Re: Counting characters in command output?

["Mark J. Reed" <markjreed@gmail.com>]

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I think splitting vs not splitting could go either way; there are pros and
cons to each default, and as long as there's a mechanism to do the other
one, it's workable.

But I do think it's a bit counterintuitive that parameter expansion is not
split by default, while command expansion is. I would in general have
expected *foo $(bar) * to behave identically to *baz=$(bar); foo $baz*.  In
bash or ksh, that's true (probably absent some edge cases I'm not thinking
of), but not in zsh. Specifically, I would have thought you'd need *${=$(...)}
*to get the word spitting that you instead get by default.

On Wed, Feb 14, 2024 at 1:49 AM Roman Perepelitsa <
roman.perepelitsa@gmail.com> wrote:

> On Wed, Feb 14, 2024 at 1:46 AM Mark J. Reed <markjreed@gmail.com> wrote:
> >
> > On Tue, Feb 13, 2024 at 3:41 PM Roman Perepelitsa <
> roman.perepelitsa@gmail.com> wrote:
> >>
> >> Process substitution gives you an array. Quote it if you need a scalar.
> >
> > You mean command substitution, but thanks! It wouldn't have
> > occurred to me to put quotes around it to change what the `#` was
> > counting!
>
> Right, command substitution. In my opinion, it's unfortunate that
> command substitution in zsh splits on IFS by default. I wish this
> wasn't the case. It's not even common that one wants to split command
> output into words. Lines -- perhaps, but not words. So the default
> behavior in this case is rarely what is desired and has to be actively
> turned off.
>
> Roman.
>


-- 
Mark J. Reed <markjreed@gmail.com>

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