I think what's going on is that since @_ are aliases, this actually clears out the argument as well.
That would be the behaviour you'd get if local simply backed up the current value of a variable and restored it on scope exit.
As you discovered, that's not what local does. Instead, it creates a new variable, aliases the name to the new variable, and aliases the name back to the original variable on scope exit.
$ perl -e' use feature qw( say ); say 0+\$x; { local $x; say 0+\$x; } say 0+\$x; ' 94031376424552 94031376229560 94031376424552
That means the fact that $_[0] is an alias for the localized variable shouldn't matter. And it normally doesn't.
$ perl -e' use feature qw( say ); $x = 2; say 0+\$x; say $x; sub { local $x = 3; say 0+\$x; say $x; say $_[0]; }->( $x ); say 0+\$x; say $x; ' 94300947464040 Address of pre-local $x 2 Value of pre-local $x 94300947268840 Address of post-local $x 3 Value of post-local $x 2 Value of pre-local $x 94300947464040 $x is restored 2 Value of pre-local $x
But $! is magical. And more specifically, it's a proxy for a C variable. local copies the magic onto the new scalar it creates. So you end up with two variables (the $! passed as argument and the new $! created by local) which both use the same C variable for storage.
$ perl -e' use feature qw( say ); $! = 2; say 0+\$!; say 0+$!; sub { local $! = 3; say 0+\$!; say 0+$!; say 0+$_[0]; }->( $! ); say 0+\$!; say 0+$!; ' 93839899443048 Address of pre-local $! 2 Value of pre-local $! (aka value of errno) 93839899247848 Address of post-local $! 3 Value of post-local $! (aka value of errno) 3 Value of pre-local $! (aka value of errno) 93839899443048 $! is restored 3 Value of pre-local $! (aka value of errno)
The net effect is that local $! doesn't really localize $! at all!
$? is similarly a proxy for a C variable. $@ isn't magical.
In reply to Re^2: Carp: $! vs "$!" in croaking (updated)
by ikegami
in thread Carp: $! vs "$!" in croaking
by cavac
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