I don't suppose your knowledge of the internals can confirm that the example in the OP is indeed just passing a pointer to $vector, and NOT copying the entire byte sequence somewhere else at the same time (even just as a side-effect)?

Yes. I can confirm that. No copying is done.

When you define an XS argument as SV* sv_vec, you are asking for a pointer to the SV. When you operate via that pointer, you are changing the original SV.

As with ordinary perl subs, the subroutines receives aliases to the actual variables passed:

sub x{ 
    ++$_  for @_ 
};;

( $a, $b, $c) = 12345..12347;;
x( $a, $b, $c );;

print $a, $b, $c;;
12346 12347 12348
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No copying occurs unless the programmer assigns them to local vars:

sub x{ 
    my( $a, $b, $c ) = @_;
    ++$_  for $a, $b, $c;
}
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If I am defining perl subs to operate upon large scalars, and they are more complex than a couple of lines--at which point the $_[0], $_[1] nomenclature can become awkward--then I will use scalar refs:

sub xyz (\$) {
    my $rStr = shift;
    substr $$rStr, ...;
    vec $$rStr, ...;
    ...
}
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Which achieves the benefit of named variables without the cost of copying.

BTW: You still haven't mentioned what the "complex processing" you are performing in XS is?

I ask because my instinctual reaction that if you are performing boolean operations on whole pairs or more of large bit vectors, it is almost certainly quicker doing it in Perl.