我正在开发一个基于UDP的客户机/服务器,我想从服务器向客户机发送不同的消息为每个消息定义了不同的C结构。
我想知道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;
}

用于反序列化数据的接收器端代码
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1:
为什么反序列化值与预期值不同?
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
}

存储在数据库中的序列化数据
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

最佳答案

OP在serializeTask()中有两个问题

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循环每4次序列化一次OP当然想序列化连续的cId[]
只有cId[]字符串的长度被序列化。OP当然想序列化所有这些和一个NUL终止字节。
发布的缓冲区中的数据更有可能是下面的数据,这与序列化代码不匹配。这意味着填充data的高级代码是错误的我确信在字段Task* tmType中看到的值要么是指针,要么是tkType,同样,在序列化之前float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
...

09-27 00:25
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