Commodore PET User Manual page 221

Method 2: Multiple Integer Subtraction
The alternative method to numeric string subtraction is multiple integer
subtraction. Recall from the previous discussion of multiple integer addition that
the multiple integer method divides a large number into smaller segments. calcu-
lates the segments separately, and joins the answers into one string. This method
evades the 9-digit length limit.
The total process of multiple integer subtraction is as follows:
1. Input the minuend and subtrahend as two positive numeric strings.
2. Determine which string has the larger value.
3. Divide the numbers into parts: high, low.
4. Calculate the difference of low- and high-order digits.
5. Concatenate the differences into a one-string answer.
6. Truncate leading zeros.
7. Print the answer string.
At the end of the subtraction section is the sample program used to demonstrate
multiple integer subtraction.
Step 1: Input the minuend and the subtrahend as two positive numeric
strings:
113 PRINT":")IE.lIEMULTIPLE INTEGER SUBTRACTIONlIElIElIt":PRINT
2(1
INPUT A$, B$
.1IE1IEMULTIPLE INTEGER SUBTRACTIONliElIElIE
?12345678ge12
??57943572
A$, the minuend, and B$, the subtrahend. are entered as strings to avoid the
9-digit length limit.
Like multiple integer addition, A$ and B$ are divided into smaller segments.
The maximum input length is arbitrarily set at 16 digits, so that we can divide the
largest possible string into equal segments of eight digits each.
Step 2: Determine which input string has the larger value. If A$ is equal
to B$ then the program drops down to line 1190 to print a zero answer. If B$ is
larger than A$ the difference is negative and extra steps are needed.
If the answer is to be negative. the contents of the two strings are switched
to put the larger value in A$ and the smaller value in B$. They are then subtracted.
and a negative sign (" -") is concatenated onto the front of the difference (C$) as
was demonstrated in line 70 of "Numeric String Subtraction" (page 204). Line 30
is used here to direct the program past the switching routine if switching is not
needed.
30
IF VALCA$»VALCB$) THEN
1800
413 X$=A$:A$=B$:B$=X$
58
$=1
208