Re: set -u not working as expected

From	Oğuz <oguzismailuysal@gmail.com>
Newsgroups	gnu.bash.bug
Subject	Re: set -u not working as expected
Date	2020-08-02 11:01 +0300
Message-ID	<mailman.500.1596355268.2739.bug-bash@gnu.org> (permalink)
References	(1 earlier) <F4821044-5F8D-4B6D-A266-BAC88AE3DC2C@larryv.me> <1867D8FC-85DD-4406-A239-3002913493AB@larryv.me> <CAH7i3Lr7ZCYJVpPZdc9xQMuF_yc_C2q0S9px+NY4prEUkU+24w@mail.gmail.com> <61B2CB9D-3978-4872-B88A-1542CF95B5B9@larryv.me> <CAH7i3LqAXNdCGjNqTGmnkimnXpz_kGnhQsiR2m7a5ezWtqe42A@mail.gmail.com>

Show all headers | View raw

2 Ağustos 2020 Pazar tarihinde Lawrence Velázquez <vq@larryv.me> yazdı:

> > On Aug 2, 2020, at 2:51 AM, Oğuz <oguzismailuysal@gmail.com> wrote:
> >
> > `u' has no members, so there's nothing to expand. If you use `${u[0]}'
> for
> > example, you'll see an error, I think how bash and ksh behave is
> perfectly
> > reasonable.
>
> Agreed. Their behavior logically follows from POSIX's carveout for $@.
>
> >> % bash -c 'set -u; typeset -i v; printf "<%s>\\n" "$v"'
> >> bash: v: unbound variable
> >> % ksh -c 'set -u; typeset -i v; printf "<%s>\\n" "$v"'
> >> ksh: v: parameter not set
> >> % zsh -c 'set -u; typeset -i v; printf "<%s>\\n" "$v"'
> >> <0>
> >>
> >>
> > `typeset -i v' doesn't assign `v', just gives it the integer attribute.
> > Again, I can't see any problem with bash and ksh here.
>
> Also agreed, but I was more interested in the next part...
>
> >> % bash -c 'set -u; typeset -i v; v+=1; printf "<%s>\\n" "$v"'
> >> <1>
> >> % ksh -c 'set -u; typeset -i v; v+=1; printf "<%s>\\n" "$v"'
> >> <1>
> >> % zsh -c 'set -u; typeset -i v; v+=1; printf "<%s>\\n" "$v"'
> >> <1>
>
> ...which contrasts with the behavior of let. Someone else will have
> to explain this, as I don't know what to make of it.
>
>
Yeah, that's interesting. See this one:

    $ set -u
    $ unset foo bar
    $ typeset -i foo bar
    $
    $ foo+=foo+1
    $
    $ foo+=bar+1
    bash: bar: unbound variable

Only referencing `bar' triggers the _unbound variable_ error, it makes
sense that the name being assigned is immune to that.

Concerning `let v+=1', let is not a declaration utility, it's arguments are
arithmetic expressions to be evaluated. And in arithmetic evaluation
context:

> Shell variables are allowed as operands; parameter expansion
> is performed before the expression is evaluated.


> vq



-- 
Oğuz

Back to gnu.bash.bug | Previous | Next | Find similar

Thread

Re: set -u not working as expected Oğuz <oguzismailuysal@gmail.com> - 2020-08-02 11:01 +0300

csiph-web