Thanks. So, I have to dereference the variable I assigned the typeglob to in order to turn it back into a typeglob? For instance,

$val = *x;
print "@{ *{ $val } }";
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Based on the output here:

print $val;

--output:--
*main::x
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...it's not obvious that $val needs dereferencing.

But I did $var = *x, which is not the same:

$what = *x;
print $what, "\n";

$what = \*x;
print $what, "\n";

*main::x
GLOB(0x10082ad88)
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It seems like in the first case, I wouldn't have to dereference $what into a typeglob, where I would expect to do so in the second case.

Also, why can I do this:

$fh = *STDOUT;

print $fh 'hello', "\n";  

--output:--
hello
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...and not this:

$STDOUT = 10;
$fh = *STDOUT;

print ($fh + 2);

--output:--
2
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But I did $var = *x

So did I. I said \*x also works.

it's not obvious that $val needs dereferencing.

I showed that it doesn't need dereferencing.

$gv = *x; @$gv  =>  @x
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Of course, one needs to use the sigil to access the components of the glob.

$gv = *x;
$gv{ARRAY}    # Obviously wrong.
*$gv{ARRAY}   # Ok
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...and not this:

Do you really want print($count + 2) to use $count as a file handle?

print(STDOUT + 2)   =>  print( { STDOUT } 2 )
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Barewords will be handled as expected, but anything else followed by what could be an infix operator will be treated as an argument to that operator.

print(*STDOUT + 2)  =>  print( { select() } *STDOUT+2 )
print(*$x + 2)      =>  print( { select() } *$x+2 )
print($x + 2)       =>  print( { select() } $x+2 )
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I showed that it doesn't need dereferencing.

$gv = *x; @$gv  =>  @x
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perlreftut says that the syntax @$gv is a shortcut for @{$gv}, and that looks like the very definition of dereferencing to me (see perlreftut, Use Rule 1). As far as I can tell, the only time you don't have to dereference the variable is when you use it as a filehandle--in that case it appears that perl uses the context to determine that it should extract the filehandle part of the typeglob.