# Gsub! and quoting question

**URL:** <https://www.ruby-forum.com/t/gsub-and-quoting-question/114247>\
**Category:** Ruby\
**Created:** [September 10, 2007, 1:56pm UTC](https://www.ruby-forum.com/t/gsub-and-quoting-question/114247 "2007-09-10T13:56:49Z")\
**Posts on this page:** 6\
**Page:** 1

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**Author:** ![ronaldfischer](https://www.ruby-forum.com/letter_avatar_proxy/v4/letter/r/67e7ee/32.png) [@ronaldfischer](https://www.ruby-forum.com/u/ronaldfischer)\
**Post date:** [September 10, 2007, 1:56pm UTC](https://www.ruby-forum.com/t/gsub-and-quoting-question/114247/1 "2007-09-10T13:56:49Z")

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I thought this would be a trivial task, but it seems to be more  
difficult than expected:

I have a variable, data, containing some string. I would like to  
preprend every  
single apostrophe in this string by a backslash. Here is my solution:

data.gsub!(/’/,%q(\’))

Strangely, this does not work. I tested it with data containing a string  
consisting  
of a single quote solely, and surrounded the code by “puts” like this:

data=%q(’)  
puts “:”+data+" replace quotes by "+%q(\’)  
data.gsub!(/’/,%q(\’))  
puts “data length now #{data.length}”

This produced as output:

:’ replace quotes by ’  
data length now 0

From this I conclude that the gsub! had shortened the string to length  
zero.  
Any explanation for this? How do I solve my problem in a proper way?

Ronald

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**Author:** ![ronaldfischer](https://www.ruby-forum.com/letter_avatar_proxy/v4/letter/r/67e7ee/32.png) [@ronaldfischer](https://www.ruby-forum.com/u/ronaldfischer)\
**Post date:** [September 10, 2007, 2:57pm UTC](https://www.ruby-forum.com/t/gsub-and-quoting-question/114247/2 "2007-09-10T14:57:05Z")

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2007/9/10, Ronald F. [removed\_email\_address@domain.invalid](mailto:removed_email_address@domain.invalid):

> consisting  
> data length now 0
> 
> From this I conclude that the gsub! had shortened the string to length  
> zero.  
> Any explanation for this? How do I solve my problem in a proper way?

This comes up frequently. You need to be aware that there are several  
levels of interpretation involved and so you need multiple levels of  
escaping. This is sometimes obscured by the fact that ‘\1’ works  
although it should read ‘\1’. These levels are: 1. string escaping,  
2. regexp replacement string escaping. All these variants do work:

str.gsub /‘/, ‘\\\&’  
str.gsub /’/, ‘\\'’  
str.gsub /‘/, "\\’"

Kind regards

robert

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<div class="post-metadata">

**Author:** ![ronaldfischer](https://www.ruby-forum.com/letter_avatar_proxy/v4/letter/r/67e7ee/32.png) [@ronaldfischer](https://www.ruby-forum.com/u/ronaldfischer)\
**Post date:** [September 10, 2007, 3:46pm UTC](https://www.ruby-forum.com/t/gsub-and-quoting-question/114247/3 "2007-09-10T15:46:15Z")

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> > From this I conclude that the gsub! had shortened the  
> > string to length  
> > zero.  
> > Any explanation for this? How do I solve my problem in a proper way?
> 
> This comes up frequently.

I’m sorry if I posted a question which has been answered already ☹

> You need to be aware that there are several  
> levels of interpretation involved and so you need multiple levels of  
> escaping. This is sometimes obscured by the fact that ‘\1’ works  
> although it should read ‘\1’. These levels are: 1. string escaping,  
> 2. regexp replacement string escaping.

Hmmm… I would understand your argument if I had a digit following  
a backslash (which would mean a backreference to the n-th group in  
the pattern), but here we have ', and this is not a back reference,  
isn’t it? At least the discussion of String#gsub at  
[http://www.ruby-doc.org/core/](http://www.ruby-doc.org/core/)  
mentions only \1,\2,… etc. as back references.

