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iOS Generics in Swift Generic Functions, Parameters and Constraints Protocol Based Type Constraints

NIKOLA RUSEV
NIKOLA RUSEV
5,293 Points
Posted by NIKOLA RUSEV
NIKOLA RUSEV
5,293 Points

second chalenge??

i try to solve second chalenge but i cant. and i find the code that pass in comunity but i still have some confusions.

  1. every func that will return a square of some int or double will be return also int or double and requarement in chalenge was (T)->U which means T and U should be defferent type

so this func have no sense. func map<T, U>(array: [T], transformation: (T) -> U) -> [U] { var newArray = U for values in array{ newArray.append(transformation(values)) }

return newArray

}

is there something that i missed??

Claire Worrmald
Claire Worrmald
6,442 Points

The two generic types are T and U, they can be any type and the same if you want but I believe they were trying to make you use different ones to prove a point that you can have as many type parameters as you wish. What they were trying to show you is the flexibility of generics, so rather having to change the map function to cater for every type you can pass in any closure/function that conforms to transformation parameter and it will produce the required result. Here is an example of a function returning square numbers that are of type Double and another one summing the values. Its a bit of a confusing question but hopefully I've helped lift some of the mystery behind it.

func map<T, U>(array: [T], transformation:(T) -> U) -> [U]  {
    var transformedArray = [U]()

    for item in array {
        transformedArray.append(transformation(item))
    }

    return transformedArray
}

func square(number: Int) -> Double   {
    return Double(number * number)
}

func sum(number: Int) -> Int    {
    return number + number
}

let numbers = [1,2,3,4,5]

map(array: numbers, transformation: square)
map(array: numbers, transformation: sum)

Its a weird one to wrap your head around, I could be wrong but I believe this challenge is actually mirroring the map closure that Apple provides which I have included below as a reference.

let stringNumbers = numbers.map { String($0) }
print(stringNumbers)

Hopefully I haven't confused you :)

2 Answers

Xavier D
PLUS
Xavier D
Courses Plus Student 5,840 Points
Xavier D
Xavier D
Courses Plus Student 5,840 Points

used a while instead of for

func map<T,U>(array:[T],transformation:(T)->U)->[U]
{
    var repeatwhile=0;var returnArray:[U]=[]
    repeat{returnArray.append((transformation(array[repeatwhile])))
    repeatwhile+=1}while repeatwhile<array.count
    return returnArray
}

func squaring(_ int:Int)->Int{let squared=int*int;return squared}

map(array: [1,2,3,4,5], transformation: squaring)