tl;dr: on 64-bit little-endian machines, does
x := math.Float64frombits(binary.LittleEndian.Uint64(byteSlice[:8]))
get optimized to a single move instruction from byteSlice[:8] to x?
Background info:
I'm currently writing a bunch of code that unmarshals binary data. A small
example is this that unmarshals eight float64s
<https://github.com/twpayne/go-shapefile/blob/9d0389566e789784d6f977e2da00c33b98b52ab2/shapefile.go#L307-L314>,
and a longer example is this that unmarshals an arbitrary number of float64s
<https://github.com/twpayne/go-shapefile/blob/9d0389566e789784d6f977e2da00c33b98b52ab2/byteslicereader.go#L41-L44>.
In theory, when the endianness of the binary data matches the endianness of
the architecture, these should become simple memory copies. Does Go 1.19 do
this?
Although this is a performance question, performance is not a concern in my
use case. This is just an idle "can Go do this yet?" question and I'm
curious about the answer.
Thanks for any insight,
Tom
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