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how does escaping in "`...`" work?

From: Matthew Woehlke
Subject: how does escaping in "`...`" work?
Date: Mon, 07 Jun 2010 17:47:12 -0500
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(See also https://bugs.kde.org/show_bug.cgi?id=237675)

How should bash interpret escapes in constructs like "`...`"?

For example, this do what I would expect:

$ echo "`echo "you don't   say"`"
you don't   say

(That is, the contents in ``'s are parsed independent of any other context, except that additional `'s would need to be escaped for further nesting.)

But this does not:
$ echo "`echo \"you don\'t   say\"`"
you don\'t   say

I expected:
"you don't say"

That is, I expected that \" -> " substitution would not happen inside the ``'s outside of the context of the command substitution, since the quotes do not need to be escaped in that context to be seen as quotes in the substitution context.

Is this how it is supposed to work?

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