Re: [PATCH 3/5] configure: Don't use bash-specific string-replacement syntax

[Eric Blake]

On Wed, Jul 20, 2022 at 06:32:22PM +0100, Peter Maydell wrote:
> > > +    # backslash escape any '\' and '"' characters
> > > +    printf "%s" "$e" | sed -e 's/\([\"]\)/\\\1/g'
> >
> > You've fixed the bashism, but at the expense of a non-POSIX use of
> > sed.  POSIX says the input to sed must be a text file (ending in a
> > newline; but $e does not), and as a result it always outputs a newline
> > (but you don't want a newline before the closing """).  GNU sed
> > happens to do what you want for input not ending in a newline, but I
> > don't remember off-hand whether BSD sed does, and I know that Solaris
> > sed does not.
> 
> I just copied the approach we already take in quote_sh:
> 
> quote_sh() {
>     printf "%s" "$1" | sed "s,','\\\\'',g; s,.*,'&',"
> }
> 
> Is that also relying on this non-portable sed use?

Yep. And since no one has complained, BSD sed probably handles it the
way we want (input without a trailing line getting substituted and
re-output without a trailing newline).

If we cared, this one could also be fixed:

quote_sh() {
    printf "'%s'" "$(printf '%s\n' "$1" | sed "s,','\\\\'',g")"
}

Also note that we are depending on $1 never containing newlines, because
this change would alter the existing:

$ quote_sh "a
c"
'a'
'c'

into:

$ quote_sh "a
c"
'a
c'

which is probably more what you wanted.

> 
> > If this passes on BSD, then I'm okay with it; but if we want to avoid
> > non-POSIX altogether, this should work (using the shell's $() to strip
> > the trailing newline we added to keep sed happy):
> >
> > # backslash escape any '\' and '"' characters
> > printf '"""%s""",' "$(printf "%s\n" "$e" | sed -e '/s/\([\"]\)/\\\1/g')"
> 
> Mmm.

Given that no one has complained about quote_sh, insisting on POSIX
compliance (with its resulting increase in line noise and length) is
not winning any maintainability arguments ;)

-- 
Eric Blake, Principal Software Engineer
Red Hat, Inc.           +1-919-301-3266
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