In the org-mode the outline-regexp variable is set to "[*\f]+".
I am not able to figure out what it stands for, referred this without success.
Edit:- [*\f]+ changed to "[*\f]+"
The default value matches asterisks and page breaks:
"[*\f]+"
http://www.emacswiki.org/emacs/OutlineMode#toc2
\f stands for form feed, which is a special character used to instruct the printer to start a new page.
[*\f]+ Then means any sequence entirely composed of * and form feed, arbitrarily long.
C-hig (elisp) Regexp Special RET
Note also that the usual regexp special characters are not special
inside a character alternative. A completely different set of
characters is special inside character alternatives: `]', `-' and `^'.
So [*\f]+ matches any sequence which is at least one character long, and contains (only) any number and combination of asterisks, backslashes, and the letter 'f'.
EDIT:
Ah, you actually meant "[*\f]", did you? That's not the same thing as the regexp [*\f] (as the latter would be represented in string read syntax as "[*\\f]").
Make sure you quote appropriately.
If you did mean "[*\f]" then the \f is indeed a form-feed, as indicated by the other answers.
Related
I am struggling with a little problem concerning regular expressions.
I want to replace all odd length substrings of a specific character with another substring of the same length but with a different character.
All even sequences of the specified character should remain the same.
Simplified example: A string contains the letters a,b and y and all the odd length sequences of y's should be replaced by z's:
abyyyab -> abzzzab
Another possible example might be:
ycyayybybcyyyyycyybyyyyyyy
becomes
zczayybzbczzzzzcyybzzzzzzz
I have no problem matching all the sequences of odd length using a regular expression.
Unfortunately I have no idea how to incorporate the length information from these matches into the replacement string.
I know I have to use backreferences/capture groups somehow, but even after reading lots of documentation and Stack Overflow articles I still don't know how to pursue the issue correctly.
Concerning possible regex engines, I am working with mainly with Emacs or Vim.
In case I have overlooked an easier general solution without a complicated regular expression (e.g. a small and fixed series of simple search and replace commands), this would help too.
Here's how I'd do it in vim:
:s/\vy#<!y(yy)*y#!/\=repeat('z', len(submatch(0)))/g
Explanation:
The regex we're using is \vy#<!y(yy)*y#!. The \v at the beginning turns on the magic option, so we don't have to escape as much. Without it, we would have y\#<!y\(yy\)*y\#!.
The basic idea for this search, is that we're looking for a 'y' y followed by a run of pairs of 'y's,(yy)*. Then we add y#<! to guarantee there isn't a 'y' before our match, and add y\#! to guarantee there isn't a 'y' after our match.
Then we replace this using the eval register, i.e. \=. From :h sub-replace-\=:
*sub-replace-\=* *s/\=*
When the substitute string starts with "\=" the remainder is interpreted as an
expression.
The special meaning for characters as mentioned at |sub-replace-special| does
not apply except for "<CR>". A <NL> character is used as a line break, you
can get one with a double-quote string: "\n". Prepend a backslash to get a
real <NL> character (which will be a NUL in the file).
The "\=" notation can also be used inside the third argument {sub} of
|substitute()| function. In this case, the special meaning for characters as
mentioned at |sub-replace-special| does not apply at all. Especially, <CR> and
<NL> are interpreted not as a line break but as a carriage-return and a
new-line respectively.
When the result is a |List| then the items are joined with separating line
breaks. Thus each item becomes a line, except that they can contain line
breaks themselves.
The whole matched text can be accessed with "submatch(0)". The text matched
with the first pair of () with "submatch(1)". Likewise for further
sub-matches in ().
TL;DR, :s/foo/\=blah replaces foo with blah evaluated as vimscript code. So the code we're evaluating is repeat('z', len(submatch(0))) which simply makes on 'z' for each 'y' we've matched.
I was working on regex and studying the applications of character sets.
I tried the regex /[64-bit]/g, but the highlighted answer was contradictory; it highlighted uppercase letters, numbers and certain operators.
Why is that?
It's obvious that you're not using the right construct. Once you fix that, everything falls into place.
It doesn't make sense to use a character class if you want to match 64-bit literally. You should just use /64-bit/g as your regex in this case.
Character classes (specified by []) have different rules than the rest of the regex. They match a single character listed within (or not listed, if it's a negated char class).
A range of characters can also be specified to match, and that is where you have your problem. According to any online ASCII chart, 4 is #52 in the table, and b is #98. (Note that [4-bit] is actually an equivalent regex.) Between those two points, there are many characters, including the uppercase letters. That is why you are getting unexpected matches.
How to rewrite the [a-zA-Z0-9!$* \t\r\n] pattern to match hyphen along with the existing characters ?
The hyphen is usually a normal character in regular expressions. Only if it’s in a character class and between two other characters does it take a special meaning.
Thus:
[-] matches a hyphen.
[abc-] matches a, b, c or a hyphen.
[-abc] matches a, b, c or a hyphen.
[ab-d] matches a, b, c or d (only here the hyphen denotes a character range).
Escape the hyphen.
[a-zA-Z0-9!$* \t\r\n\-]
UPDATE:
Never mind this answer - you can add the hyphen to the group but you don't have to escape it. See Konrad Rudolph's answer instead which does a much better job of answering and explains why.
