In general, if you're careful to not introduce unwanted non-deterministic
constructions, then performance comes down to the code style that you choose
and what's in your actions
The scanner construction is super-regular and can indroduce an extra cost
beyond a single pass because it backtracks. In the best case, the backtracking
introduces 0 additional cost, yeilding O(n) running time overall. At worst you
can have O(n^2) running time.
-----Original Message-----
From: Michael Lachmann <lachm...@eva.mpg.de>
Sender: ragel-users-boun...@complang.org
Date: Sat, 3 Nov 2012 23:34:30
To: <ragel-users@complang.org>
Reply-To: ragel-users@complang.org
Subject: [ragel-users] Hints about ragel for performance
Hi,
I'm starting to learn ragel because I'd like to write a very fast
parser to a fairly simple file structure.
I'd like to learn some of the tricks of increasing the performance of
the resulting program. So,
here are a few questions:
1. Is there a good sample program in terms of performance? I
downloaded awkemu - is that a good example?
2. Often, one can use **, or one can find a terminating character.
For example, awkemu has:
line = ( blineElements** '\n' )
I think here just * would have been enough, because there is the
terminating \n - is that right? Does it matter?
Should ** be avoided if possible?
3. Is there a disadvantage of using the lex-like scanner with
\*
pat =>
pat =>
etc., vs just specifying the full machine?
4. Is there a disadvantage of using intersection? For example, I think
the above line handling can written as:
line = something & [^\n]* '\n'
where something doesn't care about handling end-of-line. Is it just as
fast as writing expressions that also handle end-of line?
5. awkemu uses the following:
--
/* Find the last newline by searching backwards. This is where
* we will stop processing on this iteration. */
p = buf;
pe = buf + have + len - 1;
while ( *pe != '\n' && pe >= buf )
pe--;
pe += 1;
/* fprintf( stderr, "running on: %i\n", pe - p ); */
%% write exec;
/* How much is still in the buffer. */
have = data + len - pe;
if ( have > 0 )
memmove( buf, pe, have );
--
Is the first running backward to find the last eol necessary? It seems
to run part of the file through two parsers.
Thanks!
Michael
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