BUGFIX-66

Bug 55: Float32 To Float8 With A Fault

This converts a 32-bit IEEE 754 binary32 (single-precision) float to an 8-bit non-standard E4M3 float.
An E4M3 float has (from most- to least-significant): 1 sign bit, 4 exponent bits, and 3 mantissa bits.
E4M3 cannot encode infinities. E4M3 has exactly two NaN bit patterns: all exponent/mantissa bits are 1.
E4M3 exponent bias is 7. Unbiased binary32 exponents -7 and below yield E4M3 signed zero or subnormals.
Unbiased binary32 exponents -6 to 8 (inclusive) yield E4M3 normals, avoiding E4M3 NaN bit patterns.
Higher binary32 exponents (and infinities) clamp to the maximum non-NaN E4M3 exponent/mantissa (126).
E4M3 NaN is generated if and only if the input is a binary32 NaN; the NaN sign bit is preserved.
Rounding follows the IEEE 754 standard: To nearest, with ties to even.

Fix The Tiny Bug In This Go Code:

type E4M3 uint8 func NewE4M3(from float32) E4M3 { var ( all = math.Float32bits(from) sign = all >> 31 exp = all >> 23 & 255 mant = all & (1<<23 - 1) ret = sign << 7 ) switch { case exp <= 127-7: shift := 127 - 7 - exp + 1 if shift > 4 { return E4M3(ret) } mant |= 1 << 23 mant |= mant & 15 << 4 mant >>= shift ret |= mant >> 20 case exp <= 127+8: ret |= (exp - 127 + 7) << 3 ret |= mant >> 20 if ret&127 >= 126 { return E4M3(ret &^ 1) } case exp != 255 || mant != 0: return E4M3(ret | 126) default: return E4M3(ret | 127) } switch { case mant&(1<<19) == 0: case mant&(3<<19-1) != 0: ret++ } return E4M3(ret) }

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