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[Note: parts of this message were removed to make it a legal post.]

hi,
trying to solve someones mailinglist posted problem, I just created my own.
Why do 2 backslashes produce the same result as 4 and 6 the same as 8?

014:0> 'ab'.gsub('a', '\\')
=> "\\b"
015:0> 'ab'.gsub('a', '\\\\')
=> "\\b"
016:0> 'ab'.gsub('a', '\\\\\\')
=> "\\\\b"
017:0> 'ab'.gsub('a', '\\\\\\\\')
=> "\\\\b"
018:0> RUBY_DESCRIPTION
=> "ruby 1.8.7 (2009-06-12 patchlevel 174) [i686-linux]"

Doesn't that strike anyone as odd?
Greetz!

On Mon, 2009-09-07 at 17:36 +0900, Fabian Streitel wrote:
> hi,
> trying to solve someones mailinglist posted problem, I just created my own.
> Why do 2 backslashes produce the same result as 4 and 6 the same as 8?
>
> 014:0> 'ab'.gsub('a', '\\')
> => "\\b"
> 015:0> 'ab'.gsub('a', '\\\\')
> => "\\b"
> 016:0> 'ab'.gsub('a', '\\\\\\')
> => "\\\\b"
> 017:0> 'ab'.gsub('a', '\\\\\\\\')
> => "\\\\b"
> 018:0> RUBY_DESCRIPTION
> => "ruby 1.8.7 (2009-06-12 patchlevel 174) [i686-linux]"
>
> Doesn't that strike anyone as odd?
> Greetz!



Be careful with gsub, it does a lot of magic. From the documentation:

If a string is used as the replacement, special variables from the
match (such as +$&+ and +$1+) cannot be substituted into it, as
substitution into the string occurs before the pattern match
starts. However, the sequences +\1+, +\2+, and so on may be used to
interpolate successive groups in the match.

To fix, put the replacement into a block:

irb(main):001:0> s = '\\'
=> "\\"
irb(main):002:0> s.size
=> 1
irb(main):003:0> s = s * 10
=> "\\\\\\\\\\\\\\\\\\\\"
irb(main):004:0> s.size
=> 10
irb(main):005:0> 'a'.gsub('a', s).size
=> 5
irb(main):006:0> 'a'.gsub('a') { s }.size
=> 10
irb(main):007:0> 'a'.gsub('a') { s }
=> "\\\\\\\\\\\\\\\\\\\\"
irb(main):008:0> 'a'.gsub('a', s)
=> "\\\\\\\\\\"



Martin

2009/9/7 Fabian Streitel <>:
> hi,
> trying to solve someones mailinglist posted problem, I just created my ow=
n.
> Why do 2 backslashes produce the same result as 4 and 6 the same as 8?
>
> 014:0> 'ab'.gsub('a', '\\')
> =A0 =A0=3D> "\\b"
> 015:0> 'ab'.gsub('a', '\\\\')
> =A0 =A0=3D> "\\b"
> 016:0> 'ab'.gsub('a', '\\\\\\')
> =A0 =A0=3D> "\\\\b"
> 017:0> 'ab'.gsub('a', '\\\\\\\\')
> =A0 =A0=3D> "\\\\b"
> 018:0> RUBY_DESCRIPTION
> =A0 =A0=3D> "ruby 1.8.7 (2009-06-12 patchlevel 174) [i686-linux]"
>
> Doesn't that strike anyone as odd?

Well, not me anyway. You can find a lot of threads about this in
the archives... (hint, hint)

The short story: you need to keep in mind that there are multiple
levels of escaping going on which unfortunately use the same character
("\") as meta character AND output in IRB is done via #inspect.

To get a single backslash in a string you need to type two:

irb(main):001:0> puts '\\'
\
=3D> nil

In order to make gsub use a backslash literally (and not interpret it
as meta character) you need to have two of them in the string:

irb(main):002:0> puts '\\\\'
\\
=3D> nil

Now, there are some corner cases: your first example uses a
replacement string with just a single backslash so it looses its
metaness (because there is nothing behind it to escape). This is true
for all cases where there is nothing behind it which can be escaped:

irb(main):004:0> puts '\1', '\\1'
\1
\1
=3D> nil

HTH

Kind regards

robert

--=20
remember.guy do |as, often| as.you_can - without end
http://blog.rubybestpractices.com/

[Note: parts of this message were removed to make it a legal post.]

>
> Well, not me anyway. You can find a lot of threads about this in
> the archives... (hint, hint)
>
>
ah, yeah, there was something... I guess I'm gonna hand myself a quick
JFGI here...
sry...

well... thanks anyways, I think I get it now. that magic is reaaaally mean.
Greetz!