On Sun, 20 Aug 2000, Tony Olekshy wrote:
> That would be nice. But does this mean that in the following
> case:
>
> try { fragile(); }
> catch { my $caught = 1; }
> finally { $caught and ... }
>
> storage for $caught is allocated and initialized to undef at the
> beginning of the *try* block, even though you're not allowed to
> see it there? Finally has to see it, as undef, even if the catch
> block isn't entered.
If all those pieces were in the same scope I think it would still work
like this (in Perl5-ish code):
{
try { fragile(); # It must be Italian }
my $caught;
catch { $caught = 1; }
finally { $caught and ... }
}
It's the same as in perl5 with a block:
{
print $foo;
my $foo = 1;
}
This is a compile time error because $foo isn't in scope til its declared.
Of course, what I know about the Perl internals fits on the head of a pin
so someone more knowledgeable should feel free to jump in.
-dave
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