Re: Reviving planned ldso changes

[Szabolcs Nagy <nsz@port70.net>]

* Rich Felker <dalias@libc.org> [2017-02-25 20:39:26 -0500]:
> On Sun, Feb 26, 2017 at 02:04:30AM +0100, Szabolcs Nagy wrote:
> > * Rich Felker <dalias@libc.org> [2017-01-15 12:44:38 -0500]:
> > >  static void do_init_fini(struct dso *p)
> > >  {
> > >  	size_t dyn[DYN_CNT];
> > > -	int need_locking = libc.threads_minus_1;
> > > -	/* Allow recursive calls that arise when a library calls
> > > -	 * dlopen from one of its constructors, but block any
> > > -	 * other threads until all ctors have finished. */
> > > -	if (need_locking) pthread_mutex_lock(&init_fini_lock);
> > > -	for (; p; p=p->prev) {
> > > -		if (p->constructed) continue;
> > > +	pthread_mutex_lock(&init_fini_lock);
> > > +	/* Construct in dependency order without any recursive state. */
> > > +	while (p && !p->constructed) {
> > > +		/* The following loop descends into the first dependency
> > > +		 * that is neither alredy constructed nor pending
> > > +		 * construction due to circular deps, stopping only
> > > +		 * when it reaches a dso with no remaining dependencies
> > > +		 * to descend into. */
> > > +		while (p->deps && p->deps[p->next_dep]) {
> > > +			if (!p->deps[p->next_dep]->constructed &&
> > > +			    !p->deps[p->next_dep]->next_dep)
> > > +				p = p->deps[p->next_dep++];
> > > +			else
> > > +				p->next_dep++;
> > > +		}
> > >  		p->constructed = 1;
> > >  		decode_vec(p->dynv, dyn, DYN_CNT);
> > >  		if (dyn[0] & ((1<<DT_FINI) | (1<<DT_FINI_ARRAY))) {
> > > @@ -1233,17 +1246,19 @@ static void do_init_fini(struct dso *p)
> > >  			size_t *fn = laddr(p, dyn[DT_INIT_ARRAY]);
> > >  			while (n--) ((void (*)(void))*fn++)();
> > >  		}
> > > -		if (!need_locking && libc.threads_minus_1) {
> > > -			need_locking = 1;
> > > -			pthread_mutex_lock(&init_fini_lock);
> > > -		}
> > > -	}
> > > -	if (need_locking) pthread_mutex_unlock(&init_fini_lock);
> > > +		/* Revisit "parent" dso which caused the just-constructed
> > > +		 * dso to be pulled in as a dependency. On the next loop
> > > +		 * iteration we will either descend to construct a sibling
> > > +		 * of the just-constructed dso, or finish constructing the
> > > +		 * parent if no unfinished deps remain. */
> > > +		p = p->needed_by;
> > > +	}
> > 
> > i think with
> > 
> > a.deps: b c
> > b.deps: c d
> > b.needed_by: a
> > c.needed_by: a
> > 
> > the visiting order starting from a is
> > a
> > b
> > c
> > a
> > 
> > and d never gets constructed.
> 
> Are you sure? My understanding of what it does is:
> 
> 1. Descend a->b->c, construct c, and back up to b.

you did not explain how you get back to b after c
without a stack of visited dsos or modified c->needed_by.

> 2. Descend b->d, construct d, and back up to b.
> 3. Find all of b's deps constructed, construct b, and back up to a.
> 4. Find all of a's deps constructed, construct a, and end.
> 
> I think you have a misunderstanding of "visit order". Nodes are not
> visited while descending, only when reaching a point where no further
> descent into a non-constructed dep is possible.
> 
> Rich