I imagine there's a slick way to do this, but I'm not seeing it.
I have a string of hex digits which I'd like to convert to an
array of 8 ubytes:
0123456789abcdef --> [0x01, 0x23, 0x45, 0x67, 0x89, 0xAB, 0xCD,
0xEF]
I'm looking at std.format.formattedRead, but the documentation
is...lightish. First of all, it seems there's no format
specifier except %s on reads and type information is gleaned from
the args' types. I was able to experiment and show that %x
works, but no documentation on exactly how.
Second, array syntax seems to work only if there's some
delimiter. With:
void main(string[] args)
{
ubyte[8] b;
formattedRead(args[1], "%(%s%)", &b);
}
I get
std.conv.ConvOverflowException@C:\Tools\D\dmd2\windows\bin\..\..\src\phobos\std\
conv.d(2006): Overflow in integral conversion
at least once. :) But that makes sense--hard to tell how many
input chars to assign to one byte versus another (although it
seems to me a hungry algorithm would work--saturate one type's
max and move to the next.)
There doesn't seem to be any support for field sizes or counts in
formatted read, similar to old C "%16x". This barks at me right
away--"%1 not supported."
I know I could read (in this case) as two longs or a uint16, but
I don't want to deal with endianess--just data.
Is there some trick to use the fact that b is fixed size 8 bytes
and know that requires 16 hex digits and converts automatically?
Is there some other suggestion for how to do this eloquently? I
can play around with split and join, but it seemed like there is
probably some way to do this directly that I'm not seeing.
Thanks!
Jason
