Maybe instead of checking with 3 different ways whether the service is lazy, we could obtain that information from the service definition?

Unfortunately there is no way to detect if object itself is natively lazy. It's not possible to detect it at service container compile time by reading the container definition either. But, we can use the version of the Symfony DI bundle to determine if the lazy services are represented as lazy native objects, because since version 7.3 that's exactly the case. Prior version 7.3 the lazy services are LazyObjectInterface.

I reconsidered the usage of $config->isNativeLazyObjectsEnabled() because it's not meant for that purpose.

After analyzing the changes between 7.2 and 7.3 for:

symfony/dependency-injection 7.2-7.3 diff

I think it's best to check if the following method exists in order to determine if symfony version is >= 7.3:

Symfony\Component\DependencyInjection\ContainerBuilder->findTaggedResourceIds

Then:

For PHP < 8.4 or symfony < 7.3 the logic remains as is:

      if ((! $manager instanceof LazyLoadingInterface && ! $manager instanceof LazyObjectInterface) || $manager->isOpen()) {
          $manager->clear();

          return;
      }

      $this->resetManager($managerName);

Else:

    $r = new ReflectionClass($manager);
    if ($r->isUninitializedLazyObject($manager)) {
        return;
    }

    if ($manager->isOpen()) {
        $manager->clear();

        return;
    }

    $this->resetManager($managerName);

Unfortunately there is no way to detect if object itself is natively lazy.

I'm confused… isn't that exactly what https://www.php.net/manual/de/reflectionclass.isuninitializedlazyobject.php is supposed to achieve?

IIRC it only checks if it's lazy uninitialized object. If it's initialized then you are out of luck.

Yes, that's correct.

After initialization, the object is indistinguishable from an object that was never lazy.

-- PHP RFC: Lazy Objects