How does mongoid generate BSON ObjectId - ruby-on-rails

Question

This is with reference to this answer: https://stackoverflow.com/a/5694803/762747
So the BSON Object ID consists of:
[4 bytes seconds since epoch, 3 bytes machine hash, 2 bytes process ID, 3 bytes counter]
What logic does mongoid/moped (http://mongoid.org/) use to generate the 3 bytes counter - do they increment or generate a random number?
If effect, can we be almost certain that the BSON ObjectId generated by mongoid will be unique (and avoid point 2 in the referenced answer)?

Answer 1

We can be almost certain that the ObjectId will be unique. :)
When not using the native extensions (which are now included in the bson gem apparently which is used by mongoid/moped), it's using a counter.
def next(time = nil)
#mutex.lock
begin
count = #counter = (#counter + 1) % 0xFFFFFF
ensure
#mutex.unlock rescue nil
end
generate(time || ::Time.new.to_i, count)
end
It's using a counter as you can see in the Generator class.
When using the native C code for generating an ObjectId, it too is using a counter:
static VALUE rb_object_id_generator_next(int argc, VALUE* time, VALUE self)
{
char bytes[12];
unsigned long t;
unsigned short pid = htons(getpid());
if (argc == 0 || (argc == 1 && *time == Qnil)) {
t = rb_current_time_milliseconds();
}
else {
t = htonl(NUM2UINT(rb_funcall(*time, rb_intern("to_i"), 0)));
}
memcpy(&bytes, &t, 4);
memcpy(&bytes[4], rb_bson_machine_id, 3);
memcpy(&bytes[7], &pid, 2);
memcpy(&bytes[9], (unsigned char*) &rb_bson_object_id_counter, 3);
rb_bson_object_id_counter++;
return rb_str_new(bytes, 12);
}
Addendum (thanks to #ChrisHeald for the tip)
Assuming the Ruby implementation being used is using a GIL (Global Interpreter Lock) and using the C code implementation above, the C code would also be safe when it increments the counter, rb_bson_object_id_counter, as there would be an outer lock on the call to this code.