CODEWITHSUNDEEP
arraysswiftgenericsclosures
XmasRights
XmasRights

Reputation: 1509

Generics in Sequence Extensions

I have some code which I'm trying to make generic, and can't quite figure out how.

func randomInt() -> Int { return Int(arc4random_uniform(42)) }

let closures = [randomInt, randomInt, randomInt, randomInt]

let result = closures.map { $0() }
print(result)

In essence, I want to take an array of closures, that return some value of type T, and return an array of type [T] with all the results.

This is my attempt:

extension Sequence where Element == () -> T
{
    func callAll() -> [T]
    {
        return self.map { $0() }
    }
}

However, this clearly doesn't work, as T is undefined. How would I go about writing this extension?

Upvotes: 1

Views: 63

Answers (1)

David Pasztor
David Pasztor

Reputation: 54735

You can make the generic callAll function work only on Sequences, where Element is of type ()->T, but as far as I know, you cannot declare a generic type in a where clause to an extension.

extension Sequence {
    func callAll<T>() -> [T] where Self.Element == ()->T{
        return self.map { $0() }
    }
}

func randomInt() -> Int { return Int(arc4random_uniform(42)) }

let closures = [randomInt, randomInt, randomInt, randomInt]

let result = closures.callAll()
print(result) //one sample output: [41, 19, 15, 36]

Upvotes: 1

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