This should help a bit, although there's probably some room for improvement by
replacing the hex(byte, 2) calls with something that doesn't allocate an
intermediate string object:
function bytes2hex(bytes::Vector{Uint8})
io = IOBuffer()
for byte in bytes
write(io, hex(byte, 2))
end
return takebuf_string(io)
end
bytes2hex([0x01, 0x23, 0xab, 0xff]) # => "0123abff"
-- John
On Nov 30, 2014, at 9:39 PM, Ronald L. Rivest <[email protected]> wrote:
> Suppose you have a length-n array x of Uint8's = [1,5,32,7,...] , and you
> want to convert this
> to a long string of hex digits (two per x[i]). The code
> y = reduce(string, [hex(xi,2) for xi in x])
> ==> "01052007..."
> will do the trick. (Or, you could use map_reduce or one of the fold
> operations.)
>
> But I would like this operation to run in linear time.
>
> Would any of the reduce, map_reduce, or fold operations run in linear time?
> With a strict abstract implementation, each intermediate result string would
> need to
> be created, and each such intermediate result string is immutable, so the
> running
> time would be Theta(n^2).
>
> Or is the compiler smart about this case? (A small experiment suggests not.)
>
> Perhaps the right approach is:
> x2 = [ hex(xi,2) for xi in x]
> y = string(x2)
> ??
>
> Cheers,
> Ron
