in reply to regexp causes segfault

Do it C style using pos and substr. This is laser fast and tested to 40MB 

# proof it behaves right, uncomment to see
# $_ = "  '==\\'==' '==5=='  '\\'\\'' '\\'3' '\\'' '1'  '' " x 2;

$n = 40000000;
$_ = "'" . "=" x $n . "'";

my @pos;
while ( /(?<!\\)'/gc ) {
    push @pos, pos;
}

for ( my $i= 0; $i <@pos; $i +=2 ) {
    my $begin = $pos[$i];
    my $end   = $pos[$i+1]-1;
    my $str   = substr $_, $begin, ($end -$begin);
    # check what we have found using test string commented out
    #print "$begin $end '$str'\n";
    print length($str), "\n";
}

[download]

cheers

tachyon

s&&rsenoyhcatreve&&&s&n.+t&"$'$`$\"$\&"&ee&&y&srve&&d&&print

Replies are listed 'Best First'.
Re: Re: regexp causes segfault
by Anonymous Monk on Mar 06, 2003 at 01:29 UTC
    tachyon,
    Should a beast of a regular expression like this be put into an EVAL block?
    Just wondering.


    Shirkdog
[reply]
Re: Re: regexp causes segfault
by shirkdog_perl (Beadle) on Mar 06, 2003 at 01:30 UTC
    That was me, forgot I was not logged in:-)
[reply]

      I have no idea what you mean. All this regex does is walk the string (char by char) with a 1 char buffer (the last char). When it gets a match (last char not \\ and char eq ') it gets a match and we record the position. This is hardly a regex at all!

      Here it is completely C-ified - no regexes in sight. Possibly faster than the original post to boot but I can't be bothered to test. 

      $str  = "  '==\\'==' '==5=='  '\\'\\'' '\\'3' '\\'' '1'  '' ";

my $pos = 0;
my $len = length $str;
my $last = '';
my $char;
while ( $pos < $len ) {
   $char = substr $str, $pos, 1;
   push @pos, $pos if $char eq "'" and $last ne "\\";
   $pos++;
   $last = $char;
}

for ( my $i= 0; $i <@pos; $i +=2 ) {
    my $begin = $pos[$i]+1;
    my $end   = $pos[$i+1];
    my $str   = substr $str, $begin, ($end -$begin);
    print "$begin $end |$str|\n";
}

      [download]

      cheers

      tachyon

      s&&rsenoyhcatreve&&&s&n.+t&"$'$`$\"$\&"&ee&&y&srve&&d&&print

[reply]
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