| ??? 08/06/07 17:49 Read: times |
#142825 - Like this Responding to: ???'s previous message |
Salaam Salim,
It helps if you understand that there is a 1 to 1 correspondence between a nibble of binary (4 bits) and the 16 hexits of hexadecimal numbers. 0b0000 = 0x0 0b0001 = 0x1 0b0010 = 0x2 0b0011 = 0x3 0b0100 = 0x4 0b0101 = 0x5 0b0110 = 0x6 0b0111 = 0x7 0b1000 = 0x8 0b1001 = 0x9 0b1010 = 0xA 0b1011 = 0xB 0b1100 = 0xC 0b1101 = 0xD 0b1110 = 0xE 0b1111 = 0xF Now if you want to set bits 4 and 6 (remembering that the bits are numbered 0 - 7), you get 0101 0000, which by the table above can be seen to translate into 0x50. I hope this helps. Good luck, Joe |
| Topic | Author | Date |
| in 'mov SCON, #50h' where does the 50 come from? | 01/01/70 00:00 | |
| Simple hex/binary arithmetics | 01/01/70 00:00 | |
| or, rather than add | 01/01/70 00:00 | |
| Comment help the author, too! | 01/01/70 00:00 | |
| That would be the Standards Committe | 01/01/70 00:00 | |
| this is how I do it | 01/01/70 00:00 | |
As Jan Said | 01/01/70 00:00 | |
| Like this | 01/01/70 00:00 | |
| Thanks for the hlp | 01/01/70 00:00 |