I'm trying to implement RFC2217 in my code but I can't understand how the last parity bit (46H and 28H) is generated.

I'm using RS485 to Ethernet device.

What will be the code, if I'm using 2400,E,8,1?

Is it: 55 AA 55 09 60 1B XX?

• Is 1B right?
• What will be XX?

User manual: page 42 in https://www.sarcitalia.it/file_upload/prodotti//USR-N520-Manual-EN-V1.0.4.pdf

CodePudding user response：

1. In the field for the baud rate you missed the MSByte. This field shall be 00 09 60.

2. Yes, 1B for "E,8,1" is correct. BTW, the table lists 2 bits for the 1-bit fields of "stop bit" and "parity enable", which is quite irritating.

3. The field "parity" is actually just a sum, without the header and the MSBit cleared. (I don't grasp the text of the explanation, but the document seems to be low quality anyway.)

   01 C2 00 03: 0x01 0xC2 0x00 0x03 = 0xC6; without bit 7 = 0x46.

   00 25 80 03: 0x00 0x25 0x80 0x03 = 0xA8; without bit 7 = 0x28.

   Your telegram 00 09 60 1B: 0x00 0x09 0x60 0x1B = 0x84; without bit 7 = 0x04. So XX is 04.