[Haskell-cafe] Re: Has anybody replicated =~ s/../../ or even
something more basic
for doing replacements with pcre haskell regexen?
haskell at list.mightyreason.com
Mon Mar 16 08:50:29 EDT 2009
Thomas Hartman wrote:
> testPcre = ( subRegex (mkRegex "(?<!\n)\n(?!\n)") "asdf\n \n\n\nadsf"
> "" ) == "asdf \n\n\nadsf"
quoting from the man page for regcomp:
> REG_NEWLINE Compile for newline-sensitive matching. By default, newline is a completely ordinary character with
> no special meaning in either REs or strings. With this flag, `[^' bracket expressions and `.' never
> match newline, a `^' anchor matches the null string after any newline in the string in addition to
> its normal function, and the `$' anchor matches the null string before any newline in the string in
> addition to its normal function.
This is the carried over to Text.Regex with
> mkRegexWithOpts Source
> :: String The regular expression to compile
> -> Bool True <=> '^' and '$' match the beginning and end of individual lines respectively, and '.' does not match the newline character.
> -> Bool True <=> matching is case-sensitive
> -> Regex Returns: the compiled regular expression
> Makes a regular expression, where the multi-line and case-sensitive options can be changed from the default settings.
Or with regex-posix directly the flag is "compNewline":
> The defaultCompOpt is (compExtended .|. compNewline).
You want to match a \n that is not next to any other \n.
So you want to turn off REG_NEWLINE.
> import Text.Regex.Compat
> r :: Regex
> r = mkRegexWithOpts "(^|[^\n])\n($|[^\n])" False True -- False is important here
The ^ and $ take care of matching a lone newline at the start or end of the
whole text. In the middle of the text the pattern is equivalent to [^\n]\n[^\n].
When substituting you can use the \1 and \2 captures to restore the matched
non-newline character if one was present.
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