How does one use mutually defined Streams in Idris? - idris

Question

The issue I'm having may be more general than stated in the question. I'm trying to get the following program to work:
module Main
main: IO ()
process: Int -> Int
process req = req+1
server: Stream Int -> Stream Int
client: Int -> Stream Int -> Stream Int
server (req :: reqs) = process req :: server reqs
client initreq (resp :: resps) = initreq :: client resp resps
mutual
reqsOut: Stream Int
respsOut: Stream Int
-- This fixes the segfault:
-- reqsOut = cycle [1, 2, 3, 4]
reqsOut = client 0 respsOut
respsOut = server reqsOut
main = do
printLn(take 5 reqsOut)
It will run if I swap out the definition of reqsOut with the commented version, but generates a segmentation fault as is. My guess is I'm using mutual incorrectly, but I'm not sure how.

Answer 1

Note that in function call the arguments gets evaluated beforehand, in particular case splits. Here in client the stream gets case splitted with client initreq (req :: reqs), so respsOut in client 0 respsOut gets evaluated before the delayed tail:
reqsOut =
client 0 respsOut =
client 0 (case respsOut of (req :: reqs) => ...) =
client 0 (case (server regsOut) of (req :: regs) => ...) =
...
You can delay the split with
client initreq stream = initreq :: client (head stream) (tail stream)
But then you'd still have the infinite loop through server:
reqsOut =
client 0 respsOut =
client 0 (server regsOut) =
client 0 (case regsOut of (req :: reqs) => ...) =
...
You can delay the computation of respsOut by making the argument Lazy:
client : Int -> Lazy (Stream Int) -> Stream Int
client initreq stream = initreq :: client (head stream) (tail stream)
And now client can finally construct a Stream Int without evaluating its arguments:
client 0 respsOut =
0 :: Delay (client (head (Force respsOut)) (tail (Force respsOut))) : Stream int