Viacheslav Dubeyko <Slava.Dubeyko@xxxxxxx> wrote: > > + rreq->buffer.iter = *iter; > > The struct iov_iter structure is complex enough and we assign it by value to > rreq->buffer.iter. So, the initial pointer will not receive any changes > then. Is it desired behavior here? Yes. The buffer described by the iterator is going to get partitioned across a number of subrequests, each of which will get a copy of the iterator suitably advanced and truncated. As they may run in parallel, there's no way for them to share the original iterator. David
