cc: [email protected]
Subject: Re: [ast-users] ksh read byte
--------
> The builtin manual of ksh's "read" says:
> "-N nbyte Read exactly nsize characters.
> For binary fields size will be in bytes."
>
> I've tried to read single bytes with "read -N1". The following, using
> "od", works as expected and thus confirms that "read -N1" is working:
>
> #!/bin/ksh
>
> exec 3< hexfile ## file of 256 bytes: 0, 1, 2, 3, ... 255
>
> while read -u3 -r -N1 B
> do
> H=$(printf "$B" | od -tx1 -An)
> print ${H:=" 00"}
> done | less
>
> exec 3<&-
>
> But I want to compare $B to constant values, and print $B as ascii-hex,
> and I'd like to avoid using external application "od". If I declare B
> with "typeset -is", or "typeset -i", or "typeset -b", the result is
> "arithmetic syntax error" or nonsense. Behavior is the same for locale
> "C" and "en_ZA.utf8". Of course it can be done (ksh can do anything!)
> but please tell me, "How?".
>
>
Assuming that the data you read in is a valid character in that locale,
I would expect that in the C locale
exec 3< hexfile ## file of 256 bytes: 1, 2, 3, ... 127
while read -u3 -r -N1 B
do print -r -- $(('$B'))
done | less
exec 3<&-
would work. However, for 0 or a character not in the current locale,
this will not work. read return the Empty string for the 0 byte.
You could use
printf "%d\n" "'$B'"
instead and shorten it to
while read -u3 -r -N1 B
do printf "%d\n" ${B+"'$B'"}
done < hexfile | less
which also handles 0.
read will return 1 for an invalid character.
David Korn
[email protected]
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