Thread: FPN math

Originally Posted by awsdert

No not trying to convert a simple integer, trying to construct an fpn from 2 seperate integers, one for the whole number part and another for the decimal part

Fixed point to floating point then? If you already have the bits isn't its just a matter of shifting then to the correct position?

Of course, you have to recalculate the fractional part... if you are dealing the integral and fractional parts as 32 bits values, 2^32 (fractional) is the same as 1.0, so:

n = (2^32*f)/(10^(log10(f) + 1)).

Taking 3.14 as your example... f=14 can be encoded as (2^32*14)/(10^2) = 601295421 (0b00100011110101110000101000111101 in 32 bits binary - of course this calculation must be done with enough precision to avoid overflows). So 3.14 can be encoded as 0b11.[00100011110101110000101000111101]. Shifting the binary point 1 position to the left we get 0b1.100100011110101110000101000111101 and e=1. Now we have our "inplicit" one and the fractional part that will satisfy the floating point float format if restricted to 23 bits: [0b1.100_1000_1111_0101_1100_0010]_1000111101.

So, M=0b10010001111010111000010 (0x48f5c2 - 23 bits), E=128 (0x80) (E=e+127) and S=0. Almost exactly what is expected for a floating point (float) value... The only difference is about rounding. Notice the _10001111101 final part, if this msb is 1 we need to add 1 to M, getting exactly the correct value:

To be sure:

v = (1 + 0x48f5c3 / 2^23) * 2^1 = (1+4781507/2^23)*2 = 3.14000010490417480468