Commodore PET User Manual page 219

Once we have determined the number of leading zeros in the answer, we
separate the leading zeros from the remainder of the answer C$. At line 1130, the
RIGHT$ function takes the LEN(C$)-L rightmost digits and stores them in the
answer variable C$.
C$ =10101*1315171
_1_ MID$(C$.1,1)
1 12357
2 0 12357
3 00 2357
7
=0
=0
< >
0
LEN(C$) = 7
L = 1
L=2
1
= LEN(C$)
1
=
7 drop out of loop
1130 C$=RIGHT$(C$,LEN(C$)-L)
C$=RIGHT$ (\0101112131517\.7-2)
C$=RIGHT$ 001011121315171.5)
C$-11121315171
Step 7: Print the answer string C$. But before we print C$, we check to
see if the answer is to be negative by testing variable
S
at line 1140. If
S=
1, that
means that originally A$
<
B$, and the final answer is to be negative, so a nega-
tive sign is added to C$. If
S=O,
the answer is positive, so nothing is added. Line
1150 prints C$:
1140 IF :::;=1 THEN
(:$="_"
+(:$
J.15€1 F'F.:HJT:
F'F.:INT"A;·jS~·JEF:=".;(:$:
F'RINT
The la st lines, 1160 through 1180, clear ail strings and variables to zero, or
to null mode and return the program to the beginning for the next input numbers.
The total program listing, complete with a RUN, is listed below.
206