CODEWITHSUNDEEP
functional-programmingocamlmutable
Faisal Abid
Faisal Abid

Reputation: 9140

Mutable Data in OCaml

I've created a mutable data structure in OCaml, however when I go to access it, it gives a weird error,

Here is my code 

type vector = {a:float;b:float};;
type vec_store = {mutable seq:vector array;mutable size:int};;

let max_seq_length = ref 200;;

exception Out_of_bounds;;
exception Vec_store_full;;

let vec_mag {a=c;b=d} = sqrt( c**2.0 +. d**2.0);;


let make_vec_store() = 
    let vecarr = ref ((Array.create (!max_seq_length)) {a=0.0;b=0.0}) in
         {seq= !vecarr;size=0};;

When I do this in ocaml top-level 

let x = make _ vec _store;;

and then try to do x.size I get this error

Error: This expression has type unit -> vec_store
       but an expression was expected of type vec_store

Whats seems to be the problem? I cant see why this would not work.

Thanks, Faisal

Upvotes: 3

Views: 1820

Answers (3)

chollida
chollida

Reputation: 7894

As a follow up to the excellent answere provided. You can tell that your example line:

# let x = make_vec_store;;
val x : unit -> vec_store = <fun>

returns a function as the repl will tell you this. You can see from the output that x is of type <fun> that takes no parameters unit and returns a type vec_store.

Contrast this to the declaration

# let x = 1;;
val x : int = 1

which tells you that x is of type int and value 1.

Upvotes: 2

Chuck
Chuck

Reputation: 237110

make_vec_store is a function. When you say let x = make_vec_store, you are setting x to be that function, just like if you'd written let x = 1, that would make x the number 1. What you want is the result of calling that function. According to make_vec_store's definition, it takes () (also known as "unit") as an argument, so you would write let x = make_vec_store ().

Upvotes: 12

swegi
swegi

Reputation: 4112

try x = make_ vec_store()

Upvotes: 4

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