Paul Tremblay wrote at Mon, 15 Jul 2002 18:45:21 +0200:
> On Mon, Jul 15, 2002 at 10:26:25AM +0200, Janek Schleicher wrote:
>
>
>> To increase speed, we can make also a lookahead statement:
>>
>> my @tokens = split / ( \\ (?=\S) # there's never a whitespace
>> (?: [^\s{}]+ |
>> [^\s\\}]+ |
>> [\\}] )
>> | }
>> )/x
>> => $line;
>>
>>
> Can you explain the lookahead statement to me, or better yet, point me to some good
>documentation?
> It is not explained in *Perl Cookbook.*
The Camel Book and Mastering Regular Expressions.
You can say (?= ... ) in a regex,
to say that this ... has to come.
But the regex machine stays where it it.
E.g. /(?=\w)(?=\d)(?=[^39])/ matches still a "4".
A typical example is something like
my $keyword =~ /(for|while|do|nothing|die)/;
what is quite slow as alternations are always slow.
But you can tell perl not to go into the alternation
when there's a comma or so with
my $keyword =~ /(?=\w)(for|while|do|nothing|die)/;
as it will only go into the alternation when the next char is a word character,
what is a quick standard test - implemented directly in C.
In your case,
the next character can be a lot,
but it is definitly not a whitespace.
That's why I proposed to use (?=\S).
> As John pointed out, your solution leaves off the leading open bracket. However, it
>is about twice
> as quick as the previous solution. You solution tokenizes this line:
A tried something different,
look at the answer to John's response.
Perhaps (?!) that works :-)
Best Wishes,
Janek
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