId;
    char data[128];
};

序列化/反序列化函数 

unsigned char * serialize_int(unsigned char *buffer, int value)
{
    buffer[0] = value >> 24;
    buffer[1] = value >> 16;
    buffer[2] = value >> 8;
    buffer[3] = value;
    return buffer + 4;
}

unsigned char * serialize_char(unsigned char *buffer, char value)
{
    buffer[0] = value;
    return buffer + 1;
}

int deserialize_int(unsigned char *buffer)
{
    int value = 0;

    value |= buffer[0] << 24;
    value |= buffer[1] << 16;
    value |= buffer[2] << 8;
    value |= buffer[3];
    return value;

}

char deserialize_char(unsigned char *buffer)
{
    return buffer[0];
}

用于序列化结构的发送方代码 

unsigned char* serializeTask(unsigned char* msg, const Task* t)
{
    msg = serialize_int(msg,t->mType);
    msg = serialize_int(msg,t->tkType);
    msg = serialize_int(msg,t->cCnt);
    for(int i=0; i<t->cCnt; i++)
            msg = serialize_int(msg,t->cId[i*4]);

for(int i=0; i<strlen(data); i++)
    msg = serialize_char(msg,t->data[i]);

    return msg;
}

用于反序列化数据的接收端代码 

printf("Msg type:%d\n", deserialize_int(message) );
printf("Task Type:%d\n", deserialize_int(message+4) );
printf("Task Count:%d\n", deserialize_int(message+8));

Output 
Msg type:50364598         //Expected value is 3
Task Type:-2013036362     //Expected value is 1
Task Count:1745191094     //Expected value is 3

问题一：
为什么反序列化的值和预期的不一样？

问题2：
序列化/反序列化方法与memcpy有何不同？ 

Task t;
memcpy(&t, msg, sizeof(t));  //msg is unsigned char* holding the struct data

编辑

调用的代码serializeTask 

void addToDatabase(unsigned char* message, int msgSize, Task* task)
{
    message = new unsigned char[2*msgSize+1];
    unsigned char* msg = message;  //To preserve start address of message
    message = serializeTask(message, task); //Now message points to end of the array

//Insert serialized data to DB
//msg is inserted to DB 
}

存储在 DB 中的序列化数据 

Message:
00 
03 70 B6 88 03 70 B6 68 05 70 B6 68 05 70 B6 00 
00 00 00 00 00 00 00 A8 05 70 B6 AC 05 70 B6 B4 
05 70 B6 C9 05 70 B6 DE 05 70 B6 E6 05 70 B6 EE 
05 70 B6 FB 05 70 B6 64 65 66 00 63 6F 68 6F 72 
74 73 00 70 65 6E 64 69 6E 67 5F 61 73 73 69 67 
6E 5F 74 61 73 6B 73 00 70 65 6E 64 69 6E 67 5F 
61 73 73 69 67 6E 5F 74 61 73 6B 73 00 6D 65 73 
73 61 67 65 00 6D 65 73 73 61 67 65 00 3F 00 FF 
FF 00 00 FC 90 00 00 00 00 00 00 00 C9 2D B7 00 
00 00 00 10 06 70 B6 00 00 00 00 00 00 00 00 30 
06 70 B6 34 06 70 B6 3C 06 70 B6
4

1 回答 1

0

OP有2个问题 serializeTask()

for(int i=0; i<t->cCnt; i++)
  msg = serialize_int(msg,t->cId[i*4]); [i*4]
...
for(int i=0; i<strlen(data); i++)
  msg = serialize_char(msg,t->data[i]); strlen(data)

应该是（假设i<strlen(data)应该是i<strlen(t->data)

for(int i=0; i<t->cCnt; i++)
  msg = serialize_int(msg,t->cId[i]);  // [i]
...
for(int i=0; i<strlen(t->data); i++)   // strlen(data) + 1
  msg = serialize_char(msg,t->data[i]);

第一个 for 循环每 4 次序列化一次cId[]。OP当然想序列化连续cId[]的 . 只有字符串
的长度被序列化。dataOP 当然想序列化所有这些和一个 NUL 终止字节。

发布缓冲区中的数据更可能是以下，与序列化代码不匹配。这意味着填充更高级别的代码Task* t是错误的。我相信在字段中看到的值mType或者tkType是指针或者float，Task* t在序列化之前很可能有问题。

0xb6700300 or -3.576453e-06
0xb6700388 or -3.576484e-06
0xb6700568 or -3.576593e-06
0xb6700568 or -3.576593e-06
0x000000 or 0.000000e+00
0x000000 or 0.000000e+00
0xb67005a8 or -3.576608e-06
0xb67005ac or -3.576609e-06
0xb67005b4 or -3.576611e-06
0xb67005c9 or -3.576615e-06
0xb67005de or -3.576620e-06
0xb67005e6 or -3.576622e-06
0xb67005ee or -3.576624e-06
0xb67005fb or -3.576627e-06
def\0cohorts\0pending_assign_tasks\0pending_assign_tasks\0message\0message\0?\0
...
于 2013-09-09T18:49:15.307 回答