On 01/02/2017 20:17, Rasmus Lerdorf wrote:
You probably think of $var as being a local scope variable here,
but really it isn't. It is an imported variable that happened to not exist
in the outer scope and it was assigned a value in the inner scope, but that
assignment can't leak back out so it is still unassigned in the outer
scope.
I'm very confused by your terminology here. You definitely talk about
two distinct scopes, "the inner scope" and "the outer scope", but then
you assert that there is no "local scope". Surely the "inner scope" is
"local" to the closure?
In actual fact, it's local to *each invocation* of the closure, because
the function is initialised with the original values each time. For
instance:
$a = 1;
$fn = fn() => ++$a;
$a = 42;
echo $fn(), ' ', $fn, ' ', $fn();
This echoes 2 2 2, not 2 3 4, because the $a being incremented is new
for each invocation. Nor does it echo 43 43 43, because it is only the
value of $a that is captured, not its reference.
I think the best description is this: the closure has its own scope,
which is initialised on each invocation with a copy of another scope
taken when it was defined.
To go back to Bruce's question, the original statement was:
you don't typically need a local scope at all for these short closures
The important point here is that within a single expression there is
very little you can meaningfully do with a local scope other than read
from it. It's generally a bad idea to alter the value of a variable and
expect another part of the same expression to see the new value, because
you can't always predict how the compiler will rewrite the expression (I
think an example of this came up on the list a while back, but can't
think how to search for it).
So for most purposes you can consider all the automatically imported
variables to be immutable values, which makes their scope much less
relevant.
Regards,
--
Rowan Collins
[IMSoP]
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