From: Gui Jianfeng <[email protected]>
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.
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