Hi,
After trying to split a string into fields delimited with colons and
spaces, I found this bug in how ksh(1) does substitution. The actual
behavior contradicts what other shells like bash and mksh do and also
contradicts its own manual.
Running the following on other shells (say, bash) prints "/foo/bar/".
This command splits the string " foo : bar " into two fields: "foo"
and "bar", considering colon and space as delimiters.
echo " foo : bar " | {
IFS=": "
read -r a b
printf -- "/%s/%s/\n" "$a" "$b"
}
However, running the same command in OpenBSD ksh(1) (or sh(1)) splits
the string into "foo" and ": bar".
The manual ksh(1) provides the following, similar example:
> Example: If IFS is set to “<space>:”, and VAR is set to
> “<space>A<space>:<space><space>B::D”, the substitution for $VAR
> results in four fields: ‘A’, ‘B’, ‘’ (an empty field), and ‘D’.
> Note that if the IFS parameter is set to the NULL string, no field
> splitting is done; if the parameter is unset, the default value of
> space, tab, and newline is used.
Let's try it:
echo " A : B::D" | {
IFS=" :"
read -r arg1 arg2 arg3 arg4
printf -- '1st: "%s"\n' "$arg1"
printf -- '2nd: "%s"\n' "$arg2"
printf -- '3rd: "%s"\n' "$arg3"
printf -- '4th: "%s"\n' "$arg4"
}
bash(1) splits the line into the following fields:
1st: "A"
2nd: "B"
3rd: ""
4th: "D"
This is actually the expected output, as described in the manual.
However, running the same command in OpenBSD ksh, prints this:
1st: "A"
2nd: ""
3rd: "B"
4th: ":D"
A completelly different thing.
The same occurs with OpenBSD sh(1).
I could not understand how OpenBSD does the spliting, but the way it
does is clearly a bug: it does not only contradicts its own manual,
but also differs from other implementations.
Thank you,
Lucas de Sena.
