rir has asked for the wisdom of the Perl Monks concerning the following question:

In the below, I don't see why the x matches differently from the * ? 
#!/home/rir/rakudo/parrot_install/bin/perl6
use v6;

grammar Calc {
    rule TOP { <term> <op> <term>  }
#    rule TOP { <term>\s?<op>\s?<term>  }   # a fix
    token term { \d+(\.\d*)? }
    token op { '*' | 'x' }
}

my @t = (
    [ 44, " 8.8 * 5.0 " ],
    [ 44, " 8.8 x 5.0 " ],
    [ 44, " 8.8*5.0 " ],
    [ 44, " 8.8x5.0 " ],        # no match
);

for @t -> $i {
    my $m = Calc.parse( $i[1]);
    print $m ?? "  " !! "no";
    say " match: $i[1]"
}

[download]
Be well,
rir

Replies are listed 'Best First'.
Re: P6: parsing whitespace
by moritz (Cardinal) on Jun 09, 2010 at 18:01 UTC
    If you have whitespace in a rule, it is internally replaced by a call to the ws token.

    The default ws matches at least one space (\s+) if the left and right are word characters (ie matching \w), and matching zero or more (\s*) otherwise.

    In code: 

    class Grammar {
   token ws {
       <!ww> \s*
   }
}

    [download]

    Where ww is within word, and could be defined as 

    token ww {
    <after  \w>
    <before \w>
}

    [download]

    If you want 8.8x5.0 to match, you need to override the ws rule in your grammar: 

    grammar Calc {
    token ws { \s* }
    # rest of your code here without any modifications

    [download]

    (Update: fixed ww rule after comment from TimToady++)

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