c++ - 通过 Indy UDP 正确发送和接收结构

Question

考虑以下使用 Borland 2007 和 Indy UDP 服务器和客户端的最小程序:

struct DATA_PACKAGE
{
    int t;
    int x;
    int y;
};

void __fastcall TForm1::Button1Click(TObject *Sender)
{
   DATA_PACKAGE a;
   a.t = 3;
   a.x = 2;
   a.y = 1;
   Form1->Memo1->Lines->Add("sent " + IntToStr(sizeof(DATA_PACKAGE)));
   Form1->UDPClient1->SendBuffer(server,port,RawToBytes(&a, sizeof(DATA_PACKAGE)));
}

void __fastcall TForm1::UDPServer1UDPRead(TObject *Sender, TBytes AData,
      TIdSocketHandle *ABinding)
{
    DATA_PACKAGE r;
    Form1->Memo1->Lines->Add("received " + IntToStr(sizeof(AData)));
    BytesToRaw(AData, &r, sizeof(AData));
    Form1->Memo1->Lines->Add(IntToStr(r.t) + " " + IntToStr(r.x) + " " + IntToStr(r.y));
}

输出:

sent 12
received 4
3 4717901 0 

首先，为什么发送了12个字节，却只收到了4个字节？

其次，x 和 y 发生了什么？

当我将 t,x,y 的数据类型更改为 short 时，我得到:

sent 6 
received 4 
3 2 0

环顾四周，我发现结构的打包(可能还有字节顺序？)很重要，但我找不到如何正确设计它的明确指南。

Answer 1

TBytes 是动态字节数组，它由 RTL 实现为指针，这就是 sizeof(AData) 返回 4 的原因。不要使用 sizeof(AData)，请改用 AData.Length 属性:

void __fastcall TForm1::UDPServer1UDPRead(TObject *Sender, TBytes AData,
      TIdSocketHandle *ABinding)
{
    DATA_PACKAGE r;
    Form1->Memo1->Lines->Add("received " + IntToStr(AData.Length));
    BytesToRaw(AData, &r, AData.Length);
    Form1->Memo1->Lines->Add(IntToStr(r.t) + " " + IntToStr(r.x) + " " + IntToStr(r.y));
}