Re: Apparent inconsistency in f/z expansion flags behavior

[Pablo Lalloni <plalloni@gmail.com>]

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Thanks for your responses guys.

Both variations works as you said.

Better yet, I understand why.

Cheers!

On Sun, Dec 18, 2016 at 11:08 PM, Bart Schaefer <schaefer@brasslantern.com>
wrote:

> On Dec 18,  8:02pm, Pablo Lalloni wrote:
> }
> } words=(${(z)$(</proc/meminfo)})
> }
> } Which set words with the array of all the words in the file and that's
> } great.
> }
> } lines=(${(f)$(</proc/meminfo)})
> }
> } But then I get an array with just one string containing all the lines
> } concatenated (no NLs).
>
> That's happening because $(</proc/meminfo) has replaced all the NLs with
> spaces before (z) or (f) begin working.  (z) doesn't care because it
> splits on shell-syntax whitespace, but (f) has nothing to split on.
>
> To get what you want, you have to quote the $(...) substitution so the
> NLs are preserved:
>
>     lines=(${(f)"$(</proc/meminfo)"})
>
> That would be "more correct" in the (z) use as well, just in case the
> line breaks were e.g. inside quoted strings that would change the shell
> parse.  (Not an issue with /proc/meminfo, of course, but in general.)
>
> } Note that if you split the last assignment in 2 steps, it works as
> expected:
> }
> } lines=$(</proc/meminfo)
> } lines=(${(f)lines})
>
> There you've done first an assignment to a scalar, which preserves the
> NLs as if $(...) were quoted.
>