Re: ${Schwartzian transform} ?

[BrowserUk]

At a first glance, I think your creating extra work rather than saving any. $_ is an alias, so when you include it in the anon. array, you are only passing a reference to the original value (ie. a scalar).

Your code is actually creating an extra reference to the scalar, (ie. copying the alias?), which it then has to dereference, hence doing slightly more work.

I *think*.

---

Examine what is said, not who speaks.

"Efficiency is intelligent laziness." -David Dunham
"When I'm working on a problem, I never think about beauty. I think only how to solve the problem. But when I have finished, if the solution is not beautiful, I know it is wrong." -Richard Buckminster Fuller
If I understand your problem, I can solve it! Of course, the same can be said for you.

[Abigail-II]

While $_ in a map is an alias, it doesn't mean that $_ in rvalue context magically resolves to the alias instead of the value. Watch:

perl -wle '@a = 1; @b = map {++ $_ -> [0]} map {[$_]} @a; print "@a"; 
+print "@b"'
1
2

No alias passes out of the right-most map.

Or a shorter example:

perl -wle '@a = 1; @b = map {$_} @a; $a [0] ++; print "@a @b"'
2 1

@b isn't an array whose elements are aliases for the elements of @a.

Abigail