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Re^3: closure clarity, please

by vitoco (Friar)
on Nov 24, 2009 at 19:28 UTC ( #809175=note: print w/replies, xml ) Need Help??

in reply to Re^2: closure clarity, please
in thread closure clarity, please

Well, without reading the explanation for the warning, I would expect ()(BBB)() or (AAA)(BBB)(AAA) or just ()()() like the case of no parameters.

It seems that sub f (the second one) compiles in runtime, and internal $a is not asigned the first time it is called, but when called from g, it glues the first value it receives.

Let me show an improved example:

#!perl use v5.10; use strict; use warnings; my $a = shift; sub f { say "global f ($a)"; } f(); sub g { my ($a) = @_; say "global g ($a)"; sub f { say "local f ($a)"; # line 17 } f(); } f(); g(shift); say "main1 $a"; $a = shift; say "main2 $a"; f(); g(shift); f(); __END__ C:\Temp> AAA BBB CCC DDD Variable "$a" will not stay shared at C:\Temp\ line 17. Subroutine f redefined at C:\Temp\ line 16. Use of uninitialized value $a in concatenation (.) or string at C:\Tem +p\ line 17. local f () Use of uninitialized value $a in concatenation (.) or string at C:\Tem +p\ line 17. local f () global g (BBB) local f (BBB) main1 AAA main2 CCC local f (BBB) global g (DDD) local f (BBB) local f (BBB)

As you can see, the internal variable is empty (uninitialized) the first two times f is called, but as soon as it is defined, it keeps that value forever. Weird...

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Re^4: closure clarity, please
by JadeNB (Chaplain) on Nov 25, 2009 at 03:01 UTC

    Thanks for the clarification. Indeed, your example shows that my statement about the function being compiled at run time is completely wrong!

    Now that I understand it, I think your example offers a great insight into scoping issues, and the difference between compile- and run-time. Let's trace through what happens (or at least what seems to me to be happening).

    1. The original definition of f is compiled.
    2. The definition of g is compiled.

      When we read the my ( $a ) = @_ declaration, we set aside space for $a, but do not yet assign anything to it. This pre-allocated space will be used only in our first call to g.

      Next, we compile the re-definition of f, triggering both the “Subroutine redefined” and the “Variable "$a" will not stay shared” warnings. At this point, we're still in the compilation phase.

      Notice that f is closing over the pre-allocated copy of $a—that is, the one that will be used in the first call to g. This is the weirdness that you're observing.

      To see that all this is happening at compile time, you can add the code block

      BEGIN { $a = "Hi from compile time" }
      before the re-definition of f and observe the changes at run time.

    3. We finish compilation, and assign 'AAA' to $a on line 6. However, this is the ‘outer’ $a, not the one * in the scope of g, so it has no effect on the now-re-defined f.
    4. We run the call to f on line 11. Since f has already been re-defined, we call the re-defined subroutine. Since $a still hasn't been initialised, we get the “Use of uninitialised value” warning. (I think ikegami might prefer “Use of unitialised variable”. :-) )
    5. Nothing happens at run time as we pass over the definition of g—that was dealt with already at compile time.
    6. We call f again on line 22. Since nothing has happened since the last call to f, we get the same result.
    7. We call g with argument 'BBB'. This 'BBB' gets assigned to the pre-allocated $a, which is the one that f closed over, so f ‘sees’ this new value. The re-definition of f is ignored at run time; that happened at compile time.
    8. We say the existing value of the ‘outer’ $a (it's 'AAA', because we assigned it back on line 6), assign a new value, 'CCC', to the ‘outer’ $a, and say that new value. None of this affects the variable $a that f closed over.
    9. We call f again on line 27. Since the $a that f closed over, as opposed to the ‘outer’ one to which we just assigned, has and forever will have the value 'BBB', that's the value that f uses.
    10. We call g again. As g runs, it allocates a new lexical variable $a that has nothing to do with the one that f closed over. (This behaviour is precisely of what the warning about variables not staying shared is—well—warning you.) As a result, the assignment to this new $a has no effect on f. We again skip over the re-definition of f at run time, since we already took care of it at compile time.
    11. We call f again on line 29. Since nothing has changed with its squirreled-away $a, the result is the same as when we called it on line 27.

    * I should be careful referring to “the one in the scope of g”, since, as we shall see, each run of g gets its own fresh lexical $a. However, “one of the ones in the scope of g” is a mess to read, so the imprecision stays.

      Impressive! I could follow you step by step. After a second read (starting from the last step), I realized how the interpreter seems to work.

      I think that I could simplify your work by excluding the "outer f" subroutine from the improved example, but as it is, is a great pedagogical example and explanation.

      Thank you very much! And more thanks goes to ikegami.

      BTW, this is not the way I program. I just wanted to understand what the OP was trying to explain. Everyday I learn something new of Perl's world. I'm not sure if I'm ready for Perl 6.

        Here's another way of demonstrating the effect in your sample code, with hopefully a bit less “magic gluing”:
        sub g { my ( $a ) = @_; sub f { $a }; return sub { $a = $_[0] }; } push my @setters, g("Hi\n"), g("Bye\n"); print f; # => Hi $setters[0]("Bye\n"); print f; # => Bye $setters[1]("Hi\n"); print f; # => Bye (still)

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