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hi-v-rocknroll · 2 years ago
Formats, plural.

sign:explicit leading mantissa bit:mantissa bits:exponent bits

1:0:(P-1):(8-P) where P ∈ [1,7]

P ∈ [3,5] appears to be more useful

binary8p3 -> 1:0:2:5

binary8p4 -> 1:0:3:4

binary8p5 -> 1:0:4:3

Edge cases:

P = 8 would disallow all exponent bits, leaving a sign bit and 7 explicit mantissa bits of precision

P = 0 would be unsigned and 8 exponent bits with only an implicit precision bit but no explicit mantissa bits

avianes · 2 years ago
Also, note that:

- There is only one Zero (encoded as 0x00), no negative-zero

- There is only one NaN (encoded as 0x80)

hi-v-rocknroll · 2 years ago
-0, qNANs, and sNANs of the world unite to raise an exception to this travesty!

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