CODEWITHSUNDEEP
ocaml
mak
mak

Reputation: 13

This expression has type string but an expression was expected of type 'a list

I want to count how many "u"s are in a list. An example of the function call would be

     countu(["u"; "a"; "x"], 0);;
        :-int = 1

The code I have is 

let rec countu = fun(x, y) ->
    if List.length (List.hd x) == 0 then y
    else if List.hd x == "u" then countu(List.tl x, y+1)
    else countu(List.tl x, y);;

I understand why I'm getting the problem. Ocaml wants a 'a list, but because I am comparing it to a string, it makes it a string. It will also get mad when I recall countu with y+1, because again, it wants a non-specified unit instead of an integer. How do I get around this. I want the signature to look like

val countups : string list * int -> int = <fun>

Upvotes: 1

Views: 2858

Answers (1)

Jeffrey Scofield
Jeffrey Scofield

Reputation: 66818

You have:

if List.length (List.hd x) == 0 then ...

But this only makes sense if x is a list of lists. Your x is a list of strings.

Possibly what you want is just this:

if List.length x = 0 then ...

As a side comment, the equality comparison operator in OCaml is =. The == operator is for special cases, and you shouldn't use it without having a good reason.

Upvotes: 1

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