When you say, "I expect it to print nothing", I'm guessing you're thinking your first element is undefined? So you're thinking that the shift was done twice, once for what you think is an undefined element and once for an element with the value.

This interpretation is incorrect: you're passing a 4-element array to your function, and it's as if the @3 didn't exist at all. There is no "implicit shift" going on, unless you think of the my ($a)=@_; as a "shift". If you merge two arrays, and one has no elements, you don't end up with an undefined element as a placeholder for the empty array (I don't know if I can put it more clearly).

sub f1{
    
    my ($zeroelement, $oneelement, $twoelement, $threeelement) =@_;

    print scalar(@_);  ## prints number of elements -->4
                       ## each element is in the appropriate value, an
+d $zeroelement is 9
    $cp=1;
    $LOCAL_CODEPAGE =10;
    if ($a==1){
        $LOCAL_CODEPAGE = $cp;
        break;
    }
    return $LOCAL_CODEPAGE = $a;
    print "\n";
}
@3=();

print "\$LOCAL_CODEPAGE=$LOCAL_CODEPAGE: f1(@3,9,3,4,5): ".f1(@3,9,3,4
+,5)."\n";
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