Re: shift implicit dereference

[LanX]

if you don't really need references use aliasing ¹!

  DB<122> $a
 => 5

  DB<123> sub alias_add { $_[0]+=2 }

  DB<124> alias_add $a
 => 7

  DB<125> $a
 => 7

But to fully answer the original question:

  DB<126> sub ref_add { ${ shift() }+=2 }

  DB<127> ref_add \$a
 => 9

  DB<128> $a
 => 9

  DB<129> sub ref_add2 { ${ +shift }+=2 }

  DB<130> ref_add \$a
 => 11

  DB<131> $a
 => 11

update

or using parens

  DB<102> $a=11
 => 11

  DB<103> sub ref_add3 { ${ (shift) }+=2 } 

  DB<104> ref_add3 \$a; $a
 => 13

1) from perlsub:

       Any arguments passed in show up in the array @_.  Therefore, if
+ you
       called a function with two arguments, those would be stored in 
+$_[0]
       and $_[1].  The array @_ is a local array, but its elements are
+ aliases
       for the actual scalar parameters.  In particular, if an element
+ $_[0]
       is updated, the corresponding argument is updated (or an error 
+occurs
       if it is not updatable).