Re: Combination of "eval set -- ..." and $() command substitution is slow

From	Ilkka Virta <itvirta@iki.fi>
Newsgroups	gnu.bash.bug
Subject	Re: Combination of "eval set -- ..." and $() command substitution is slow
Date	2019-07-16 12:18 +0300
Message-ID	<mailman.1492.1563268736.2688.bug-bash@gnu.org> (permalink)
References	<bcd08f6c-1c13-0eb4-92b2-4e904b19a0ce@e-nautia.com> <8091.1563212947@jinx.noi.kre.to> <08d234bc-5dca-38f5-cb38-02b0c7fee542@iki.fi>

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On 15.7. 20:49, Robert Elz wrote:

> 			printf '%s\n' "`printf %s "$i"`"
> 			printf '%s\n' "$(printf %s "$i")"
> 
> aren't actually the same.   In the first $i is unquoted, in the second it is
> quoted.   

Huh, really? It looks to me like the first one treats $i as quoted too:

  $ touch file.txt; i='123 *'
  $ printf '%s\n' "`printf :%s: "$i"`"
  :123 *:

But not here, of course:

  $ printf '%s\n' "`printf :%s: $i`"
  :123::file.txt:

I tried with Bash and some other shells, but couldn't find one where the 
result was different. Did I miss something?

-- 
Ilkka Virta / itvirta@iki.fi

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Re: Combination of "eval set -- ..." and $() command substitution is slow Ilkka Virta <itvirta@iki.fi> - 2019-07-16 12:18 +0300

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