Antoine,
Thank you very much for your reply. Adding type sigs did help me think
about it. I got it to work.
I replaced:
eol = char '\n'
textLines = endBy eol
with:
textLine :: Parser String
textLine = do
x - many (noneOf \n)
char '\n'
return x
textLines :: Parser [String]
On Wednesday 15 September 2010 23:01:34, Peter Schmitz wrote:
textLine :: Parser String
textLine = do
x - many (noneOf \n)
char '\n'
return x
textLines :: Parser [String]
textLines = many textLine
And it can probably be coded more succinctly that that (suggestions
Daniel,
Thanks much; the more I learn Haskell and Parsec, the more I like them.
-- Peter
On Wed, Sep 15, 2010 at 4:02 PM, Daniel Fischer
daniel.is.fisc...@web.de wrote:
On Wednesday 15 September 2010 23:01:34, Peter Schmitz wrote:
textLine :: Parser String
textLine = do
x - many
Simple Parsec example, question
I am learning Parsec and have been studying some great reference and
tutorial sites I have found (much thanks to the authors), including:
http://legacy.cs.uu.nl/daan/download/parsec/parsec.html#UserGuide
Hi Peter,
On Tue, Sep 14, 2010 at 8:23 PM, Peter Schmitz ps.hask...@gmail.com wrote:
Simple Parsec example, question
I am learning Parsec and have been studying some great reference and
tutorial sites I have found (much thanks to the authors), including:
