26 Jul
2010
26 Jul
'10
2:44 p.m.
Hi,
I am playing with PetitParser (which looks super-cool!).
Following Lukas' blogpost [1], I defined my identifier as:
<snip>
identifier := #letter asParser , #word asParser star.
</snip>
then, I am using it to find the matches in a given string:
<snip>
identifier matchesIn: 'foo 123 bar12'.
</snip>
which returns:
<snip>
an OrderedCollection(
#($f #($o $o))
#($o #($o))
#($o #())
#($b #($a $r $1 $2))
#($a #($r $1 $2))
#($r #($1 $2))
)
</snip>
Even though the result is correct and makes much sense, would it
be possible to simply have the longest matches?
In that case I would expect something like:
<snip>
an OrderedCollection(
#($f #($o $o))
#($b #($a $r $1 $2))
)
</snip>
Thank you!
Cheers,
Alberto
[1] http://www.lukas-renggli.ch/blog/petitparser-1
26 Jul
26 Jul
4:24 p.m.
I am playing with PetitParser (which looks
super-cool!).
Following Lukas' blogpost [1], I defined my identifier as:
<snip>
identifier := #letter asParser , #word asParser star.
</snip>
then, I am using it to find the matches in a given string:
<snip>
identifier matchesIn: 'foo 123 bar12'.
</snip>
which returns:
<snip>
an OrderedCollection(
#($f #($o $o))
#($o #($o))
#($o #())
#($b #($a $r $1 $2))
#($a #($r $1 $2))
#($r #($1 $2))
)
</snip>
Even though the result is correct and makes much sense, would it
be possible to simply have the longest matches?
Sure.
As you can see in the implementation and in the comment of
#matchesIn:do: which is called by #matchesIn: the implementation tries
the parser at each input position. You can skip over matches by using
the following code instead:
PPParser>>matchesSkipIn: anObject do: aBlock
"Search anObject repeatedly for the matches of the receiver. Evaluate
aBlock for each match with the matched parse-tree as the argument.
Skip over the matched part."
(self ==> aBlock / #any asParser) star parse: anObject
Maybe that should be added? Maybe somebody has a better name :-)
Lukas
--
Lukas Renggli
www.lukas-renggli.ch
4:47 p.m.
Even though
the result is correct and makes much sense, would it
be possible to simply have the longest matches?
Sure.
As you can see in the implementation and in the comment of
#matchesIn:do: which is called by #matchesIn: the implementation tries
the parser at each input position. You can skip over matches by using
the following code instead: Maybe that should be added? Maybe somebody has a
better name :-)
What about matchesLongestIn: , or longestMatchesIn: ?
Alberto
5:48 p.m.
Another fast question :) :
since the "not" method parses p and succeeds when it fails, but does
not consume its input.
What is the best way to define something negated? For example to define:
"everything but a word or a space"
Thanks,
Alberto
6:05 p.m.
On 26 July 2010 17:48, Alberto Bacchelli <alberto.bacchelli(a)usi.ch> wrote:
Another fast question :) :
since the "not" method parses p and succeeds when it fails, but does
not consume its input.
What is the best way to define something negated? For example to define:
"everything but a word or a space"
If
wordOrSpace := #word asParser / #space asParser.
is given you can write the negation using the helper #negate like
wordOrSpace negate
which simply consumes one character if the receiver does not parse:
wordOrSpace not , #any asParser ==> #second
Lukas
--
Lukas Renggli
www.lukas-renggli.ch
5263
days inactive
5263
days old
4 comments
2 participants
participants (2)
-
Alberto Bacchelli -
Lukas Renggli