How can I implement the modulus operator in Commodore 64 Basic?

Question

How can I implement the modulus operator in Commodore 64 Basic? There are not a lot of math functions to work with, and modulus would be very handy.

Answer 1

Using this expression

A-INT(A/B)*B

Answer 2

For a power-of-two modulus in the range 2 to 32768, nothing will beat using the AND operator with a value of (modulus-1). While the version of Microsoft BASIC on the Apple won't allow such usage, the one in 8-bit Commodore machines will. One could compute "Y=X mod 32" via Y = X AND 31. While this approach only works for powers of two, it is definitely worth using for scenarios involving such values.

Answer 3

While I'd be inclined to put it into a DEF FN() statement, the problem is that functions only take one argument and for modulo you need two.

As others noted, you can write a subroutine for this - easy enough. However, if you want to be a bit more clever and you're not using the USR() function for anything else, you could follow the code listed on this page which described how to implement a USR() call to machine language code on a C64 and they've specifically implemented MOD with it.

Answer 4

The following implements Y % X, populating RE with the remainder.

Below I set up 103 % 10 on lines 10 and 20 as an example use.

10 x = 10
20 y = 103
30 gosub 1000
40 print re
50 end
1000 rem mod operation
1010 re = y - ((y/x)*x)
1020 return