Given:
sub foo {
    my (undef, $x) = @_;
    print $_[0], $x;
}

my $big = 'a' x 99999999;

foo($big, 'bar');

[download]

Does the list assignment of $_[0] to undef incur any significant runtime cost?

I expect that it doesn't, and would like to know what goes on under the hood.

I can't tell what happens in an aassign:
$ perl -MO=Concise -e '(undef) = $x'
8  <@> leave[1 ref] vKP/REFC ->(end)
1     <0> enter ->2
2     <;> nextstate(main 1 -e:1) v:{ ->3
7     <2> aassign[t1] vKS ->8
-        <1> ex-list lK ->5
3           <0> pushmark s ->4
-           <1> ex-rv2sv sK/1 ->-
4              <$> gvsv(*x) s ->5
-        <1> ex-list lK ->7
5           <0> pushmark s ->6
6           <0> undef sP ->7
-e syntax OK

$ perl -MO=Concise -e '($y) = $x'   
8  <@> leave[1 ref] vKP/REFC ->(end)
1     <0> enter ->2
2     <;> nextstate(main 1 -e:1) v:{ ->3
7     <2> aassign[t1] vKS ->8
-        <1> ex-list lK ->5
3           <0> pushmark s ->4
-           <1> ex-rv2sv sK/1 ->-
4              <$> gvsv(*x) s ->5
-        <1> ex-list lK ->7
5           <0> pushmark s ->6
-           <1> ex-rv2sv sKPRM*/1 ->-
6              <$> gvsv(*y) s ->7
-e syntax OK

[download]

In reply to Is undef list assignment a no-op? by repellent

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