CODEWITHSUNDEEP
cfixed-point
Ryou
Ryou

Reputation: 57

How do you convert the remainder of a division operation to a fixed point in C?

I understand the concept of fixed point pretty well at this point, but I'm having trouble making a logical jump.

I'm working with M68000 CPUs using gcc with no standard libraries of any sort. Using DIVU/DIVS opcodes, I can obtain the quotient and the remainder. Given a Q16.16 fixed point value stored in an unsigned 32bit memory space, I know I can put the quotient in the upper 16 bits. However, how does one convert the integer remainder into the fractional portion of the fixed point value?

I'm sure this is something simple and I'm just missing it. Any help would be greatly appreciated.

Upvotes: 0

Views: 579

Answers (1)

Chris Dodd
Chris Dodd

Reputation: 126203

The way to think about it is that fixed point numbers are actually integers hold the value of your number times some fixed multiplier. You want to build you fixed point operations out of the integer operations you have available in your hardware.

So for a 16.16 fixed-point format, your multiplier is 65536 (216), so if you want to do a divide c = a/b, the numbers (integers) you have to work with are actually a' = a * 65536 and b' = b * 65536 and you want to find c' = c * 65536. So substituting into the desired c = a/b, you have

c'/65536 = (a'/65536) / (b'/65536) = a'/b'
c' = 65536 * a' / b'

So you actually want to first (integer) mulitply the fixed-point value of a by 65536 (left shift by 16), then do an integer divide by the fixed point value of b, and that will give you the fixed point value of c. The issue is that the first multiply will almost certainly overflow 32 bits, so you need a 64 bit (actually only 48 bit) intermediate. So if you're using a 68020+ with a 64/32 DIVS.L instruction (divides a 64 bit value in a pair of registers by a 32 bit value), you're fine. You don't need the remainder at all.

If you're using a pure 68000 that doesn't have the wide divide, you'll need to do 16-bit long division on the values (where you use 16 bit numbers as "digits", so you're dividing a 3-"digit" number by a 2-"digit" one)

Upvotes: 1

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