Just continuing the performance question.
I tried to write a simple wc in ragel.
The main part looks like this:
--
%%{
machine wc;
word = (^space)+ > { nwords++; };
line = (space* . (word . space+)* word?) &
(^'\n')* > { nlines++; };
main := (line . '\n')** (line)? ;
}%%
--
(NOTE: I think this is still not totally correct. How do I handle a
line without '\n' at the end of the file correctly?)
Then I coded it in C:
--
switch( *B.cur )
{
case '\n':
n_lines++;
case ' ':
case '\t':
in_word = false ;
break ;
default:
if( !in_word )
{
in_word = true ;
nwords++ ;
}
}
--
The ragel parser, compiled with g++ -O3, runs at ~35MB/s on a big
file, and the hand-crafted parser at least at 130MB/s.
Did I do some errors in the ragel coding? Could I have done it more efficiently?
Or, will hand coding always be a fairly bit faster than ragel generated code?
Thanks!
Michael
On 3 November 2012 23:34, Michael Lachmann <lachm...@eva.mpg.de> wrote:
> 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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