在F#中编写相互递归函数的问题

Question

我正在翻译Little Mler中运行此数据类型的函数

type sexp<'T> = 
    An_Atom of 'T
    | A_slist of slist<'T>
and 
    slist<'T> = 
    Empty
    | Scons of sexp<'T> * slist<'T>

功能

// occurs_in_slist : aVal slist -> int
// checks the number of occurrence for aVal in slist

let rec occurs_in_slist =
    function
    _, Empty-> 0
   | (aVal : 'T), Scons(aSexp, (aSlist : 'T)) -> 
    occurs_in_sexp (aVal, aSexp) + occurs_in_slist (aVal, aSlist)
and
   aVal, An_Atom (bVal) ->  if (aVal = bVal) then 1 else 0
   |  (aVal , A_slist(aSlist)) -> occurs_in_slist (aval, aSlist)

但是,我得到第二个函数的这个错误

error FS0010: Unexpected symbol '->' in binding. Expected '=' or other token.

Answer 1

在函数定义中,您已使用and关键字来定义相互递归的函数集,但是您只为第一个函数指定了名称.它期待其他函数的名称,and这就是你得到错误的原因.不幸的是你已经把它遗弃了.

我相信这是你想要做的:

let rec occurs_in_slist = function
  | _        , Empty -> 0
  | aVal : 'T, Scons(aSexp, aSlist : slist<'T>) -> 
        occurs_in_sexp (aVal, aSexp) + occurs_in_slist (aVal, aSlist)
and occurs_in_sexp = function
  | aVal : 'T, An_Atom(bVal) -> if (aVal = bVal) then 1 else 0
  | aVal     , A_slist(aSlist) -> occurs_in_slist (aVal, aSlist)

虽然我觉得这里更合适的回归类型应该是一个bool.

let rec occurs_in_slist = function
  | _        , Empty -> false
  | aVal : 'T, Scons(aSexp, aSlist : slist<'T>) -> 
        occurs_in_sexp (aVal, aSexp) || occurs_in_slist (aVal, aSlist)
and occurs_in_sexp = function
  | aVal : 'T, An_Atom(bVal) -> aVal = bVal
  | aVal     , A_slist(aSlist) -> occurs_in_slist (aVal, aSlist)