Getting the fractional part of a float without using modf()

I would recommend taking a look at how modf is implemented on the systems you use today. Check out uClibc's version.

http://git.uclibc.org/uClibc/tree/libm/s_modf.c

(For legal reasons, it appears to be BSD licensed, but you'd obviously want to double check)

Some of the macros are defined here.

As I suspected, modf does not use any arithmetic per se -- it's all shifts and masks, take a look here. Can't you use the same ideas on your platform?

If I understand your question correctly, you just want the part after the decimal right? You don't need it actually in a fraction (integer numerator and denominator)?

So we have some number, say 3.14159 and we want to end up with just 0.14159. Assuming our number is stored in float f;, we can do this:

f = f-(long)f;

Which, if we insert our number, works like this:

0.14159 = 3.14159 - 3;

What this does is remove the whole number portion of the float leaving only the decimal portion. When you convert the float to a long, it drops the decimal portion. Then when you subtract that from your original float, you're left with only the decimal portion. We need to use a long here because of the size of the float type (8 bytes on most systems). An integer (only 4 bytes on many systems) isn't necessarily large enough to cover the same range of numbers as a float, but a long should be.