| ??? 01/16/09 05:31 Read: times |
#161539 - How will you do that? Responding to: ???'s previous message |
Rishiraj Singh Ahuja said:
Am developing a project in which an 89C51 (A) sends a byte (from P3.1,(Txd)) to P3.0(Rxd) of an 89C2051 (B) (Serial comm. in Mode2, using timer1). I undertand that data is transmitted serially with LSB (least significant bit) first. Correct?
In that case if I load "01234567" in the SBUF of A it will appear as "76543210" in the SBUF of B. How do I convert the bits back to their original order? B needs the original byte to process it. Just how do you propose to load SBUF with "01234567"? It only holds one character. If you send '0', it will send the '0' right away, provided you've set up the UART and timer/baud-rate-generator correctly. If you send the next character too soon, neither will arrive properly. The character will arrive in precisely the order in which you transmit them, provided, of course, you do things correctly. If you don't, you needn't worry about the character order, as it will still be the same at both ends, but may be unintelligible. Might seem like a trivial problem , but am foxed.
Rishi If you spend a bit more time with the datasheet and "bible" as references, you'll see how all this works. RE |
| Topic | Author | Date |
| Decoding SBUF contents | 01/01/70 00:00 | |
| How will you do that? | 01/01/70 00:00 | |
| Refers to the order of the bits | 01/01/70 00:00 | |
| Not the way UART works | 01/01/70 00:00 | |
| Correct! | 01/01/70 00:00 | |
| No It does not | 01/01/70 00:00 | |
| Reveiving is the reverse process of Transmitting... | 01/01/70 00:00 | |
| Thanks ! | 01/01/70 00:00 | |
RE: totally on the wrong track | 01/01/70 00:00 |