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Message-ID: <af54caaf0709231228m6d527d52q797c168cdc0a2242@mail.gmail.com>
Date: Sun, 23 Sep 2007 20:28:14 +0100
To: "Stanislav Malyshev" <stas@zend.com>
Cc: internals@lists.php.net
In-Reply-To: <46F6B7BE.4020205@zend.com>
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References: <af54caaf0709230846o762fb56doc34c324d5ac4942a@mail.gmail.com>
	 <46F6B7BE.4020205@zend.com>
Subject: Re: [PHP-DEV] Compiled variables and backpatching
From: paul.biggar@gmail.com ("Paul Biggar")

On 9/23/07, Stanislav Malyshev <stas@zend.com> wrote:
> > What is a compiled variable?
> > How does it differ from a non-compiled-variable?
>
> When engine encounters variable like $a what it usually does is it looks
> up current symbol table for an entry named 'a', and then takes zval*
> sitting there. However, since during the same scope the entry for $a
> will be the same (though not a value sitting there!), once found in can
> be stored so future calls to $a will not require looking up in the hash
> table. That's what CVs do. The reason, of course, is performance :)

OK, so its a bit of caching. So it sounds like it stores a pointer
into the symbol table, so that if $a is redefined it will still be
correct. What happens if the symbol table must be rehashed? Does it
still work, or does it look it up again?

Is there a single CV for each $a in a scope, or is there one per
bytecode operand? In other words, in a function which uses $a a lot,
does it look it up for each bytecode which uses $a, or only once for
that function?


Thanks for the quick response,
Paul


-- 
Paul Biggar
paul.biggar@gmail.com