From: Paul Lutus <nospam@...>
Date: 2006-09-16T08:20:56+09:00
Subject: Re: No regex backreference with four backslashes

gabriel.birke@gmail.com wrote:

> 
> Paul Lutus schrieb:
> 
>>
>> To find out how your strings are being parsed, print them out.

/ ...

> I've done that already, the test was only to show the problem: I could
> not escape chars in the numbers string with a backslash.
> 
> Anyway, I found the solution, it's five backslashes instead of four.
> That's a bit counter-intuitive, maybe someone can explain it.
> Especially when these two are compared:
> 
> numbers.gsub(/(2|4)/,'\\ \\1')
> numbers.gsub(/(2|4)/,'\\\\\1')
> 
> I expected that when I remove the space from the first expression, that
> my characters would get quoted. instead, the four backslashes get
> interpreted as two escaped backslashes and the 1 as a literal
> character. Can somebdoy shed some light on the how and why of this
> case?

Sure. Parsing these strings is a trivial exercise. Each adjacent pair of
backslashes collapses into one literal backslash, and any orphan
backslashes are associated with the character to its immediate right.

> Especially, why the solution with the five backslashes doesn't 
> yield double backlashes in the result string?

To sort out how Ruby is parsing your strings, *print* *them* *out.*

puts '\\\\1'

\\1 # meaning: a backslash and an escaped '1'

puts '\\ \\1'

\ \1 # meaning a backslash, a space, and an escaped '1'

puts '\\\\\1'

\\\1 # meaning two backslashes and an escaped '1'

Oh, by the way. You haven't said what you are trying to accomplish.

-- 
Paul Lutus
http://www.arachnoid.com