Question

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/反序列化功能

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];
}

序列化结构的发件人端代码

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;
}

反序列化数据的接收方代码

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1：
为什么反序列化的值与预期的不一样？

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serializeTask

的代码

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

Answer 1

OP在serializeTask()

中有2个问题

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循环序列化每隔4 cId[]。 OP当然希望连续cId[]序列化仅序列化data字符串的长度。 OP当然希望将所有和序列化为NUL终止字节。

发布的缓冲区中的数据更可能是下面的，不匹配序列化代码。这意味着填充Task* t的更高级代码是错误的。我相信，字段mType和tkType中显示的值可以是指针，也可以是float，再次Task* t可能在序列化之前出错。

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
...

在socket上发送struct时序列化问题

1 个答案: