Re: One IFS non-whitespace is stripped at the end of a word during word splitting

[Lawrence Velázquez]

On Sun, Dec 8, 2024, at 3:27 AM, Yuri Kanivetsky wrote:
> $ bash -c 'IFS=x; a=ax; f() { for arg; do echo "($arg)"; done; }; f $a'
> (a)
>
> $ bash --version
> GNU bash, version 5.2.37(1)-release (x86_64-pc-linux-gnu)
>
> I.e. IFS non-whitespaces are not stripped at the beginning of a word,
> but if there's one such non-whitespace at the end, it is stripped.

It's not being stripped.  IFS characters *terminate* fields, so
"ax" is split to the single field "a".

        The shell treats each character of $IFS as a delimiter, and
        splits the results of the other expansions into words using
        these characters as field terminators.

(https://www.gnu.org/software/bash/manual/html_node/Word-Splitting.html)


> This looks like a bug, unless I'm missing something.

It's not a bug.  It is how most (but not all) other shells behave:

        $ cat /tmp/ifs.sh
        IFS=x
        a=ax
        f() {
                for arg
                do
                        echo "($arg)"
                done
        }
        f $a
        $ bash /tmp/ifs.sh
        (a)
        $ dash /tmp/ifs.sh
        (a)
        $ ksh /tmp/ifs.sh
        (a)
        $ mksh /tmp/ifs.sh
        (a)
        $ oksh /tmp/ifs.sh
        (a)
        $ yash /tmp/ifs.sh
        (a)
        $ zsh --emulate sh /tmp/ifs.sh
        (a)
        ()


-- 
vq