Question

当我做转换时，我会做以下编码，

Dim stx As integer = &H82

我知道ASCii通常是0x02，但在我的情况下，实际数据只有7位，MSB总是由＆＃39; 1＆＃39;填充，这就是为什么= 0x02 - ＆GT;为0x82。这会导致以下麻烦。

tx_buff = Chr(stx) 'put stx into buffer prepare to send out

'convert stx back to integer
rx_buff = Asc(Chr(stx))

＆＃39;当转换回来时，rx_buff实际上是0x3F而不是0x82，我想这是因为Chr（）函数将＆amp; H82转换为UTF-16编码，这是BREAK PERMITTED HERE（U + 0082），但是转换回来时，使用Asc（）使用ASCii表，0x3F是？（问号）。

我的问题是如何解决这个问题？谢谢！

Answer 1

作为如何发送数据的示例：

Option Infer On
Module Module1

    Const STX As Byte = &H2
    Const ETX As Byte = &H3

    Sub SendByte(b As Byte)
        ' set the msb before sending the byte
        ' code here to send (b OR &H80)
    End Sub

    Sub SendMessage(msgBytes() As Byte)
        ' assumes all messages to be sent start with STX and end with ETX
        SendByte(STX)
        For i = 0 To msgBytes.Length - 1
            SendByte(msgBytes(i))
        Next
        SendByte(ETX)

    End Sub

    Sub SendMessage(msg As String)
        ' For characters with the msb set, this will give &H3F -
        ' so don't use such characters.
        Dim bytesToSend = System.Text.Encoding.ASCII.GetBytes(msg)
        SendMessage(bytesToSend)

    End Sub

    Sub Main()
        ' if the data to send should be represented in ASCII
        Dim messageToSend = "J123"
        SendMessage(messageToSend)

        ' if the "J" should be sent as its ASCII representation but
        '  the following data should be bytes
        Dim bytesMessage() As Byte = {Asc("J"), 1, 2, 3}
        SendMessage(bytesMessage)

    End Sub

End Module

当您收到数据时，请使用

receivedByte = receivedByte And &H7F

取消设置msb。

Answer 2

根据您的定义;

Dim stx As Byte = &H82   ' 7-bit encoded
Dim etx As Byte = &H83   ' ditto

我认为你应该使用Byte数组传送给++;使用Dim tx_buff() = {stx, &H1, &H2, &H3, etx}数组;

Byte

当它返回时，再次作为Dim rx_buff(4) As String For vIndex = 0 To tx_Buff.Length - 1 rx_buff(vIndex) = tx_Buff.ToString("X2") Next Console.WriteLine(String.Join(" ", rx_Buff))数组，它只需要转换为int / asc / hex或任何要显示的内容（对于这个例子，我只是再现Tx缓冲区）;

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注意：所有未经测试！随意扩展您的问题，我会相应地更新此答案。

将char转换回整数

2 个答案: