Oh, wait, I posted the wrong function. Here's the one I meant:
(defn pr-pattern [pat]
(pr (re-pattern (.replaceAll (re-matcher (re-pattern "\n")
(.toString pat))
"\\\\n"))))
On 1 April 2013 10:00, Michał Marczyk <[email protected]> wrote:
> On 1 April 2013 07:53, Mark Engelberg <[email protected]> wrote:
>> Yeah, my goal is simply to get (re-pattern #"a\nb") to print (or more
>> precisely, pr) as #"a\nb" without affecting the semantics of printing other
>> regular expressions, but that seems to be impossible to achieve. Sigh...
>
> Could you just preprocess the strings passed to re-pattern (or
> patterns if you're getting those as input) to replace literal newlines
> with escape sequences? I'm assuming you don't care about ?x given the
> result you wish to achieve.
>
> Otherwise, I haven't done much testing on this, but perhaps you could
> see whether it would work for you:
>
> (defn pr-pattern [pat]
> (print (str "#\""
> (.replaceAll (re-matcher (re-pattern "\n")
> (.toString pat))
> "\\\\n")
> "\"")))
>
> user=> (pr-pattern (re-pattern "a\nb"))
> #"a\nb"nil
> user=> (pr-pattern #"a\nb")
> #"a\nb"nil
>
> (The nils are of course the return values. There are indeed four
> backslashes in the replacement string literal, since
> Matcher.replaceAll has its own escaping for $ (captured subsequence
> reference when not escaped) and \.)
>
> Cheers,
> Michał
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