Re: ${Schwartzian transform} ?

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.

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Comment on Re: ${Schwartzian transform} ?

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Re: ${Schwartzian transform} ? by Abigail-II (Bishop) on Sep 18, 2003 at 12:14 UTC
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` [download] 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` [download] @b isn't an array whose elements are aliases for the elements of @a. Abigail	[reply] [d/l] [select]

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Re: ${Schwartzian transform} ?
by Abigail-II (Bishop) on Sep 18, 2003 at 12:14 UTC

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

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
[download]

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

Abigail

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