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Re: Bug Report concerning backslash in bash5

From: Dale R. Worley
Subject: Re: Bug Report concerning backslash in bash5
Date: Wed, 29 Jul 2020 01:35:32 -0400

Ralph Beckmann <rb@rbx.de> writes:
> I found this misbehaviour in Bash 5 (e.g. GNU bash, version 
> 5.0.16(1)-release (x86_64-pc-linux-gnu)):
> $ BLA="1\.2"; echo 'x/'$BLA'/y/'
> \x/1\.2/\y/
> I don't see any reasonable reason for the generated backslashes here.

My guess is that you're running into the fact that there are two types
of quoting character.  One quotes *any* character that follows it, and
thus it never appears in "the output" unless it was doubled in the
input.  The other type *only* quotes characters that are somewhow
special in that particular context.  Reading the manual page:

       Enclosing  characters  in  double quotes preserves the literal value of
       all characters within the quotes, with the exception of $, `,  \,  and,
       when  history  expansion  is enabled, !.  The characters $ and ` retain
       their special meaning within double quotes.  The backslash retains  its
       special  meaning only when followed by one of the following characters:
       $, `, ", \, or <newline>.

So backslash-inside-double-quotes-in-bash is of the second type, it only
quotes things that would otherwise be special.  So the value of $BLA is
1-\-.-2, whereas if the period was replaced by $, $BLA would only have 3

    $ BLA="1\$2"; echo 'x/'$BLA'/y/'


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