CODEWITHSUNDEEP
.netf#functional-programming
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qewqew qdqw d

Reputation: 189

How do you define types in a tuple in a function

I have just started with F# and functional programming. I want to know how I can make a function taking a tuple where I define that the first value have to be a string, instead of the standard int.

Example:

A function which replicates a string, s, n times and returns it. What i have right now is this:

let rec pow2 = function
    | (s:string,0) -> ""
    | (s:string,n) -> s + pow2(s,n-1)

This works, but I think that there is a better way than defining s:string every case.

(I know String.replicate, this is not for the sake of the effect, but learning)

Upvotes: 2

Views: 191

Answers (3)

Thorkil Holm-Jacobsen
Thorkil Holm-Jacobsen

Reputation: 7676

You can also do it like this, where the type of the tuple is given in the function definition:

let rec pow2 ((s, n) : string * int) =
    match n with
    | 0 -> ""
    | _ -> s + pow2 (s, n - 1)

Note that elements of a tuple are seperated by * in the explicit type annotation. Also, as Stephen pointed out, the type inference system will, in this example, be smart enough to figure out the type of tuple because we match n with an integer and because we add a string ("") to s.

Upvotes: 4

Jack P.
Jack P.

Reputation: 11525

You only need the type annotation on the first case -- the F# compiler can infer that the first element of the tuple must be a string in the rest of the cases.

let rec pow2 = function
    | (s:string,0) -> ""
    | (s,n) -> s + pow2(s,n-1)

It's just a matter of style, but I think it's a little easier to read if you write the function like this:

let rec pow2 (s, n) =
    match n with
    | 0 -> ""
    | _ -> s + pow2(s, n-1)

Upvotes: 4

Stephen Swensen
Stephen Swensen

Reputation: 22297

In fact, no type annotations are required here. The "" return value in the the first pattern match is a sufficient enough hint for the compiler:

> let rec pow2 = function
    | (s,0) -> ""
    | (s,n) -> s + pow2(s,n-1);;

val pow2 : string * int -> string

Upvotes: 7

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