CODEWITHSUNDEEP
.netf#
John Oriely
John Oriely

Reputation: 43

Passing array into Function

I'm trying to write a simple F# function that I can pass an array into and then print the values, but I'm having trouble. Here is what I have so far:

let a = [| a; b; c; d |];;

let f arrayFunction (string[] array) = function
    for b=0 to array.Length
      Console.WriteLine(array.[]);;

Upvotes: 2

Views: 2928

Answers (3)

Gary.S
Gary.S

Reputation: 7121

Jack's answer is exactly correct however there are built-in functions in F# to do these kinds of tasks. In this instance we can send the array to Array.iter which will iterate over each item and pass the item into a string -> unit function.

So an example might look like this:

let a = [| "a"; "b"; "c"; "d" |];;
let f arrayFunction (array : string[]) =
    array |> Array.iter arrayFunction;;

a |> f Console.WriteLine;;

Upvotes: 3

Kit
Kit

Reputation: 2136

In addition to the other answers - you didn't need to specify the type of the array argument explicitly. Type inference normally handles it fine (depending on the wider context). So for example this works:

let a = [| "a"; "b"; "c"; "d" |]
let f arrayFunction array =
    array |> Seq.iter arrayFunction

let printme s = printfn "%s" s
f printme a

Upvotes: 1

Jack P.
Jack P.

Reputation: 11525

The F# syntax for defining parameters is backwards from the C# syntax; in F#, the name of the parameter comes first, then the type (with a colon to separate the two).

You also don't need the function keyword here, just the normal let binding -- function is for creating anonymous pattern-matching functions. You do, however, need to add a do at the end of the line in your for loop. Finally, the value after the to in an F# for loop is inclusive -- so you need to subtract one from the array length or you'll end up raising an IndexOutOfRangeException.

Your function should be written like this:

let a = [| a; b; c; d |];;

let f arrayFunction (array : string[]) =
    for b = 0 to array.Length - 1 do
        Console.WriteLine (array.[b]);;

Upvotes: 6

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