Dim? uout？ As? different

uout？ =? “& amph "？ & amp？ Hexadecimal (Val(Text 1. Text)) Here uout is actually a string, which is a hexadecimal string of 16.

Note that I said the hexadecimal string of 16, not the hexadecimal number of 16.

MSComm 1。 Output? =? Chr(uout) So what is sent here? Chr has a string. Actually, the parameter of Chr should be long.

Actually, this is how I understand what you mean.

Get the value in the text box (Text 1. Text), not more than one byte expression range (greater than 0 and less than 256).

Send this value via MSComm 1

If this idea is correct, you can refer to the following code.

? With code? Buffer? There is only one length.

? Because this code is only used to send one byte at a time.

? Variable declaration

Dim? uOut？ As? Double? Save the entered value.

Dim? Buffer (0)? As? Bytes? ? Buffering of communication output

? Get? Text 1。 Words? The value of

uOut？ =? Val(Text 1。 Text)

? Range check of byte (unsigned 1 byte) type data

What if? uOut？ & gt？ & ampHFF？ then what uOut？ =? & ampHFF

What if? uOut？ & lt？ & ampH00？ then what uOut？ =? & ampH00

? Output to VB in 16 format? Text 2

Text 2. Words? =? “& amph "？ & amp？ CInt (output)

? Write the value to be output to the buffer.

Buffer (0)? =? Val (text 2. Text)

? Write the value in the buffer to the terminal.

MSComm 1。 Output? =? If you send a byte instead of a byte, you need to send an&; A string starting with H.

Then refer to the following code:

? Variable declaration

Dim? uOut？ As? Double? Save the entered value.

Dim? Buffer ()? As? Bytes? Buffering of communication output

? Get? Text 1。 Words? The value of

uOut？ =? Val(Text 1。 Text)

? Output to VB in 16 format? Text 2

Text 2. Words? =? “& amph "？ & amp？ CInt (output)

? Writes the string to be output to the buffer.

Buffer? =? Text 2. text

? Write the value in the buffer to the terminal.

MSComm 1。 Output? =? buffer