| ??? 12/17/08 20:56 Read: times |
#161003 - Just happens? Responding to: ???'s previous message |
This works because 0xD0 XOR 0x07 happens to give the correct address ;-) But if SFR addresses evenly divisable by 8 are also bit-adressable, that means that all SFR that are bit-adressable have the last 3 bits zero. xxxxx000 ^ 000 => xxxxx000 xxxxx000 ^ 001 => xxxxx001 xxxxx000 ^ 010 => xxxxx010 xxxxx000 ^ 011 => xxxxx011 xxxxx000 ^ 100 => xxxxx100 xxxxx000 ^ 101 => xxxxx101 xxxxx000 ^ 110 => xxxxx110 xxxxx000 ^ 111 => xxxxx111 So writing <sfr_addr> ^ <bit_number> will always produe the correct bit address. Maybe a reason why Keil are using this way to define sbit variables? And the Intel engineers also wanted the simplest possible mapping for bit variables to keep down the number of transistors. |
| Topic | Author | Date |
| Keil vs SDCC again | 01/01/70 00:00 | |
| that's going backwards | 01/01/70 00:00 | |
| Not the whole story | 01/01/70 00:00 | |
| Outdated too | 01/01/70 00:00 | |
| Just happens? | 01/01/70 00:00 | |
I doubt it | 01/01/70 00:00 |