CODEWITHSUNDEEP
swiftgenerics
Carl Hung
Carl Hung

Reputation: 555

overload operator as generic function

i saw an example on "Pro swift" book.

it overloaded the operator, the first parameter "lhs" is a function that takes T -> U.

but "generateRandomNumber" function is Int -> Int

How can it work on >>> operator?

how does it work?

thanks. 

import Foundation
infix operator >>> { associativity left }
func >>> <T, U, V>(lhs: T -> U, rhs: U -> V) -> T -> V { 
    return { rhs(lhs($0)) }
}

func generateRandomNumber(max: Int) -> Int {
    let number = Int(arc4random_uniform(UInt32(max)))
    print("Using number: \(number)")
    return number
}
func calculateFactors(number: Int) -> [Int] {
    return (1...number).filter { number % $0 == 0 }
}
func reduceToString(numbers: [Int]) -> String {
    return numbers.reduce("Factors: ") { $0 + String($1) + " " }
}

let combined = generateRandomNumber >>> calculateFactors >>>
reduceToString
print(combined(100))

Upvotes: 1

Views: 205

Answers (1)

user3441734
user3441734

Reputation: 17582

see the documentation about generics.

let combined = generateRandomNumber >>> calculateFactors >>> reduceToString
print(generateRandomNumber.dynamicType)
print(calculateFactors.dynamicType)
print(reduceToString.dynamicType)
print(combined.dynamicType)
/*
 Int -> Int
 Int -> Array<Int>
 Array<Int> -> String
 Int -> String
 */

Upvotes: 1

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