VB.net 串口发送数据
send_bytes(0)=&HAAsend_bytes(1)=&H55send_bytes(2)=&HDsend_bytes(3)=&H73send_bytes(4)=...
send_bytes(0) = &HAA
send_bytes(1) = &H55
send_bytes(2) = &HD
send_bytes(3) = &H73
send_bytes(4) = &H68
send_bytes(5) = &H69
send_bytes(6) = &H73
send_bytes(7) = &H68
send_bytes(8) = &H69
send_bytes(9) = &H73
send_bytes(10) = &H68
send_bytes(11) = &H75
send_bytes(12) = &H6A
send_bytes(13) = &H75
send_bytes(14) = &H3A
send_bytes(15) = &H31
send_bytes(16) = &H5
send_bytes(17) = &H22
send_buffer = System.Text.Encoding.Default.GetString(send_bytes)
正常发送是这个顺序,但是监视串口发送是这个顺序
3F 0D 73 68 69 73 68 69 73 68 75 6A 75 3A 31 05 22 00
后面多出来个00 前面几位也没对上 展开
send_bytes(1) = &H55
send_bytes(2) = &HD
send_bytes(3) = &H73
send_bytes(4) = &H68
send_bytes(5) = &H69
send_bytes(6) = &H73
send_bytes(7) = &H68
send_bytes(8) = &H69
send_bytes(9) = &H73
send_bytes(10) = &H68
send_bytes(11) = &H75
send_bytes(12) = &H6A
send_bytes(13) = &H75
send_bytes(14) = &H3A
send_bytes(15) = &H31
send_bytes(16) = &H5
send_bytes(17) = &H22
send_buffer = System.Text.Encoding.Default.GetString(send_bytes)
正常发送是这个顺序,但是监视串口发送是这个顺序
3F 0D 73 68 69 73 68 69 73 68 75 6A 75 3A 31 05 22 00
后面多出来个00 前面几位也没对上 展开
1个回答
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你可以这样操作：
Form1.BeginInvoke(Sub()
'一些操作
End Sub)
这样主线程就会放下手里的事情并执行Sub里的操作了
Form1.BeginInvoke(Sub()
'一些操作
End Sub)
这样主线程就会放下手里的事情并执行Sub里的操作了
追问
这是什么意思啊 我问的是数据传输的问题
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