Recursive functions in computation expressions

问题

Some background first. I am currently learning some stuff about monadic parser combinators. While I tried to transfer the 'chainl1' function from this paper (p. 16-17), I came up with this solution:

let chainl1 p op = parser {
  let! x = p
  let rec chainl1' (acc : 'a) : Parser<'a> =
      let p' = parser {
          let! f = op
          let! y = p
          return! chainl1' (f acc y)
          }
      p' <|> succeed acc
  return! chainl1' x
}

I tested the function with some large input and got a StackOverflowException. Now I am wondering, is it posible to rewrite a recursive function, that uses some computation expression, in a way so it is using tail recursion?

When I expand the computation expression, I can not see how it would be generally possible.

let chainl1 p op =
    let b = parser
    b.Bind(p, (fun x ->
    let rec chainl1' (acc : 'a) : Parser<'a> =
        let p' =
            let b = parser
            b.Bind(op, (fun f ->
            b.Bind(p, (fun y ->
            b.ReturnFrom(chainl1' (f acc y))))))
        p' <|> succeed acc
    b.ReturnFrom(chainl1' x)))

回答1:

In your code, the following function isn't tail-recursive, because - in every iteration - it makes a choice between either p' or succeed:

// Renamed a few symbols to avoid breaking SO code formatter
let rec chainl1Util (acc : 'a) : Parser<'a> = 
  let pOp = parser { 
    let! f = op 
    let! y = p 
    return! chainl1Util (f acc y) } 
  // This is done 'after' the call using 'return!', which means 
  // that the 'cahinl1Util' function isn't really tail-recursive!
  pOp <|> succeed acc

Depending on your implementation of parser combinators, the following rewrite could work (I'm not an expert here, but it may be worth trying this):

let rec chainl1Util (acc : 'a) : Parser<'a> = 
  // Succeeds always returning the accumulated value (?)
  let pSuc = parser {
    let! r = succeed acc
    return Choice1Of2(r) }
  // Parses the next operator (if it is available)
  let pOp = parser {
    let! f = op
    return Choice2Of2(f) }

  // The main parsing code is tail-recursive now...
  parser { 
    // We can continue using one of the previous two options
    let! cont = pOp <|> pSuc 
    match cont with
    // In case of 'succeed acc', we call this branch and finish...
    | Choice1Of2(r) -> return r
    // In case of 'op', we need to read the next sub-expression..
    | Choice2Of2(f) ->
        let! y = p 
        // ..and then continue (this is tail-call now, because there are
        // no operations left - e.g. this isn't used as a parameter to <|>)
        return! chainl1Util (f acc y) }

In general, the pattern for writing tail-recursive functions inside computation expressions works. Something like this will work (for computation expressions that are implemented in a way that allows tail-recursion):

let rec foo(arg) = id { 
  // some computation here
  return! foo(expr) }

As you can check, the new version matches this pattern, but the original one did not.

回答2:

In general it is possible to write tail-recursive computation expressions (see 1 and 2), even with multiple let! bindings thanks to the 'delay' mechanism.

In this case the last statement of chainl1 is what gets you into a corner I think.

来源：https://stackoverflow.com/questions/3141669/recursive-functions-in-computation-expressions

标签

tail-recursion

computation-expression