From: Rob Biedenharn <Rob@...>
Date: 2010-01-26T03:35:01+09:00
Subject: Re: gsub problem

On Jan 25, 2010, at 12:00 PM, lalawawa wrote:

> Actually, I oversimplified my example.  I want to escape single  
> quotes and question marks.
>
> irb(main):001:0> a = "Can't we?"
> => "Can't we?"
> irb(main):002:0> a.gsub(/([\'\?])/, '\\1')
> => "Can't we?"
> irb(main):003:0> a.gsub(/([\'\?])/, '\\\1')
> => "Can\\1t we\\1"
> irb(main):004:0> a.gsub(/([\'\?])/, '\\\\1')
> => "Can\\1t we\\1"
> irb(main):005:0> a.gsub(/([\'\?])/, '\\\\\1')
> => "Can\\'t we\\?"


(Well, Mike's answer steals my thunder a bit, but I'll put this out  
there for anyone else that finds this thread.)


You just have to remember that there's two levels of confusion, uh, I  
mean *escaping* going on here.

Within single-quoted strings, the only backslash-escape is for a '  
and, so you can have a \, you also need to escape the escape character:

'\1' means a backslash and a digit (because the character following  
the \ is not ' or \ it doesn't have its special meaning)
'\\1' means an escaped backslash and a digit (yes, the same as above,  
but this time the first \ escapes the second \ so that it is  
interpreted literally)
'\\\1' means an escaped backslash, a backslash, and a digit (you see  
where this is going, right?)

Now, within the replacement string for a gsub, you can have a  
backslash-digit to mean "the n-th parenthesized group where the digit  
is n"

So you want to replace with "a backslash and a backslash-digit for the  
first group".  You need the final interpretation of the replacement to  
end up as \ \1 (no quotes here to confuse things)

'\\' is a literal \
'\\\\' is a literal \\ so gsub will see a real \
'\1' is \1 because 1 isn't a special character
SO...
'\\\\\1' is seen as \\\1 (as the argument) and gsub interprets it as  
"literal backslash, first group"

'\\\\\\1' is seen the same way; the fifth backslash escapes the sixth,  
then the 1

You might consider using the block form of gsub replacement.

irb> a = "Can't we?"
=> "Can't we?"
irb> a.gsub(/(['?])/, '\\\\\\1')
=> "Can\\'t we\\?"
irb> puts _
Can\'t we\?
=> nil
irb> a.gsub(/['?]/) {|m| '\\' + m}
=> "Can\\'t we\\?"
irb> puts _
Can\'t we\?
=> nil

In your case, the clarity that results from the simplification of the  
replacement should be obvious.

-Rob

Rob Biedenharn		http://agileconsultingllc.com
Rob@AgileConsultingLLC.com