It’s less confusing to always use an escaped hyphen, so that it doesn't have to be positionally dependent. That’s a \- inside the bracketed character class.
But there’s something else to consider. Some of those enumerated characters should possibly be written differently. In some circumstances, they definitely should.
This comparison of regex flavors says that C♯ can use some of the simpler Unicode properties. If you’re dealing with Unicode, you should probably use the general category \p{L} for all possible letters, and maybe \p{Nd} for decimal numbers. Also, if you want to accomodate all that dash punctuation, not just HYPHEN-MINUS, you should use the \p{Pd} property. You might also want to write that sequence of whitespace characters simply as \s, assuming that’s not too general for you.
All together, that works out to apattern of [\p{L}\p{Nd}\p{Pd}!$*] to match any one character from that set.
I’d likely use that anyway, even if I didn’t plan on dealing with the full Unicode set, because it’s a good habit to get into, and because these things often grow beyond their original parameters. Now when you lift it to use in other code, it will still work correctly. If you hard‐code all the characters, it won’t.
[-a-z0-9]+,[a-z0-9-]+,[a-z-0-9]+ and also [a-z-0-9]+ all are same.The hyphen between two ranges considered as a symbol.And also [a-z0-9-+()]+ this regex allow hyphen.
use "\p{Pd}" without quotes to match any type of hyphen. The '-' character is just one type of hyphen which also happens to be a special character in Regex.
Is this what you are after?
MatchCollection matches = Regex.Matches(mystring, "-");
Trying to match ONLY the first character in the sample below.
Sample string: C/C++/Objective C/Objective-C/ObjectiveC/objectiveC
My faulty regex: (?![O|o]bjective[ |-]?)C(?!\+\+)
Doh.
Try this:
(?<![Oo]bjective[ -]?)C(?!\+\+)
Corrections are:
Use negative lookbehind instead of negative lookahead (the (?<!...) bit).
Removed pipe character from character classes (the [...] bits).
It might also be worth adding a pair of \bs either side of the C, since your current regex will match Coconut, BBC, CFML and so on
Also worth pointing out that, inside character classes, the - is special if not the first or last character. Some people prefer to escape it even in these situations, i.e. [ \-], in case a later character is accidentally added after it.
I need to match a colon (':') in a string, but not when it's enclosed by quotes - either a " or ' character.
So the following should have 2 matches
something:'firstValue':'secondValue'
something:"firstValue":'secondValue'
but this should only have 1 match
something:'no:match'
If the regular expression implementation supports look-around assertions, try this:
:(?:(?<=["']:)|(?=["']))
This will match any colon that is either preceeded or followed by a double or single quote. So that does only consider construct like you mentioned. something:firstValue would not be matched.
It would be better if you build a little parser that reads the input byte-by-byte and remembers when quotation is open.
Regular expressions are stateless. Tracking whether you are inside of quotes or not is state information. It is, therefore, impossible to handle this correctly using only a single regular expression. (Note that some "regular expression" implementations add extensions which may make this possible; I'm talking solely about "true" regular expressions here.)
Doing it with two regular expressions is possible, though, provided that you're willing to modify the original string or to work with a copy of it. In Perl:
$string =~ s/['"][^'"]*['"]//g;
my $match_count = $string =~ /:/g;
The first will find every sequence consisting of a quote, followed by any number of non-quote characters, and terminated by a second quote, and remove all such sequences from the string. This will eliminate any colons which are within quotes. (something:"firstValue":'secondValue' becomes something:: and something:'no:match' becomes something:)
The second does a simple count of the remaining colons, which will be those that weren't within quotes to start with.
Just counting the non-quoted colons doesn't seem like a particularly useful thing to do in most cases, though, so I suspect that your real goal is to split the string up into fields with colons as the field delimiter, in which case this regex-based solution is unsuitable, as it will destroy any data in quoted fields. In that case, you need to use a real parser (most CSV parsers allow you to specify the delimiter and would be ideal for this) or, in the worst case, walk through the string character-by-character and split it manually.
If you tell us the language you're using, I'm sure somebody could suggest a good parser library for that language.
Uppps ... missed the point. Forget the rest. It's quite hard to do this because regex is not good at counting balanced characters (but the .NET implementation for example has an extension that can do it, but it's a bit complicated).
You can use negated character groups to do this.
[^'"]:[^'"]
You can further wrap the quotes in non-capturing groups.
(?:[^'"]):(?:[^'"])
Or you can use assertion.
(?<!['"]):(?!['"])
I've come up with the following slightly worrying construction:
(?<=^('[^']*')*("[^"]*")*[^'"]*):
It uses a lookbehind assertion to make sure you match an even number of quotes from the beginning of the line to the current colon. It allows for embedding a single quote inside double quotes and vice versa. As in:
'a":b':c::"':" (matches at positions 6, 8 and 9)
EDIT
Gumbo is right, using * within a look behind assertion is not allowed.
You can try to catch the strings withing the quotes
/(?<q>'|")([\w ]+)(\k<q>)/m
First pattern defines the allowed quote types, second pattern takes all Word-Digits and spaces.
Very good on this solution is, it takes ONLY Strings where opening and closing quotes match.
Try it at regex101.com