C# 小端还是大端?

C# little endian or big endian?(C# 小端还是大端?)

问题描述

在允许我们通过 UDP/IP 控制它的硬件文档中，我找到了以下片段:

In the documentation of hardware that allows us to control it via UDP/IP, I found the following fragment:

在这个通信协议中，DWORD 是一个 4 字节的数据，WORD 是一个 2 字节的数据，BYTE 是单字节数据.存储格式为little endian，即4字节(32bits)数据存储为:d7-d0、d15-d8、d23-d16、d31-d24；双字节(16bits)数据存储为:d7-d0，d15-d8.

In this communication protocol, DWORD is a 4 bytes data, WORD is a 2 bytes data, BYTE is a single byte data. The storage format is little endian, namely 4 bytes (32bits) data is stored as: d7-d0, d15-d8, d23-d16, d31-d24; double bytes (16bits) data is stored as: d7-d0 , d15-d8.

我想知道这如何转换为 C#?我必须在发送之前转换东西吗?例如，如果我想发送一个 32 位整数，还是一个 4 个字符的字符串?

I am wondering how this translates to C#? Do I have to convert stuff before sending it over? For example, if I want to send over a 32 bit integer, or a 4 character string?

推荐答案

C# 本身并没有定义字节序.但是，无论何时转换为字节，您都在做出选择.BitConverter 类有一个 IsLittleEndian 字段告诉您它的行为方式，但它没有提供选择.BinaryReader/BinaryWriter 也是如此.

C# itself doesn't define the endianness. Whenever you convert to bytes, however, you're making a choice. The BitConverter class has an IsLittleEndian field to tell you how it will behave, but it doesn't give the choice. The same goes for BinaryReader/BinaryWriter.

我的 MiscUtil 库有一个 EndianBitConverter 类，它允许您定义字节序；BinaryReader/Writer 也有类似的等价物.恐怕没有在线使用指南，但它们很琐碎:)

My MiscUtil library has an EndianBitConverter class which allows you to define the endianness; there are similar equivalents for BinaryReader/Writer. No online usage guide I'm afraid, but they're trivial :)

(EndianBitConverter 还具有普通 BitConverter 中不存在的功能，即在字节数组中就地进行转换.)

(EndianBitConverter also has a piece of functionality which isn't present in the normal BitConverter, which is to do conversions in-place in a byte array.)

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