Hi
There is no backtracking in JavaCC, just
lookahead.
One solution for your problem is to use semantic
lookahead.
Here under is an example.
Regards
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options
{
STATIC = false;
DEBUG_LOOKAHEAD = true;
}
PARSER_BEGIN(Test)
package foo.bt1;
import java.io.*;
import java.util.*;
public class
Test
{
public static void main(String args []) throws Exception
{
Reader r = new StringReader("ABAAAB");
Test interpreter =
new Test(new TestTokenManager(new
SimpleCharStream(r)));
interpreter.Entry();
}
}
PARSER_END(Test)
TOKEN :
{
< A : "A"
>
| < B : "B"
>
}
void AorBnotBandEOF() :
{}
{
(
{
// kind : < EOF > -> 0, < A > -> 1, < B >
-> 2 : see TestConstants
System.out.println("t1k : "+
getToken(1).kind);
System.out.println("t2k : "+
getToken(2).kind);
System.out.println("t3k : "+
getToken(3).kind);
}
// semantic lookahead : check that next tokens are not < B >
and < EOF >
LOOKAHEAD({ getToken(1).kind != TestConstants.B ||
getToken(2).kind != TestConstants.EOF
})
(
< A >
| < B >
)
)+
}
void Entry() :
{}
{
AorBnotBandEOF()
< B > < EOF >
}
In JavaCC one often uses semantic lookahead instead of
backtracking. Not sure if it will help with what you have or
not..
On Wed, Jul 8, 2009 at 8:38 AM, David Portabella Clotet
<david@...>
wrote:
Hello,
One quick question;
Does JavaCC implement backtracking?
According to the JavaCC FAQ
website, lookahead does not implement backtracking.
so I cannot parse this input "ABAAAB" with the following program
(see attached the full code):
void Entry() : {}
{
( LOOKAHEAD(2) (<A> | <B>)
)+
<B>
}
note: I understand that I could rewrite this program to make it work
without backtracking;
however this is a simplified program illustrating the problem that I
have with a more complex one which does need backtracking.
In the regular expression world, backtracking is so simple: this
regex would do the job "[AB]+B".
Is there a way to tell JavaCC to consider backtracking?
Many thanks,
DAvid Portabella
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- J.Chris Findlay
(c:
Test.jj (604 bytes) 