BUGFIX-66

Bug 54: Float8 To Float32 With A Fault

This converts an 8-bit non-standard E4M3 float to a 32-bit IEEE 754 binary32 (single-precision) float.
An E4M3 float has (from most- to least-significant): 1 sign bit, 4 exponent bits, and 3 mantissa bits.
If the E4M3 exponent is zero, the E4M3 is either zero or subnormal: Nonzero mantissa makes a subnormal.
E4M3 signed zero becomes binary32 signed zero. E4M3 subnormals are converted to normalized binary32.
E4M3 cannot encode infinities. E4M3 has exactly two NaN bit patterns: all exponent/mantissa bits are 1.
E4M3 exponents are rebiased by subtracting the E4M3 bias (7) and adding the binary32 bias (127).
There are only 256 E4M3 bit patterns, so conversion is performed using a precomputed lookup table.

Fix The Tiny Bug In This Go Code:

type E4M3 uint8 func (e4m3 E4M3) Float32() float32 { return lut[e4m3] } func (e4m3 E4M3) e8m23() uint32 { var ( all = uint32(e4m3) sign = all >> 7 exp = all >> 3 & 15 mant = all & 7 ret = sign << 31 ) switch { case exp == 0: if mant == 0 { return ret } exp = 127 - 7 mant <<= 1 for mant>>3 == 0 { mant <<= 1 exp-- } mant &= 7 case exp == 15 && mant == 7: return ret + 1<<31 - 1 default: exp += 127 - 7 } ret |= exp << 24 ret |= mant << 20 return ret } var lut [256]float32 func init() { for i := range &lut { e8m23 := E4M3(i).e8m23() lut[i] = math.Float32frombits(e8m23) } }

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