Regex /o modifier: what bugs?

[kcott]

G'day All,

I've just read in "perlre: Other Modifiers":

o  - pretend to optimize your code, but actually introduce bugs

It doesn't say anything further about these "bugs". There is a link to "s/PATTERN/REPLACEMENT/msixpodualngcer in perlop" which I thought might have some more information. Alas, no; all it says about /o is:

"If you want the pattern compiled only once the first time the variable is interpolated, use the /o option."

I did have a more extensive look around this area of perlop: no /o bug info found.

So, if anyone can tell me about these bugs, that would be appreciated.

I am aware that when a regex is compiled with /o, it is fixed; that is, the regex /$var/o won't change even if $var does. That's been the case since (at least) Perl3 — I wonder if that's the "bugs" to which the doco refers.

Not really part of the question, more background really, but my "/o bug" query came about as I was doing some benchmarking (born of idle curiosity). Code and results in the spoiler for those interested.

<Reveal this spoiler or all in this thread>

[ikegami]

To my knowledge, o does what it says it does. I believe the author was hyperbolically stating that using o results in surprises, hard to test code, etc. This doesn't really belong in the doc, if so.

[Marshall]

For a regex intensive project many moons ago, I was running the same regex in loop against a big array of entries. I thought that this was the ideal place for the /o option, but I found that it made no performance difference at all. I remember reading somewhere that in some situations, Perl is smart enough to do /o automatically without being asked to do so. That appeared to be the case in my test. Its been so long ago, that I don't remember the exact regex being tested - could be that there wasn't a $var in it to begin with - don't remember.

[ikegami]

Patterns that don't interpolate are always compiled at compile time. /o is useless for these.

$ perl -Mre=debug -c -e'/a/'
Compiling REx "a"
Final program:
   1: EXACT <a> (3)
   3: END (0)
anchored "a" at 0..0 (checking anchored isall) minlen 1
-e syntax OK
Freeing REx: "a"

Each instance of a regex operator (m//, s///, qr//) that interpolates caches the last pattern it compiled in string and compiled form. If that instance is evaluated again, and if the generated pattern is the same as the previous one, recompilation is skipped.

$ perl -Mre=debug -e'$x = "a"; for my $y (qw( b b a a )) { /$x/; /$y/;
+ }' 2>&1 | grep -i comp
Compiling REx "a"
Compiling REx "b"
Compiling REx "a"
Skipping recompilation of unchanged REx "a"       /$x/ same as last
Compiling REx "b"
Skipping recompilation of unchanged REx "b"       /$y/ same as last
Compiling REx "a"
Skipping recompilation of unchanged REx "a"       /$x/ same as last
Compiling REx "a"
Compiling REx "a"
Skipping recompilation of unchanged REx "a"       /$x/ same as last
Compiling REx "a"
Skipping recompilation of unchanged REx "a"       /$y/ same as last

It still needs to perform the interpolation, and it needs to perform a string comparison, so it won't be as exactly as fast as using /o. But it should be close.

[LanX]

> I remember reading somewhere that in some situations, Perl is smart enough to do /o automatically without being asked to do so.

I remember reading that this depends on the Perl version. And I think I read it here.

At some point - long ago - Perl became capable to notice that the input pattern didn't change and reused the last compilation, IIRC.

We should have extensive discussions on this in the archives.

[kcott]

G'day ikegami,

++ Thanks for the very quick response. That's good to know.

[Anonymous Monk]

Based on the git history, someone just wanted to doc the other existing flags and their comment seems to be tongue in cheek, in that perl is already smart enough to do what it does, and you will likely just make your life harder forgetting that its on. As it says "The bottom line is that using /o is almost never a good idea."