通过套接字发送任意长度的数组.字节序 [英] Sending the array of arbitrary length through a socket. Endianness

问题描述

我现在正在使用套接字编程，但是遇到了一个问题，我不知道该如何以可移植的方式解决. 任务很简单:我需要通过网络发送16个字节的数组，在客户端应用程序中接收它并进行解析.我知道，有像htonl，htons之类的函数，因此可以与uint16和uint32一起使用.但是，我应该如何处理比这更大的数据呢?

I'm fighting with socket programming now and I've encountered a problem, which I don't know how to solve in a portable way. The task is simple : I need to send the array of 16 bytes over the network, receive it in a client application and parse it. I know, there are functions like htonl, htons and so one to use with uint16 and uint32. But what should I do with the chunks of data greater than that?

谢谢.

推荐答案

您说的是16个字节的数组.那真的没有帮助.字节序仅对大于字节的内容重要.

You say an array of 16 bytes. That doesn't really help. Endianness only matters for things larger than a byte.

如果它确实是原始字节，则只需发送它们，您将收到相同的消息

If it's really raw bytes then just send them, you will receive them just the same

如果它确实是一个结构，则要发送它

If it's really a struct you want to send it

 struct msg
 {
     int foo;
     int bar;
 .....

然后，您需要遍历缓冲区以提取所需的值.

Then you need to work through the buffer pulling that values you want.

发送时，您必须将数据包组装成标准顺序

When you send you must assemble a packet into a standard order

 int off = 0;
 *(int*)&buff[off] = htonl(foo);
 off += sizeof(int);
 *(int*)&buff[off] = htonl(bar);
 ...

收到时

 int foo = ntohl((int)buff[off]);
 off += sizeof(int);
 int bar = ntohl((int)buff[off]);
 ....

我看到您想发送一个IPv6地址，它们始终按网络字节顺序-因此您可以将其原始流式传输.

I see you want to send an IPv6 address, they are always in network byte order - so you can just stream it raw.

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