Re: fib6_del_route has redundant code

[David Miller <davem@davemloft.net>]

From: Gui Jianfeng <guijianfeng@cn.fujitsu.com>
Date: Fri, 28 Dec 2007 13:58:21 +0800

> >> I think the following code in fib6_del_route in the latest kernel is useless.
> >> 1125         if (fn->leaf == NULL && fn->fn_flags&RTN_TL_ROOT)
> >> 1126                 fn->leaf = &ip6_null_entry;
> >>
> >> ip6_null_entry will never be unlinked from fn->leaf now, that is,
> >> fn->leaf == NULL will never meet.
> > 
> > I think you are right, but if it is true the next block of
> > code is dead too:
> > 
> > 	/* If it was last route, expunge its radix tree node */
> > 	if (fn->leaf == NULL) {
> > 		fn->fn_flags &= ~RTN_RTINFO;
> > 		rt6_stats.fib_route_nodes--;
> > 		fn = fib6_repair_tree(fn);
> > 	}
> > 
> 
> I think this block of code can't be removed, because just the
> root(default route) fn->leaf always has ip6_null_entry on it. The
> normal fn->leaf becomes NULL when last route has been deleted, the
> radix tree should be expunged.

But you said (still quoted above) that fn->leaf == NULL will not
occur.

Do you mean this, only in the case that the RTN_TL_ROOT flag is set?

I thought you meant always when the function is called, fn->leaf
cannot ever be NULL.