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Improve the scalaExpr parser #2

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domna opened this issue Aug 31, 2011 · 0 comments
Open

Improve the scalaExpr parser #2

domna opened this issue Aug 31, 2011 · 0 comments
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@domna
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domna commented Aug 31, 2011

The current scalaExpr Regex has the form [^%]* and is automatically converted into a parser which is used in the grammer to parse the file into blocks (whether it is a prolog or a scala part). But the problem with this Regex is, that it parses everything but the % character. This was a workaround for me that the Regex stops consuming everything in my file and recognizes the prolog block of the file.
The problem is that if there is any % character (even in strings) in the scala definiton above the prolog block the result is a parser error.

So there must be a way to stop the regex consuming everything and also have the possibility to write every character in the scala def above.

@ghost ghost assigned domna Aug 31, 2011
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