Ronald

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<div class="post-metadata">

**Author:** ![ronaldfischer](https://www.ruby-forum.com/letter_avatar_proxy/v4/letter/r/67e7ee/32.png) [@ronaldfischer](https://www.ruby-forum.com/u/ronaldfischer)\
**Post date:** [September 10, 2007, 4:32pm UTC](https://www.ruby-forum.com/t/gsub-and-quoting-question/114247/4 "2007-09-10T16:32:40Z")

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> > Hmmm… I would understand your argument if I had a digit following  
> > a backslash (which would mean a backreference to the n-th group in  
> > the pattern), but here we have ', and this is not a back reference,  
> > isn’t it? At least the discussion of String#gsub at  
> > [RDoc Documentation](http://www.ruby-doc.org/core/)  
> > mentions only \1,\2,… etc. as back references.
> 
> No. You want the backslash to be interpreted literally by the regexp  
> engine, i.e. you want to prevent the regexp engine from identifying  
> the backslash as a meta character.

OK

> This is independent from the  
> character _after_ the backslash.

Ah, that’s the point I was missing. Now things are clear!

> Btw, not only digits are meaningful in replacement patterns, but  
> single quote is as well (I believe it’s the postmatch):

I already suspected this too (knowing this concept from Perl), but  
didn’t  
find anything in the Ruby online docs where this would be discussed.  
But this explains why my string was replaced by the empty string…

Ronald

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<div class="post-metadata">

**Author:** ![ronaldfischer](https://www.ruby-forum.com/letter_avatar_proxy/v4/letter/r/67e7ee/32.png) [@ronaldfischer](https://www.ruby-forum.com/u/ronaldfischer)\
**Post date:** [September 10, 2007, 7:02pm UTC](https://www.ruby-forum.com/t/gsub-and-quoting-question/114247/5 "2007-09-10T19:02:10Z")

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On Sep 10, 7:55 am, “Robert K.” [removed\_email\_address@domain.invalid](mailto:removed_email_address@domain.invalid)  
wrote:

> although it should read ‘\1’. These levels are: 1. string escaping,  
> 2. regexp replacement string escaping. All these variants do work:
> 
> str.gsub /‘/, ‘\\\&’  
> str.gsub /’/, ‘\\'’  
> str.gsub /‘/, "\\’"

str.gsub /(?=')/, “\”

---

<div class="post-metadata">

**Author:** ![ronaldfischer](https://www.ruby-forum.com/letter_avatar_proxy/v4/letter/r/67e7ee/32.png) [@ronaldfischer](https://www.ruby-forum.com/u/ronaldfischer)\
**Post date:** [September 10, 2007, 4:20pm UTC](https://www.ruby-forum.com/t/gsub-and-quoting-question/114247/6 "2007-09-10T16:20:37Z")

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2007/9/10, Ronald F. [removed\_email\_address@domain.invalid](mailto:removed_email_address@domain.invalid):

> > > From this I conclude that the gsub! had shortened the  
> > > escaping. This is sometimes obscured by the fact that ‘\1’ works  
> > > although it should read ‘\1’. These levels are: 1. string escaping,
> > 
> > 1. regexp replacement string escaping.
> 
> Hmmm… I would understand your argument if I had a digit following  
> a backslash (which would mean a backreference to the n-th group in  
> the pattern), but here we have ', and this is not a back reference,  
> isn’t it? At least the discussion of String#gsub at  
> [RDoc Documentation](http://www.ruby-doc.org/core/)  
> mentions only \1,\2,… etc. as back references.

No. You want the backslash to be interpreted literally by the regexp  
engine, i.e. you want to prevent the regexp engine from identifying  
the backslash as a meta character. This is independent from the  
character _after_ the backslash.

First, you need to escape every backslash in a string so it is  
inserted literally (makes 2 backslashes). Then you need to escape a  
backslash because otherwise the regexp engine will interpret it (makes  
2\*2=4 backslashes). Finally you need an additional backslash to ensure  
the single quote is inserted into the string.

Btw, not only digits are meaningful in replacement patterns, but  
single quote is as well (I believe it’s the postmatch):

irb(main):012:0\> puts “abc”.gsub(/b/, ‘\'’)  
acc  
=\> nil  
irb(main):013:0\> puts “abc”.gsub(/b/, ‘\`’)  
aac  
=\> nil

Cheers

robert