# No regex backreference with four backslashes

**URL:** <https://www.ruby-forum.com/t/no-regex-backreference-with-four-backslashes/68864>\
**Category:** Ruby\
**Created:** [September 16, 2006, 12:25am UTC](https://www.ruby-forum.com/t/no-regex-backreference-with-four-backslashes/68864 "2006-09-16T00:25:31Z")\
**Posts on this page:** 8\
**Page:** 1

<div class="post-metadata">

**Author:** ![unknown](https://www.ruby-forum.com/letter_avatar_proxy/v4/letter/u/4bbf92/32.png) [@unknown](https://www.ruby-forum.com/u/unknown)\
**Post date:** [September 16, 2006, 12:25am UTC](https://www.ruby-forum.com/t/no-regex-backreference-with-four-backslashes/68864/1 "2006-09-16T00:25:31Z")

</div>

Consider the following test case:

require ‘test/unit’  
class RegexTest \< Test::Unit::TestCase  
def test\_escaping  
numbers = “12345”  
assert\_equal “12345”, numbers.gsub(/(2|4)/, ‘\1’)  
assert\_equal “12345”, numbers.gsub(/(2|4)/, “\1”)  
assert\_equal “1\ 23\ 45”, numbers.gsub(/(2|4)/, ‘\ \1’)  
assert\_equal “1\ 23\ 45”, numbers.gsub(/(2|4)/, “\ \1”)  
assert\_equal “1\23\45”, numbers.gsub(/(2|4)/, ‘\\1’)  
assert\_equal “1\23\45”, numbers.gsub(/(2|4)/, “\\1”)  
end  
end  
require ‘test/unit/ui/console/testrunner’  
Test::Unit::UI::Console::TestRunner.run(RegexTest)

The last two assertions fail (With the message \<“1\23\45”\> expected  
but was \<“1\13\15”\>.) - but why?

Is this a bug in the regex implementation or is there something wrong  
with my regular expression or substitution string?

---

<div class="post-metadata">

**Author:** ![unknown](https://www.ruby-forum.com/letter_avatar_proxy/v4/letter/u/4bbf92/32.png) [@unknown](https://www.ruby-forum.com/u/unknown)\
**Post date:** [September 16, 2006, 12:45am UTC](https://www.ruby-forum.com/t/no-regex-backreference-with-four-backslashes/68864/2 "2006-09-16T00:45:33Z")

</div>

removed\_email\_address@domain.invalid wrote:

> ```
> assert_equal "1\\23\\45", numbers.gsub(/(2|4)/, '\\\1')
> 
> ```
> 
> with my regular expression or substitution string?  
> To find out how your strings are being parsed, print them out. Then  
> print  
> out the result of the regexes directly, rather than relying on an  
> assertion.

numbers.gsub(/(2|4)/,"\\1")

“1\\0013\\0015”

numbers.gsub(/(2|4)/,"\\1")

“1\13\15”

The best “test suite” is your eyes.

---

<div class="post-metadata">

**Author:** ![unknown](https://www.ruby-forum.com/letter_avatar_proxy/v4/letter/u/4bbf92/32.png) [@unknown](https://www.ruby-forum.com/u/unknown)\
**Post date:** [September 16, 2006, 1:01am UTC](https://www.ruby-forum.com/t/no-regex-backreference-with-four-backslashes/68864/3 "2006-09-16T01:01:34Z")

</div>

Paul L. schrieb:

> “1\13\15”
> 
> The best “test suite” is your eyes.

