如何使用C＆＃x2B;＆＃x2B中的Indy TCP客户端编写原始二进制数据建筑商

发布于 2025-01-27 05:09:24 字数 477 浏览 2 评论 0 原文

使用Embarcadero C ++构建器10.3。

我有一个 dynamicArray＆lt; uint8_t＆gt; mydata 对象。我想使用 tidtcpclient 组件将其原始二进制内容（字节）发送到服务器。我是这样要做的：

TIdTcpClient tcpClient1;
// Bla Bla Bla
tcpClient1->IOHandler->Write(rawData);

其中 rawdata 应该是类型 tidbytes 或 tidstream ，

因此基本上归结为以下内容：要将 mydata 对象转换为 rawdata 类型 tidbytes 或 tidstream ？

原文

Using Embarcadero C++ Builder 10.3.

I have a DynamicArray<uint8_t> myData object. I want to send/write its raw binary content (bytes) to a server using the TIdTcpClient component. I'm going about it like this:

TIdTcpClient tcpClient1;
// Bla Bla Bla
tcpClient1->IOHandler->Write(rawData);

Where rawData should be of type TIdBytes or TIdStream

So basically, it boils down to the following: How to convert myData object to a rawData type of either TIdBytes or TIdStream?

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明媚殇 2025-02-03 05:09:24

首先， tidstream 在很长一段时间内都不是Indy的一部分，这让我想知道您是否正在使用Indy的旧版本，而不是使用C ++构建器发货的版本10.3 。 Indy很长一段时间以来支持RTL的标准 Tstream 类。

话虽如此...

tidbytes 是 system :: DynamIcarray＆lt; system :: byte＆gt; 的别名，其中 system :: byte 是一个无符号char 的别名，与 uint8_t 相同的大小和签名（根据编译器， uint8_t ，甚至可能只是一个别名 unsigned char ）。

因此，最简单的解决方案，无需必须单独的数据副本，就是简单地将其键入键入，例如：

tcpClient1->IOHandler->Write(reinterpret_cast<TIdBytes&>(myData));

这在技术上是 nordefined行为，因为 dynamicArray＆lt; uint8_t＆gt; and dynamiCarray＆lt; byte＆gt; 是无关类型（除非 uint8_t and code> byte byte 是> code>> byte byte 未签名的char ），但是在您的情况下它将起作用，因为它是两个数组背后的基础代码， uint8_t and byte 具有相同的基础内存布局。

另外，下一个最简单的解决方案，，没有复制数据或调用未定义的行为，是使用Indy的 tidreadonlymemorybufferstream in iDglobal.hpp.hpp.hpp ，EG：

TIdReadOnlyMemoryBufferStream *ms = new TIdReadOnlyMemoryBufferStream(&myData[0], myData.Length);
try {
    tcpClient1->IOHandler->Write(ms);
}
__finally {
    delete ms;
}

或：

{
auto ms = std::make_unique<TIdReadOnlyMemoryBufferStream>(&myData[0], myData.Length);
tcpClient1->IOHandler->Write(ms.get());
}

否则，最终解决方案是将数据复制到 tidbytes ，例如：

{
TIdBytes bytes;
bytes.Length = myData.Length;

memcpy(&bytes[0], &myData[0], myData.Length);
or:
std::copy(myData.begin(), myData.end(), bytes.begin());

tcpClient1->IOHandler->Write(bytes);
}

First off, TIdStream has not been part of Indy in a VERY VERY LONG time, which makes me wonder if you are using a very old version of Indy, not the one that shipped with C++Builder 10.3. Indy has supported the RTL's standard TStream class for a very long time.

That being said...

TIdBytes is an alias for System::DynamicArray<System::Byte>, where System::Byte is an alias for unsigned char, which is the same size and sign-ness as uint8_t (depending on compiler, uint8_t might even just be an alias for unsigned char).

So, the simplest solution, without having to make a separate copy of your data, is to simply type-cast it, eg:

tcpClient1->IOHandler->Write(reinterpret_cast<TIdBytes&>(myData));

This is technically undefined behavior, since DynamicArray<uint8_t> and DynamicArray<Byte> are unrelated types (unless uint8_t and Byte are both aliases for unsigned char), but it will work in your case since it is the same underlying code behind both arrays, and uint8_t and Byte have the same underlying memory layout.

Alternatively, the next simplest solution, without copying data or invoking undefined behavior, is to use Indy's TIdReadOnlyMemoryBufferStream class in IdGlobal.hpp, eg:

TIdReadOnlyMemoryBufferStream *ms = new TIdReadOnlyMemoryBufferStream(&myData[0], myData.Length);
try {
    tcpClient1->IOHandler->Write(ms);
}
__finally {
    delete ms;
}

Or:

{
auto ms = std::make_unique<TIdReadOnlyMemoryBufferStream>(&myData[0], myData.Length);
tcpClient1->IOHandler->Write(ms.get());
}

Otherwise, the final solution is to just copy the data into a TIdBytes, eg:

{
TIdBytes bytes;
bytes.Length = myData.Length;

memcpy(&bytes[0], &myData[0], myData.Length);
or:
std::copy(myData.begin(), myData.end(), bytes.begin());

tcpClient1->IOHandler->Write(bytes);
}

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