ocaml Error: This expression has type expr but is here used with type unit -> expr

Question

I can't figure out what this error means, I've never seen if before the line it is complaining about is the line that contains x -> let I was using match with before this but it was still giving me the same error.

let rec build (rand,depth) = match depth with
 | 0 -> if rand(0,2) == 0 then buildX else buildY
 | x -> let r =  rand(0, 5) in
     if r == 0 then buildSine (build (rand, x -1))
     else if r == 1 then buildCosine (build (rand, x -1))
     else if r == 2 then buildAverage (build (rand,x -1), build (rand,x-1))
     else if r == 3 then buildTimes (build (rand, x -1), build (rand, x-1))
     else buildThresh (build(rand, x-1), build(rand, x-1),               
                   build(rand, x-1), build(rand, x-1));;

each of my the builds returns a expr and this is supposed to build one big expr Any help would be greatly appreciated. Thanks in advance! :D

without the definitions, or at least the declarations of all the bindings in your code snippet, it's difficult to tell. — didierc, Jan 25 '13 at 02:18

Jeffrey Scofield · Accepted Answer · 2013-01-25T00:29:59.983

3

Perhaps buildX and buildY are supposed to take an argument?

# let f () = 14;;
val f : unit -> int = <fun>
# f ;;
- : unit -> int = <fun>
# f ();;
- : int = 14

edited Jan 25 '13 at 00:29

answered Jan 25 '13 at 00:13

Jeffrey Scofield

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They aren't :/ it has empty () – Charles Haro Jan 25 '13 at 00:26
1

In OCaml, empty () is a value. – Jeffrey Scofield Jan 25 '13 at 00:27
I guess @JeffreyScofield found out. – didierc Jan 25 '13 at 02:19
Thanks alot >.< i'm still noob at ocaml haha – Charles Haro Jan 25 '13 at 11:08

ocaml Error: This expression has type expr but is here used with type unit -> expr

1 Answers1