Date: Fri, 12 Jan 2024 07:15:35 -0500
From: Greg Wooledge <[email protected]>
Message-ID: <[email protected]>
| This was one of the things I tested:
Perhaps intended to, but didn't, or not in this example:
| { read -N1; read -r b c; } < <(printf \\nabc); declare -p REPLY a b c
Rewrite that, correctly for the purpose, as:
{ read -N1; read -r b c; } < <(printf \\\\\\nabc); declare -p REPLY b c
or a bit more cleanly as:
{ read -N1; read -r b c; } < <(printf '\\\nabc'); declare -p REPLY b c
and (both cases) you get:
declare -- REPLY="a"
declare -- b="bc"
declare -- c=""
bash is (as I assumed earlier it would) doing the correct thing, and
ignoring the elided newline completely. If it didn't, that would have
been a bug, but it does, so all is well.
In the case you tested, printf output $'\nabc' and read -N1 correctly
read the first char (\n) into REPLY (and then read abc into b, 'c' isn't
really needed for this example).
[Note: I deleted the output of 'a' as I didn't conveniently have a
variable called 'a' defined, so declare issued an error, which would
have just confused things.]
kre
ps: the use of process substitution there is just silly, it would work
just as well, and be easier to understand if written:
printf '\\\nabc' | { read -N1; read -r b c; }; declare -p REPLY b c
or { read -N1; read -r b c; } <<< $'\\\nabc' ; declare -p REPLY b c
I much prefer the former of those two.
