FormattedRead hex string

[Jason Spencer]

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 at 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