midx: teach `fill_midx_entry()` about incremental MIDXs

In a similar fashion as previous commits, teach the `fill_midx_entry()`
function to work in a incremental MIDX-aware fashion.

This function, unlike others which accept an index into either the
lexical order of objects or packs, takes in an object_id, and attempts
to fill a caller-provided 'struct pack_entry' with the remaining pieces
of information about that object from the MIDX.

The function uses `bsearch_midx()` which fills out the frame-local 'pos'
variable, recording the given object_id's lexical position within the
MIDX chain, if found (if no matching object ID was found, we'll return
immediately without filling out the `pack_entry` structure).

Once given that position, we jump back through the `->base_midx` pointer
to ensure that our `m` points at the MIDX layer which contains the given
object_id (and not an ancestor or descendant of it in the chain). Note
that we can drop the bounds check "if (pos >= m->num_objects)" because
`midx_for_object()` performs this check for us.

After that point, we only need to make two special considerations within
this function:

  - First, the pack_int_id returned to us by `nth_midxed_pack_int_id()`
    is a position in the concatenated lexical order of packs, so we must
    ensure that we subtract `m->num_packs_in_base` before accessing the
    MIDX-local `packs` array.

  - Second, we must avoid translating the `pos` back to a MIDX-local
    index, since we use it as an argument to `nth_midxed_offset()` which
    expects a position relative to the concatenated lexical order of
    objects.

Signed-off-by: Taylor Blau <me@ttaylorr.com>
Signed-off-by: Junio C Hamano <gitster@pobox.com>

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