Of course, setting $#a=$b has the (side-)effect of expanding @a to a huge size, so these examples take a few seconds (or may run out of memory even):

% perl -le '$b=[1..4]; $#a=$b; print $#{$#a}'
3
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Can this actually be doing what I think it is?? Indeed, it looks as though $#a is actually the same reference as $b$:

% perl -le '$b=[1..4]; $#a=$b; print @{$#a}; push @$b, 5; print @{$#a}
+'
1234
12345
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In fact, this also works when $b is lexical, so it's not a symbolic reference.

You can put other types of references into $#a, and dereference them in any standard way:

% perl -le '$b=[1..4]; $#a=$b; print $#a->[2]'
3

% perl -le '$b=\"foo"; $#a=$b; print ${$#a}'
foo

% perl -le '$b={foo=>1}; $#a=$b; print keys %{$#a}'
foo
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However, you cannot preserve blessed-ness of a scalar:

% perl -le 'sub foo{"foo"} $b=bless[]; $#a=$b; print $#a->foo'
Can't call method "foo" without a package or object reference at -e li
+ne 1.
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This is our first clue that this is not a fully-fledged reference.. we also see that ref is oblivioius to it:

% perl -le '$b=[1..4]; $#a=$b; print ref $#a'
[nothing]
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Also, I can't copy $#a into a variable and then use that variable as a reference:

% perl -le '$b=[1..4]; $#a=$b; $x=$#a; print @$x'
[nothing]
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In fact, after I copy $#a into a variable, I can't even use $#a itself as a reference anymore! Hello Heisenberg!

% perl -le '$b=[1..4]; $#a=$b; $x=$#a; print @{$#a}' 
[nothing]
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Very odd.

#/usr/bin/perl -l
my@a=qw[just another perl hacker];$#a=\@a;
print "@{$#{$#{$#{$#{$#{$#{$#{$#a}}}}}}}}"
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