Re: floating point representation

waltbrad wrote:
I'm posting here again seeking help on understanding floating point
numbers. I'm an absolute novice, struggling with ASM concepts, but I
am able to write the basic programs offered in AOA.

Now, with floating point I do understand how they are stored. For
instance the single precision has the low order 23 bits as a mantissa,
the next 8 bits up are the exponent, and the H.O. bit is the sign bit.

Not quite. There are 24 bits of mantissa in a normalized value, but only 23 are stored. Since the highest bit is always 1, it can be implied.


I understand the excess-bias and understand the exponent is a power of
2.

But for the life of me I cannot construct a number manually from it's
stored representation.

For instance: the number '7', has the hex representation of
40E0_0000

which translates to:

0100 0000 1110 0000 0000 0000 0000 0000

So first I take the exponent 10000001 that is 129-127=2 and 2
squared is 4.

The Mantissa? Well I guess it's (2**22)+(2**21) = 6291456. So now I
have:

1.6291456 * 4 = 6.5165824

But this number itself is stored as 40D0_87D8

Which works out to be 6.1110624


You can break down the mantissa like this:

1 (implied) + 0.5 + 0.25 = 1.75

1.75 x 4 = 7.00


.

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