CODEWITHSUNDEEP
ocaml
it doesnt mater
it doesnt mater

Reputation: 1

Why do i get syntax error at end problem with pattern matching

I have to make a function that takes list a list and returns list of pairs of first and last element,2nd and 2nd last and so forth It doesn't matter if the list has even or odd number of elements because if its odd i will just ignore the middle element.The idea i have is that make a new rec fun that takes old list and its revers as input i think i finished the code but i get Syntax error for ;;

let lip l =
    if [] then [] 
    else let l1=l l2=List.rev l in 
         let rec lp l1 l2 = match l1,l2 with 
        | [],[] ->[]
        | [],h2::t2->[]
        | h1::_,h2::_ -> 
        if (List.length l -2) >= 0 then [(h1,h2)]@ lp(List.tl l1) t2
        else [] ;;

Upvotes: 0

Views: 1075

Answers (2)

lambda.xy.x
lambda.xy.x

Reputation: 4998

The following code should at least compile:

let lip list1 =
  if list1 = [] then []
  else
    let list2=List.rev list1 in 
    let rec lp l1 l2 = match l1,l2 with 
      | [], [] ->[]
      | [], _::_->[]
      | h1::_::_, h2::t2 -> (* l1 length >= 2*)
        (h1,h2) :: lp(List.tl l1) t2
      | h1::_,h2::t2 ->     (* l1 length = 1 *)
        []
    in
    []

I have made the following changes:

  • renamed the arguments of lip to make clear they are different from the arguments of lp
  • removed the alias let l1 = l
  • changed the if condition to a term of type boolean -- there's not much to compare, so I assume you are checking list1
  • replaced the list length condition by a pattern match against two heads
  • the else path is the second match - it might be better to rewrite that one to | [h1, _] -> ...
  • the definition of lp needs to be followed with the actual body of lip - to make it compile, we just return [] at the moment but you probably would like something else there

As @Jeffrey Scofield already mentioned, you are not using lp in your code. It could help if you added a comment that explains what you'd like to achieve and what the intended role of lp is.

Upvotes: 0

Jeffrey Scofield
Jeffrey Scofield

Reputation: 66823

There are quite a few errors in your code.

I think the specific error you're seeing is caused by the fact that there is no in after let rec lp ....

Every let that's not at the top level of a module needs to be followed by in. One way to think of it is that it's a way of declaring a local variable for use in the expression that appears after in. But you need to have the in expr.

Another way to look at it is that you're defining a function named lp but you're not calling it anywhere.

As @lambda.xy.x points out, you can't say if [] then ... because [] isn't of type bool. And you can't say let x = e1 y = e2 in .... The correct form for this is let x = e1 in let y = e2 in ...

(Or you can write let x, y = e1, e2 in ..., which looks nicer for defining two similar variables to two similar values.)

Upvotes: 1

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