tl;dr Resource ordering prevents deadlock for reader-writer locks too.
Long answer:
Draw a wait-for graph like this:
First draw a vertex for all the resources from left to right in the
prescribed order.
Then, for each process, waiting for a resource, draw a vertex,
immediately preceding the vertex of the waited for resource.
If a process is not waiting on any resources, draw a vertex
on the far right, after all the resource vertices.
Then draw an edge from each process to the resource on which that
process is waiting and draw an edge from each resource to the
process, which is currently holding that resource.
Consider each edge on the graph. There are two cases:
the edge is Pi -> Ri
, i.e. from a process to a resource. Since each
process is waiting on only one resource and we have drawn the
process' vertex immediately to the left of the resource it is
waiting for, then the edge is from left to right.
the edge is Ri -> Pj
, i.e. from a resource to a process. If Pj
is
not waiting to any resource, then its vertex is on the right of
all the resources, therefore the edge is from left to right. If Pj
is waiting to Rk
, then i < k
, because processes acquire resources in
order. If i < k
, then Ri
is on the left of Rk
, and Pj
is
immediately on the left of Rk
(since we have drawn the grapgh that way),
hence Ri
is on the left of Pj
, therefore the edge is again left to
right.
Since all the edges of the graph drawn this way are from left to
right, then we have constructed a topological sorting of the graph,
hence the graph has no cycles, hence a deadlock cannot occur.
Note that what matters is the fact that a process waits, not why it
waits. Thus, it does no matter whether it waits for a mutex,
read-write lock, semaphore or whatever - this strategy for deadlock
prevention works for all.