I seem to remember that there were certain conditions under which $ would match a newline (not "just before"), but I can't find anything in the docs that would confirm that. Did that use to be true in an earlier version of Perl or am I dreaming?
views:
90answers:
4
+3
A:
If you pass the /s modifier to a regexp, then $ matches the end of the string and allows . to match \n.
This can be a bit confusing. For a string of foo\n, observe the difference between:
/.$/ # matches 'o'
/.$/s # still matches 'o'
/.+$/s # matches 'foo\n'
/.\z/ # doesn't match at all
/.\z/s # matches '\n'
szbalint
2010-08-12 15:01:23
Sorry... not relevant to my question
JoelFan
2010-08-12 15:02:08
Unless I'm misunderstanding something, you don't mean what you think you mean. `$` doesn't do any character matching, it does positional matching, so the question you might be asking is whether `$` matches before or after the newline.
szbalint
2010-08-12 15:05:29
I remember something about it consuming the newline itself
JoelFan
2010-08-12 15:17:19
@JoelFan => `$` is a zero width positional assertion, it never consumes any characters
Eric Strom
2010-08-12 15:48:33
It's `/m` that changes the meaning of the anchors (`^` and `$`); `/s` only changes the meaning of the dot. `$` *always* matches at the end of the string or before a newline at the end of the string. In `/m` mode it also matches before any *other* newline. At least, that's how it works in Perl; there could be Perl-derived flavors that don't work exactly like Perl in that regard, but I can't think of any offhand.
Alan Moore
2010-08-13 02:28:17
+5
A:
That has always been the case. See perldoc perlreref ($ is an anchor):
$Match string end (or line, if/mis used) or before newline
You should use \z if you want to match the end of the string:
\zMatch absolute string end
Sinan Ünür
2010-08-12 15:04:57
Yes. `"123\n" =~ /^\d+$/` is true. Which means all sorts of potential fun for your validation code if you get it wrong. :)
fennec
2010-08-12 15:58:29
+3
A:
You are confusing two very similar things. The $ matches at the end of a string if /m is not used, and at the end of a string or at the end of a string except for a final newline if /m is used. However, it does not actually match (consume) the newline itself. You could still match the newline character itself with something else in the regexp, for instance \n; or, if you also use /s, then /blah$./sm would theoretically match "blah\n" even though the $ is not the last thing in the regex.
(The reason this doesn't actually work is that both $. and $\ are actually interpreted as special variables instead of as $ plus ., so it is actually quite hard to put something after a $ in a regexp. Confusing, isn't it?)
Kilian Foth
2010-08-12 15:42:00
Not quite right. "at the end of a string or before a newline at the end of a string" if no /m, "at the end of a string or before any newline" if /m. Just "the end of the string" is \z.
ysth
2010-08-12 18:31:17
To use $ anywhere in a regex, without /m you can just say \Z (not \z). With /m, you can stick `$` in a `$dollar` variable and interpolate that (as `${dollar}` if needed). Or do e.g. `/@{['$']}./`
ysth
2010-08-12 18:37:59
+1
A:
Adding the 'm' modifier, '$' will match (but not consume) end of line characters. Normally '$' matches end of string.
my $sample = "first\nsecond\nthird";
$sample =~ s/$/END/mg;
print $sample;
Produces:
firstEND
secondEND
thirdEND
If you want to consume the newline, just use the \n escape character:
my $sample = "first\nsecond\nthird";
$sample =~ s/\n/END/g;
print $sample;
Now you get:
firstENDsecondENDthird
Note that the 'm' modifier also affects '^' in the same way as '$'.
Jeff
2010-08-12 15:47:03