Re: Patterns quoting in subscript (was: Re: PATCH: Assorted parameter stuff)

["Bart Schaefer" <schaefer@xxxxxxxxxxxxxxxxxxxxxxx>]

On Apr 18, 12:34pm, Andrej Borsenkow wrote:
} Subject: Re: Patterns quoting in subscript (was: Re: PATCH: Assorted param
}
} On Wed, 18 Apr 2001, Bart Schaefer wrote:
} >
} > schaefer[507] print -r $foo[(r)\\\\\?]
} > \?
} >
} > The reason you need five is that [...] is parsed as if double quoted,
} > except that \] is magic instead of \".  Then the pattern inside it is
} > tokenized by (r) before being passed to patcompile().  So it goes like:
} >
} >      parse        tokenize     pattern
} >     \\ \\ \?  -->  \\ \?   -->   \?
} >
} > (In double quotes, \\ --> \ but \? --> \?.)
} 
} O.K., any chance to explain the above in Zsh doc?

Yes, I just don't want to spend time writing doc if this change is going
to cause some other problem and therefore not be included.

} Just to make sure that I understand:
} 
} - "as if in double quotes" means, that I cannot use "..." or '...' quoting
} inside of subscript, only back-slash is possible?

It's exactly like quoting for $[...] math expressions, except that (as
of 14008) spaces are not permitted.  Which for a traditional numeric
subscript makes sense, as it *is* a math expression.
 
} - the rule for ``\\'' and ``\$'' still aplies? But ``\"'' is parsed as
} *two* characters?

Yes.

On Apr 18, 12:45pm, Andrej Borsenkow wrote:
}
} O.K., but why when I add additional backslashes I still get a match?:
} 
} bor@itsrm2:~%> print -r $foo[(r)\\\\\\\\\?]
} \?
} 
} ?? It seems to happen for any number of extra backslashes.

Well, sigh, patcompile() also apparently does a level of backslash-removal,
so i've forgotten one step in the explanation above.  It's something like:

       parse           tokenize      pattern     match
    \\ \\ \?       --> \\ \?     -->   \?    -->  \?
    \\ \\ \\ \\ \? --> \\ \\ \?  -->  \\ ?   -->  \?

Then you also have to remember that `?' as a metacharacter matches any
character, so one extra backslash will result in a pattern that matches
by wildcard rather than by a literal question-mark.  So if you want to
match 4 backslashes, you need 16 backslashes; but if you want to match
4 backslashes followed by a literal question mark, you need a whopping
34 backslashes, because you have to double all 16 of those backslashes
and then add two more to quote the `?' itself.

Clearly this is silly; it results because of the extra call to parsestr()
that happens when the `needtok' loop in getarg() detects an ispecial()
(the `?' in this example).  I'll see if that can't be made to behave a
bit more intelligibly.

-- 
Bart Schaefer                                 Brass Lantern Enterprises
http://www.well.com/user/barts              http://www.brasslantern.com

Zsh: http://www.zsh.org | PHPerl Project: http://phperl.sourceforge.net