| ??? 12/12/07 06:55 Read: times |
#148087 - Question about JBC instruction |
This is a direct quote from the description of the JBC instruction in the 80C51 family programmer's guide and instruction set (sometimes referred to in these parts as "the bible"):
Intel said:
If the indicated bit is a one, branch to the address indicated; otherwise proceed with the next instruction. The bit will not be cleared if it is already a zero. The branch destination is computed by adding the signed relative-displacement in the third instruction byte to the PC, after incrementing the PC to the first byte of the next instruction. No flags are affected. I don't understand the italicized sentence. (For the record, it's also italicized in the original.) Does anybody understand what they are trying to say there? The idea of clearing a bit that is already zero doesn't make any sense to me, and I don't see why they took such pains to emphasize that they don't do it. ??? -- Russ |
| Topic | Author | Date |
| Question about JBC instruction | 01/01/70 00:00 | |
| Don\'t think memory, think SFR ! | 01/01/70 00:00 | |
Of course. Thank you. | 01/01/70 00:00 |