CODEWITHSUNDEEP
functionocaml
FredrikA
FredrikA

Reputation: 25

function result as parameter in another function

How do I pass a function result as a parameter to another function?

Example:

let function a b = a/b;;

let anotherFunction p t = p + t;;

function 10 5;;

Can I pass the result from function (2) to a parameter - let's say t in anotherFunction?¨

Upvotes: 1

Views: 391

Answers (2)

Chris
Chris

Reputation: 36680

Late addition, but note also that for complex expressions, you may wish to use a local binding via the let ... in ... construct.

# let function1 a b = a / b
let function2 p t = p + t;;
val function1 : int -> int -> int = <fun>
val function2 : int -> int -> int = <fun>
# let result = function1 5 2 in
function2 15 result;;
- : int = 17

Where the first expression is passed as the last argument of the second function, we can also use the |> or @@ operators.

# function1 5 2 |> function2 15;;
- : int = 17
# function2 15 @@ function1 5 2;;
- : int = 17

It may be difficult to see from such a simple example, how these would be useful, but when chaining many operations together they can improve code readability substantially. Consider multiple operations on a list.

  1. Add 1 to each element.
  2. Filter out odd elements.
  3. Multiply by two.
# let lst = [1;2;3;4;5;6] in
List.(
  map (fun x -> x * 2) (filter (fun x -> x mod 2 = 0) (map (fun x -> x + 1) lst))
);;
- : int list = [4; 8; 12]

Or:

# let lst = [1;2;3;4;5;6] in
List.(
  lst 
  |> map (fun x -> x + 1)
  |> filter (fun x -> x mod 2 = 0)
  |> map (fun x -> x * 2)
);;
- : int list = [4; 8; 12]

Or:

# let lst = [1;2;3;4;5;6] in
List.(
  map (fun x -> x * 2) 
  @@ filter (fun x -> x mod 2 == 0) 
  @@ map (fun x -> x + 1) 
  @@ lst
);;
- : int list = [4; 8; 12]

I'm partial to the |> solution as it not only reduces nested parentheses (and thus mental load) but it also puts the last operation last in the reading order.

Upvotes: 0

Jeffrey Scofield
Jeffrey Scofield

Reputation: 66823

You can't have a function named function, which is a keyword in OCaml. So let's call it firstFunction.

I think what you're asking for is this:

 anotherFunction 13 (firstFunction 10 5)

Here's a toplevel session:

# let firstFunction a b = a / b;;
val firstFunction : int -> int -> int = <fun>
# let anotherFunction p t = p + t;;
val anotherFunction : int -> int -> int = <fun>
# anotherFunction 13 (firstFunction 10 5);;
- : int = 15

(For simple questions like this it might help to spend a little time typing expressions into the toplevel. It should start to make sense pretty quickly.)

Upvotes: 5

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