11 Dec
2005
11 Dec
'05
5:24 a.m.
Chuck Guzis wrote:
On 12/10/2005 at 12:09 PM Chuck Guzis wrote:
; convert a nibble in A to an ascii hex character using only
; a single register, no DAA, no branches, ; and no tables.
;
; the crux of it is to drive a wedge between the values 0-9
; and A-F. this is done by adding the right amount to always
; generate a carry for the higher group of numbers, then do
; an "adi 0" to increment only the higher numbers.
;
; we need to wedge them apart 7 times to account for the
; ascii difference between '9'+1 (0x3A) and 'A' (0x41).
ani 15 ; isolate lowest nibble
; wedge step 1
adi 246 ; 0-9 are 246-255 with cy=0
; A-F are 0-5 with cy=1
adc 0 ; A-F are now 1-6
; wedge step 2
sui 246 ; 0-9 are 0-9 again with cy=0
; A-F are now 11-16 with cy=1
adc 0 ; 0-9 are 0-9; A-F are now 12-17
; wedge step 3
adi 246
adc 0
; wedge step 4
sui 246
adc 0
; wedge step 5
adi 246
adc 0
; wedge step 6
sui 246
adc 0
; wedge step 7
adi 246
adc 0
; convert to ascii
adi '0'-246
Code-cruncher quiz: Could this be done in 8088
code without DAA or a jump
or a reference to a table in memory using only a single 8-bit register?
What? No takers?
I'll post my solution tomorrow if no one beats me to it. Uses only AL (or
A) register, no jumps, no tables and will work on x80 and x86.
Cheers,
Chuck