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Re^3: lexicals are all the same scalar and never go out of scope?

by ikegami (Pope)
on Dec 06, 2011 at 10:43 UTC ( #941995=note: print w/replies, xml ) Need Help??


in reply to Re^2: lexicals are all the same scalar and never go out of scope?
in thread lexicals are all the same scalar and never go out of scope?

So why does perl make copies instead of real references when you think your making a reference?

$s = $x . $y;

Wouldn't it be nice if concat didn't have to allocate a scalar that will just end up being copied into $s and deallocated?

Well guess what, it doesn't. That instance of the concat operator will always return the same scalar. This saves allocating a scalar and deallocating a scalar.

But of course, doing that alone would break weird code like following:

my @b = map { \($prefix.$_) } @a;

So the deref operator must be complicit and make a copy of the scalar if it comes from concat.

Why doesn't entersub clone PADTMPs into unique mortal SVs if PADTMPs are found on the stack automatically

Cause that defies the whole point of avoiding the creation of those scalars.

Shouldn't making a ref in XS be identical to making a ref in Perl?

Sure, but it's not Perl's fault your XS code isn't identical to its code.

Should there be a function for creating references that checks PADTMP? Maybe. Yes, p5p would be appropriate for this.

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