如何在 Idris 中使用相互定义的流？

Question

我遇到的问题可能比问题中所述的更普遍。我正在尝试让以下程序运行：

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)

如果我用注释版本替换 reqsOut 的定义，它将运行，但会按原样生成分段错误。我的猜测是我使用 mutual 不正确，但我不确定如何使用。

Answer 1

请注意，在函数调用中，参数会事先求值，在特定情况下会拆分。在 client 中，流用 client initreq (req :: reqs) 进行大小写拆分，因此 client 0 respsOut 中的 respsOut 在延迟尾部之前得到评估：

reqsOut =
client 0 respsOut =
client 0 (case respsOut of (req :: reqs) => ...) =
client 0 (case (server regsOut) of (req :: regs) => ...) =
...

您可以使用

延迟拆分

client initreq stream = initreq :: client (head stream) (tail stream)

但是你仍然会无限循环 server:

reqsOut =
client 0 respsOut =
client 0 (server regsOut) =
client 0 (case regsOut of (req :: reqs) => ...) =
...

您可以延迟 respsOut 的计算，方法是使参数 Lazy:

client : Int -> Lazy (Stream Int) -> Stream Int
client initreq stream = initreq :: client (head stream) (tail stream)

现在 client 终于可以构造一个 Stream Int 而无需评估其参数：

client 0 respsOut =
0 :: Delay (client (head (Force respsOut)) (tail (Force respsOut))) : Stream int

如何在 Idris 中使用相互定义的流？

How does one use mutually defined Streams in Idris?

idris