| ??? 05/13/07 21:20 Read: times |
#139191 - I don't think so... Responding to: ???'s previous message |
Frieder Ferlemann said:
while at it, there should be two mov and two xrl here:
mov a, temp1lo xrl a, temp2lo jnz proceed xrl a, temp1hi<- mov a,temp1hi xrl a, temp2hi jnz proceed I don't think you'd need the mov there. If you had 1234h in both your 16 bit words you'd get 34h xor 34h = 00h 00h xor 12h = 12h 12h xor 12h = 00h Looks good to me. As peter correctly points out though, if temp1 was 1234h and temp2 was 3412h, my original post would still make 00h, but 1234h does not equal 3412h. 34h xor 12h = 26h 26h xor 12h = 34h 34h xor 34h = 00h |
| Topic | Author | Date |
| compare two 16 bit values | 01/01/70 00:00 | |
| redundant move | 01/01/70 00:00 | |
| makes sense | 01/01/70 00:00 | |
| How about... | 01/01/70 00:00 | |
| don\'t work ! | 01/01/70 00:00 | |
| Oops. | 01/01/70 00:00 | |
| mov? | 01/01/70 00:00 | |
| I don't think so... | 01/01/70 00:00 | |
yes, I've fallen in that trap:) | 01/01/70 00:00 |