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Our tempfile struct contains an "active" flag. Long ago, this flag was
important: tempfile structs were always allocated for the lifetime of
the program and added to a global linked list, and the active flag was
what told us whether a struct's tempfile needed to be cleaned up on
exit.
But since 422a21c6a0 (tempfile: remove deactivated list entries,
2017-09-05) and 076aa2cbda (tempfile: auto-allocate tempfiles on heap,
2017-09-05), we actually remove items from the list, and the active flag
is generally always set to true for any allocated struct. We set it to
true in all of the creation functions, and in the normal code flow it
becomes false only in deactivate_tempfile(), which then immediately
frees the struct.
So the flag isn't performing that role anymore, and in fact makes things
more confusing. Dscho noted that delete_tempfile() is a noop for an
inactive struct. Since 076aa2cbda taught it to free the struct when
deactivating, we'd leak any struct whose active flag is unset. But in
practice it's not a leak, because again, we'll free when we unset the
flag, and never see the allocated-but-inactive state.
Can we just get rid of the flag? The answer is yes, but it requires
looking at a few other spots:
1. I said above that the flag only becomes false before we deallocate,
but there's one exception: when we call remove_tempfiles() from a
signal or atexit handler, we unset the active flag as we remove
each file. This isn't important for delete_tempfile(), as nobody
would call it anymore, since we're exiting.
It does in theory provide us some protection against racily
double-removing a tempfile. If we receive a second signal while we
are already in the cleanup routines, we'll start the cleanup loop
again, and may visit the same tempfile. But this race already
exists, because calling unlink() and unsetting the active flag
aren't atomic! And it's OK in practice, because unlink() is
idempotent (barring the unlikely event that some other process
chooses our exact temp filename in that instant).
So dropping the active flag widens the race a bit, but it was
already there, and is fairly harmless in practice. If we really
care about addressing it, the right thing is probably to block
further signals while we're doing our cleanup (which we could
actually do atomically).
2. The active flag is declared as "volatile sig_atomic_t". The idea is
that it's the final bit that gets set to tell the cleanup routines
that the tempfile is ready to be used (or not used), and it's safe
to receive a signal racing with regular code which adds or removes
a tempfile from the list.
In practice, I don't think this is buying us anything. The presence
on the linked list is really what tells the cleanup routines to
look at the struct. That is already marked as "volatile". It's not
a sig_atomic_t, so it's possible that we could see a sheared write
there as an entry is added or removed. But that is true of the
current code, too! Before we can even look at the "active" flag,
we'd have to follow a link to the struct itself. If we see a
sheared write in the pointer to the struct, then we'll look at
garbage memory anyway, and there's not much we can do.
This patch removes the active flag entirely, using presence on the
global linked list as an indicator that a tempfile ought to be cleaned
up. We are already careful to add to the list as the final step in
activating. On deactivation, we'll make sure to remove from the list as
the first step, before freeing any fields. The use of the volatile
keyword should mean that those things happen in the expected order.
Signed-off-by: Jeff King <peff@peff.net>
Signed-off-by: Junio C Hamano <gitster@pobox.com>
Git - fast, scalable, distributed revision control system
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