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Date: Sun, 26 Jan 2014 20:27:15 -0800
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Subject: Re: "ls" output like find ... -print0 anyhow ???
From: Kurtis Rader <krader@skepticism.us>
To: Bart Schaefer <schaefer@brasslantern.com>
Cc: Zsh Users <zsh-users@zsh.org>
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I recognize that your example is artificial but it does not support
arbitrary "ls" options. It seems to me that ~meino is asking the wrong
question.


On Sun, Jan 26, 2014 at 8:20 PM, Bart Schaefer <schaefer@brasslantern.com>wrote:

> On Jan 26,  8:59pm, Phil Pennock wrote:
> }
> } I'd use the print builtin of zsh, so that there are no argv length
> } limitations, combined with zsh's very powerful glob operators.
>
> Meino mentioned in private email that he has particular reasons for
> using "ls" instead of "find".  If the options of "ls" are important
> (rather than just the file names) and you don't need to worry too
> much about CPU consumption:
>
>   print -N -- **/*(.e:'reply=("$(ls -l "$REPLY")")':) | xargs -0 ...
>
> That's going to run "ls -l" once for each plain file (.) below the
> current directory, so it's a lot more process-intensive than a single
> "ls -lR", and it may not sort in the order you expect, but it will
> give you "ls" output with $'\0'-terminated lines.
>



-- 
Kurtis Rader
Caretaker of the exceptional canines Junior and Hank

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