Cannot convert value of type '[Int]' to expected argument type 'Int'

Question

I new in swift and now learning this hard as i can. I wrote function that got arguments: numbers and function that work with this numbers:

func anotherSum(_ numbers : Int...) -> Int { return numbers.reduce(0, +) } func makeSomething(_ numbers : Int..., f : (Int...) -> Int) { print(workFunction(numbers)) } makeSomething(1,2,3,4,5,6,7,8, f: anotherSum)

But compile gives out error cannot convert value of type '[Int]' to expected argument type 'Int' . When i tried to change argument like

workFunction : ([Int]) -> Int) and

func anotherSum(_ numbers : [Int]) -> Int

It work perfectly, but i still can't understand why realisation with Int... doesn't work and why compiler gives this error.

As far as the compiler is concerned, Int... and [Int] are completely unrelated types in this context. You need to modify anotherSum to take a [Int] instead, if you want this to work. There's no current way to "splat" arrays in Swift. — Alexander
– Alexander, Commented Dec 11, 2018 at 4:31
Int... will turn into [Int] immediately in the function body. anotherSum(Int...) is supposed to take multiple parameters, while each parameter is an Int. Calling workFunction(numbers) is essentially anotherSum([Int]), you've just passed a single parameter, which is [Int] not Int, to anotherSum. That's why types do not match. — Ricky Mo
– Ricky Mo, Commented Dec 11, 2018 at 4:37

Kamran · Accepted Answer · 2018-12-11 05:09:46Z

As the Int... is considered as an [Int] in the function body so compiler will not allow to pass [Int] in place of Int.... You better calculate the sum as below,

func makeSomething(_ numbers : Int..., workFunction : (Int...) -> Int) { let sum = numbers.map({ workFunction($0)}).reduce(0, +) print(sum) }

Or introduce another method that accepts array of Int and returns sum. Something as below,

func anotherSum(_ numbers : [Int]) -> Int { return numbers.reduce(0, +) } func makeSomething(_ numbers : Int..., workFunction : ([Int]) -> Int) { print(workFunction(numbers)) }

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