From: Brian Candler <B.Candler@...>
Date: 2007-02-22T04:50:28+09:00
Subject: Re: Weird behaviour escaping special characters in a string

On Thu, Feb 22, 2007 at 02:55:09AM +0900, Greg Hurrell wrote:
> Why do I need so many backslashes in my replacement expression?
> 
> There are five slashes in the replacement expression:
> 
>     gsub(/(\\|')/, '\\\\\1')
> 
> But I would have thought that three would work:
> 
>     gsub(/(\\|')/, '\\\1')

Because even in single quotes, blackslashes must be doubled; this in turn is
because \' is the way that you insert a single quote within a single-quoted
string.

irb(main):001:0> a='\\'
=> "\\"
irb(main):002:0> a.size
=> 1
irb(main):003:0> b='\''
=> "'"
irb(main):004:0> b.size
=> 1
irb(main):005:0> c='\x'
=> "\\x"
irb(main):006:0> c.size
=> 2

> I basically want to replace "whatever is found in the pattern" with a
> backslash (\\) followed by "whatever was found" (\1); so that's three
> slashes. But with only three slashes Ruby gives me \1foo\1 instead of
> \'foo\'. Four slashes produces the same result. Five slashes and
> suddenly everything works (funnily enough, six slashes also works).
> Two slashes and one slash have no effect (no escaping is performed).
> 
> I've got working code so it's not a huge problem, but my curiosity is
> piqued. What's going on here that I don't understand?

irb(main):009:0> a='\\\\1'
=> "\\\\1"
irb(main):010:0> a.size
=> 3
irb(main):011:0> a='\\\\\1'
=> "\\\\\\1"
irb(main):012:0> a.size
=> 4
irb(main):013:0> a='\\\\\\1'
=> "\\\\\\1"
irb(main):014:0> a.size
=> 4

In a single-quoted string:
   \'  =>  '
   \\  =>  \
   \x  =>  \x  for all other x

So '...\1' and '...\\1' are identical.

HTH,

Brian.