Commodore PET - Page 213

First the low segments AL and BL are added using the following program
statement:
1070 CL$=STR$(AL+BL)
IIUlnoer
CL$ = converted = (numeric sum of AL +BH)
to string
The sum of AL and BL is converted into numeric string form when assigned
to
CL$. It isn't necessary that the sum be in string form, but it is much simpler to
test for carry-over using the
LEN function at line 1080.
Line 1075 truncates the leading blank from the front of
CL$. Remember that
when a number
is converted into a string the leading blank is included. We do not
want this leading blank as part of CL$ when we concatenate the high and low
segments together, therefore
we truncate it with the MID$ function.
Line 1080 tests the length of sum CL$ against the segment length
F. If the
length of CL$
is greater than F, the leftmost digit is carried over and added to the
sum
CH$. (The value of D is equal to either 0 or 1.)
CH$ is obtained by adding AH, BH, and the carry D. Statements 1070 to
1090 show the process of adding the high and low numbers to obtain CH$+CL$.
1070 CL$=STR$(AL+BL)
CL$=STR$(jt.6901234561 + 1t.657943572D
CL$=STR$(Jt.6148067028\l
CL$=1161114!sI01617\ü12ls1
1075 CL$=MID$(CL$.2.LEN(CL$)-1)
CL$=MID$(1161114Is!016171012!s\'2.1 0-1)
CL$=MID$(Jt.6!114IsI0!6171012IsI.2.9)
CL$-1114IsI016171012Isl
1080 IF LEN(CL$»F THEN D=1
LEN(CL$)=9 ;F=S
9>S-D=1
1090 CH$=STR$(AH+BH+D)
CH$=STR$(j161234567SI + 116000000001 +~)
CH$=STR$(J161234567~)
CH$-@1121314151617191
1095 ÇH$=MID.(CH •• 2.LENCCH.)-1)
CH$=MID$@1ï12l314151617191.2.
1o
-
1
)
CH$=MID$qt.6~ 12131415161719\.2.9)
CH$-J1J213J41516I7J9!
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