Altera_Forum
Honored Contributor
Honored ContributorFloating point division
Hi,
does anyone has a code for floating point division? i tried using the megafunction tool for division but it keeps giving me error when i run simulation.Forum Discussion
9 Replies
Altera_Forumhow about your code?
- Altera_Forum
Floating point division in firmware is no trivial task. You could always offload that problem to a NIOS, but that seems like overkill. Are you sure you can't get by with fixed point or avoiding division entirely?
- Altera_Forum
You could try using integers.
Multiply the FP numbers by a scalar to convert it to a integer, divide the integers, then convert back to FP or divide down to get just the integer result of the multiplication. - Altera_Forum--- Quote Start --- You could try using integers. Multiply the FP numbers by a scalar to convert it to a integer, divide the integers, then convert back to FP or divide down to get just the integer result of the multiplication. --- Quote End --- Floating to integer conversion is not as simple as "multiply by a scaler", you need to do a full conversion. You may be thinking of fixed point, which is just integer arithmatic anyway.
- Altera_Forum
FP division is probably simpler than integer division - especially if you aren't worried about underflow, overflow, infinities, NaNs and denormalised small values.
Extract and subtract the exponents and then divide the mantissa's - which both start with a 1. - Altera_Forum
Hi,
For a 32 bit floating point number, i would like to know the split of signbit,exponent,mantissa. - Altera_Forum
The format is explained in the FP MegaFunction user manual. Or google for IEEE 754.
- Altera_Forum
Sign bit : Exponent (8 bits) : Mantissa (23 bits)
Keep in mind that the mantissa has the implied '1' that DSL referred to which isn't stored as part of the manatissa field. Back when I was working on floating point stuff I found this webpage very handy: http://www.h-schmidt.net/floatconverter/ieee754.html So DSL said to subtract bits [30:23] and divide bits {1'b1, [22:0]}. For the sign bit you could just XOR those bits since both positive or both negative should result in a positive result (sign bit = 0). - Altera_Forum
You are then dividing two values between 0.5 and 1.999999999 - so the exponent may need changing by 1 (or 2?) in order to normalise the resultant mantissa.
You should also worry aout the rounding rules (round to even is normally specified by default).
