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由 Al Viro 提交于
passing *any* namespace root to __d_path() as root is equivalent to just passing it {NULL, NULL}; no need to bother with finding the root of our namespace in there. Signed-off-by: NAl Viro <viro@zeniv.linux.org.uk>
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passing *any* namespace root to __d_path() as root is equivalent
to just passing it {NULL, NULL}; no need to bother with finding
the root of our namespace in there.
Signed-off-by: NAl Viro <viro@zeniv.linux.org.uk>