Commodore PET - Page 223

Therefore. we must borrow from AH to increase the value of AL so that the
difference will be positive.
Lines 1070 to 1090 borrow from AH and increase AL
before
BL is subtracted from AL:
1070 IF AL~=BL THEN 1100
1080 AL=AL+10t"F
1090 AH=AH-1
If AL is larger than BL we bypass 10S0 and 1090 and jump directly to the
subtraction. But
if BL is larger than AL we must borrow a one million value from
AH to increase the value of AL:
-1
AHlxxxxxlxl
-BH~
CHlxxxxxx\
----, +1000000
ALI 1 xxxxxxj
-BL~
CL jxxxxxxl
A ten is added to the leftmost digit of AL. The easiest way to add the ten in
the correct position
is to raise 10 to the Fth power.
AL=AL+10fF
ln our sample program. AL is smaller than BL. as tested in line 1070.
AL/~7B90 121 < BLlii9435 721
Therefore we must borrow 1000000 (101F=101 6 = 1000000) from AH to
increase the value of AL:
1080 AL=AL+10"tF
AL=AL+1üf6
AL =AL + 1000000
AL=~+ 1000000
AL=117890121
After AL is increased. AH must be decremented by 1. since we borrowed
from
it.
1090 AH=AH-1
AH-I~1234561 ~ffi]
AH-I~1234551
Once AH. AL. BH. and BL have been set up properly. the subtraction of the
segments takes place. CL$
is the difference between AL and BL. and CH$ is the
difference between AH and
BH.
Lines 1100 through 1102 determine CL$. Line 1000 changes the integer
value of AL
-BL into string form.
1100 CL$=STR$(INT(AL-BL))
C L$ =STR$ (~1 78901 2- ~9435 72)
CL$=STR$(~845340)
CL$=~345540
210