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

---

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**Author:** ![unknown](https://www.ruby-forum.com/letter_avatar_proxy/v4/letter/u/4bbf92/32.png) [@unknown](https://www.ruby-forum.com/u/unknown)\
**Post date:** [September 16, 2006, 1:21am UTC](https://www.ruby-forum.com/t/no-regex-backreference-with-four-backslashes/68864/4 "2006-09-16T01:21:35Z")

</div>

removed\_email\_address@domain.invalid wrote:

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

In the replacement string, a backreference is a backslash followed by a  
number – reference(\1) – but a double-backslash is treated as a  
literal single backslash, so \1 == literal(\1). So then, three  
backslashes and a number, \\1 is equal to literal() reference(\1).  
Four means literal(\1). Finally, five means literal(\) reference(\1),  
and thus, since backslashes must be escaped to be seen as a single  
backslash in a string, you end up with the resulting string  
“1\23\45”, meaning 1\23\45. Hope that makes sense.

Regards,  
Jordan

---

<div class="post-metadata">

**Author:** ![unknown](https://www.ruby-forum.com/letter_avatar_proxy/v4/letter/u/4bbf92/32.png) [@unknown](https://www.ruby-forum.com/u/unknown)\
**Post date:** [September 16, 2006, 1:21am UTC](https://www.ruby-forum.com/t/no-regex-backreference-with-four-backslashes/68864/5 "2006-09-16T01:21:35Z")

</div>

removed\_email\_address@domain.invalid wrote:

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

/ …

> 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.

---

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**Author:** ![unknown](https://www.ruby-forum.com/letter_avatar_proxy/v4/letter/u/4bbf92/32.png) [@unknown](https://www.ruby-forum.com/u/unknown)\
**Post date:** [September 16, 2006, 11:12am UTC](https://www.ruby-forum.com/t/no-regex-backreference-with-four-backslashes/68864/6 "2006-09-16T11:12:10Z")

</div>

Paul L. wrote:

> puts ‘\\\1’
> 
> \\1 # meaning two backslashes and an escaped ‘1’
> 
> Oh, by the way. You haven’t said what you are trying to accomplish.

I was trying to escape some characters in a string with a backslash.

When printing out ‘\\\1’ (resulting in two backslashes and and  
escaped ‘1’ like you said) I would expect the result string s  
(s=numbers.gsub(/(2|4)/, ‘\\\1’) to contain _two_ backslashes and  
then the original character. But apparently the replacement string is  
interpreted as “one backslash and a backreference (escaped with two  
backslashes).”

---

<div class="post-metadata">

**Author:** ![unknown](https://www.ruby-forum.com/letter_avatar_proxy/v4/letter/u/4bbf92/32.png) [@unknown](https://www.ruby-forum.com/u/unknown)\
**Post date:** [September 16, 2006, 11:36am UTC](https://www.ruby-forum.com/t/no-regex-backreference-with-four-backslashes/68864/7 "2006-09-16T11:36:11Z")

</div>

removed\_email\_address@domain.invalid wrote:

> But apparently the replacement string is  
> interpreted as “one backslash and a backreference (escaped with two  
> backslashes).”

After thinking a while about it I realized this is not correct.

Backslashes in a replacement string _must_ be double backslashes (four  
backslashes in the literal string) because otherwise they would be  
interpreted as escaped characters by the regex engine. Right?

---

<div class="post-metadata">

**Author:** ![unknown](https://www.ruby-forum.com/letter_avatar_proxy/v4/letter/u/4bbf92/32.png) [@unknown](https://www.ruby-forum.com/u/unknown)\
**Post date:** [September 16, 2006, 6:50pm UTC](https://www.ruby-forum.com/t/no-regex-backreference-with-four-backslashes/68864/8 "2006-09-16T18:50:50Z")

</div>

removed\_email\_address@domain.invalid wrote:

/ …

> After thinking a while about it I realized this is not correct.
> 
> Backslashes in a replacement string _must_ be double backslashes (four  
> backslashes in the literal string) because otherwise they would be  
> interpreted as escaped characters by the regex engine. Right?

Yes, if the backslashes are followed by anything other than another  
backslash (a backslash used as an escape must be followed by a target  
character other than another backslash). This is why it’s a good idea to  
print the string you intend to use. Printing the string forces it to be  
parsed, so you can see what you are getting into.