Mutable Data in OCaml
functional-programming, mutable, ocaml
Solution
`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 ()`.
Problem
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