You need to explicitly name the optional arguments you are passing,
otherwise the interpreter thinks you are trying to pass l, which should be 'a
list :
let rec enumerate ?(l'=[]) ?(n=0) l =
match l with
h::l1 -> enumerate ~l':(l'@[(n,h)]) ~n:(n+1) l1
| [] -> l'
-Pierre
Le 29 sept. 2011 à 11:07, Walter Cazzola a écrit :
> Dear all,
> waiting for some hints on my other issue for functors and generic I was
> playing with an enumerate function (with tail recursion):
>
> let rec enumerate ?(l':((int * 'a) list)=[]) ?(n=0) l =
> match l with
> h::l1 -> enumerate (l'@[(n,h)]) (n+1) l1
> | [] -> l'
>
> this should just take a list and return a list of couple whose first
> entry is the position in the list of the element; to be clearer:
>
> enumerate ['a'; 'b'; 'c'] -> [(0,'a'); (1,'b'); (2,'c')];
>
> but this doesn't work as by type mismatch:
>
> Error: This expression has type (int * 'a) list
> but an expression was expected of type 'a list
>
> Doesn't a matter if I've annotated the type (int * 'a) list
> to the optional argument l' when I try to call the function with the new
> value it says that the two things don't have the same type.
>
> I'm quite puzzled by this behavior, any explanation/help is welcome as
> usual.
>
> Walter
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>
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