No, the expression is  10, 20 and it as a whole is evaluated in scalar context

Obvious mistakes (from Useless use of %s in void context )

$foo = 1, 2;

Useless use of a constant (2) in void context at -e line 1 (#1)

assignment(=) binds tighter than comma(,) so $foo is 1

$ perl -MO=Deparse,-p -e " $foo = 1, 2; "
(($foo = 1), '???');
-e syntax OK
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$one, $two = 1, 2;

Useless use of a variable in void context at -e line 1 (#1)

$ perl -MO=Deparse,-p -e " $one, $two = 1, 2; "
($one, ($two = 1), '???');
-e syntax OK
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Another obvious one: $one, $two = foo();

Useless use of private variable in void context at -e line 1 (#1)

$ perl -MO=Deparse,-p -e " sub foo { 1, 2 } $one, $two = foo(); "
sub foo {
    (1, 2);
}
($one, ($two = foo()));
-e syntax OK
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Since the behavior of comma operator in list/scalar context is well defined,
since the behaviour of a list in scalar/list context is well defined (see If you believe in Lists in Scalar Context, Clap your Hands ),
perl doesn't warn about : $two = foo();
because it is predictable/guaranteed/you're expected to know what it means

 sub foo{1,2}  ($f)= foo(); is like  ($f) = (1,2); # no warn, $f is 1

 sub foo{1,2}  $f = foo(); is like  $f = (1,2); # no warn, $f is 2
perl won't warn you, scalar context guarantees rightest most(last), list context guarantees leftest most (first)

it's effectively

return scalar(10,20);
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so no warning

  DB<100> print scalar (10,20)
20
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UPDATE: maybe the following is clearer,

  
  DB<113> sub tst { 2..10 }
 => 0
 
  DB<114> ($a)=tst()
 => 2
  
  DB<115> $a=tst()
 => ""
 
  DB<116> sub tst { /2/../10/ }
 => 0
 
  DB<117> $_=2
 => 2
 
  DB<118> $a=tst()
 => 1
 
  DB<119> $a=tst()
 => 2
 
  DB<120> $_=10
 => 10
 
  DB<121> $a=tst()
 => "3E0"
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as you can see the range-op transforms to flip-flop-op in scalar context.

So we are not returning a list but an expression which is evaluated according to the callers context.

that makes sense! thank you very much, Rolf. -Ed