Reputation: 86
In Go, how can I encode a float into a byte array as a 16 bit unsigned float with 11 explicit bits of mantissa and 5 bits of explicit exponent?
There doesn't seem to be a clean way to do it. The only thing I can think of is encoding it as in Convert byte array "[]uint8" to float64 in GoLang and manually truncating the bits.
Is there a "go" way to do this?
Here's the exact definition:
A 16 bit unsigned float with 11 explicit bits of mantissa and 5 bits of explicit exponent
The bit format is loosely modeled after IEEE 754. For example, 1 microsecond is represented as 0x1, which has an exponent of zero, presented in the 5 high order bits, and mantissa of 1, presented in the 11 low order bits. When the explicit exponent is greater than zero, an implicit high-order 12th bit of 1 is assumed in the mantissa. For example, a floatingvalue of 0x800 has an explicit exponent of 1, as well as an explicit mantissa of 0, but then has an effective mantissa of 4096 (12th bit is assumed to be 1). Additionally, the actual exponent is one-less than the explicit exponent, and the value represents 4096 microseconds. Any values larger than the representable range are clamped to 0xFFFF.
Upvotes: 2
Views: 2083
Reputation: 3920
I am not sure whether I understand the encoding correctly (see my comment on the original question), but here is a function which may do what you want:
func EncodeFloat(seconds float64) uint16 {
us := math.Floor(1e6*seconds + 0.5)
if us < 0 {
panic("cannot encode negative value")
} else if us > (1<<30)*4095+0.5 {
return 0xffff
}
usInt := uint64(us)
expBits := uint16(0)
if usInt >= 2048 {
exp := uint16(1)
for usInt >= 4096 {
exp++
usInt >>= 1
}
usInt -= 2048
expBits = exp << 11
}
return expBits | uint16(usInt)
}
(code is at http://play.golang.org/p/G599VOBMcL )
Upvotes: 2