LSB优先还是MSB优先？

问题描述

我想用十六进制数转换一个十进制数，然后先用LSB通过串口发送。



我试过这个

I want to convert a decimal number in hexadecimal number and send through serial port with LSB first.

I tried this

Int32 decValue = 1234056789;
byte[] testBytes = new byte[4];

    testBytes [0] = (byte)(decValue & 0x000000FF);
    testBytes [1] = (byte)((decValue >> 8) & 0x000000FF);
    testBytes [2] = (byte)((decValue >> 16) & 0x000000FF);
    testBytes [3] = (byte)((decValue >> 24) & 0x000000FF);

testBytes [0]是否包含LSB中的字节第一个或MSB优先???

Is testBytes[0] contains bytes in LSB first or MSB first???

推荐答案

byte[] bytes = { 0, 0, 0, 25 };

int i = BitConverter.ToInt32(bytes, 0);
Console.WriteLine("int: {0}", i);

如果这打印25，那么你的计算机是big-endian - 你首先发送testBytes [3]。

如果它不打印25然后发送testBytes [ 0]首先。



但是有一个对endian_ness BitConverter.IsLittleEndian的测试。所以你可以写：

If this prints 25 then your computer is big-endian - you send testBytes [3] first.
If it does not print 25 then send testBytes [0] first.

There is however a test for endian_ness BitConverter.IsLittleEndian. So you can write:

byte[] bytes = { 0, 0, 0, 25 };

// If the system architecture is little-endian (that is, little end first), 
// reverse the byte array. 
if (BitConverter.IsLittleEndian)
    Array.Reverse(bytes);

int i = BitConverter.ToInt32(bytes, 0);
Console.WriteLine("int: {0}", i);
// Output: int: 25

您还应该查看BitConverter.GetBytes（Int32）以将Int32转换为字节数组。

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