2011/7/7 Lasse Reichstein <[email protected]>:
>
>
> On Thu, Jul 7, 2011 at 3:52 AM, Mike Samuel <[email protected]> wrote:
>>
>> Yes, by the extension, and whether a \<octal> is a backreference or an
>> octal escape sequence is determined by whether there are
>> parseInt(<octal>, 10) capturing groups to the left of it in the
>> regular expression.
>> So
>> /\1(foo)\1/
>> matches the same language as
>> /\u0001(foo)\1/
>
>
> I don't think thats correct.
> The \1 is a valid DecimalEscape, its value is 1, which is not greater than
> NCapturingParens in 15.10.2.9 step 7 (NCapturingParens is defined globally
> for the pattern, not just to the left of the current escape). I.e., it is
> not a Syntax Error, so the \1 must be treated as a back-reference. It will
> always be to a non-participating capture, so the regexp is equivalent to
> /(foo)\1/
> or just
> /(foo)foo/
> but never to
> /\u0001(foo)foo/
> Regards
> /Lasse
I was wrong. You're right about the spec language of course and empirically,
/\1(foo)\1/.test("\u0001foofoo")
is true, but what I should have been testing is
/^\1(foo)\1$/.test("\u0001foofoo")
which is false on all the interpreters I have installed.
I think the first test spuriously matches because group 1 is
initialized empty at the point that it matches the first \1.
One way to tell whether the group initialized to empty works on an
interpreter is to test
/^(?:\1x(y)x){2}$/.test("xyxyxyx")
which is true in most interpreters, but false in Rhino1.7 and Chrome12.
Interestingly other perl 5 interpreters
perl -e '$s = "xyxyxyx"; $m = scalar($s =~ /^(?:\1x(y)x){2}$/) ?
"true" : "false"; print "$m\n"'
yields false, as does the java
public class Foo {
public static void main(String... argv) {
System.out.println(java.util.regex.Pattern.compile(
"^(?:\\1x(y)x){2}\\z").matcher("xyxyxyx").matches());
}
}
The python
import re
re.match(r"^(?:\1x(y)x){2}$", "xyxyxyx")
fails with
sre_constants.error: bogus escape: '\\1'
but not if the \1 is after the capturing group.
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