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Re: Substitution with regex and memory consumption

by dave_the_m (Monsignor)
on Feb 29, 2020 at 20:27 UTC ( #11113586=note: print w/replies, xml ) Need Help??

in reply to Substitution with regex and memory consumption

In general after a successful match (or the match part of a substitution), the regex engine keeps a copy of the original string so that it can dynamically generate values for $1, $2, $&, $` etc on demand. This string needs to be kept for at least as long as the surrounding scope - i.e. the scope of $1 etc. The details are far more complex, but internally perl's Copy-on-Write mechanism often (but not always) avoids having to do a real copy. But it doesn't always work out for the most efficient use of memory.


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Re^2: Substitution with regex and memory consumption
by k-mx (Scribe) on Mar 01, 2020 at 09:19 UTC

    Okay, thank you! Some assumptions, please correct me if i'm wrong:

    1. Prior 5.18.0, s/// will copy original only if one of these was set: $&, $`, $'. Interpreter will set global PL_sawampersand flag that can't be disabled later. m// is also affected by this flag.
    2. Between 5.18.0 and 5.20.0, Perl can track usage of mentioned variables separately and copy only requested part of string.
    3. Perl 5.20.0+, successful s/// match always changes string, so COW mechanism always had to copy original.

    So, before 5.20 we have choice: avoid $&, $`, $' and use /p modifier to explicitly copy ${^*MATCH}. Now we can use $&, $`, $', m// don't suffer from PL_sawampersand anymore, but s/// will always copy original string, PL_sawampersand state doesn't matter, and nothing we can do with that.

      That's roughly it, yes.


        Isn't this behavior incorrect? Why we must copy variable on each substitution?

        I think we can: respect PL_sawampersand and /p flag for substitution and have only one copy for string (last one) in current scope.

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