When writing a generic function, it’s sometimes easier to approach it in 3 steps: first write the code stand-alone using a specific type. Then write the code as a function, still with a specific type. Finally, change the function to be generic.
The first part, dividing an array by 3, can be done like this:
let a = [1,2,3,4,5]
// map is run on the array of integers, and returns a new
// array with the operation performed on each element in a:
let b = a.map { $0 / 3 }
// so b will be [0,0,1,1,1]
// (don’t forget, integer division truncates)
Note the closure you provide between the { } is an operation that will be applied to each element of the array. $0 represents the element, and you divide it by 3. You could also write it as a.map { i in i / 3 }.
To put this into its own function:
func divideby3Map(source: [Int]) -> [Int] {
return source.map { $0 / 3 }
}
No need to declare a fresh array – map will create one for you. You can then return that directly (you can assign it to a temporary if you prefer, but that isn’t really necessary).
Finally, if you want to make it generic, start by adding a placeholder:
func divideby3Map<T>(source: [T]) -> [T] {
return source.map { $0 / 3 }
}
Note, there’s only a need for one placeholder, T, because you are returning the exact same type you are passed in.
Except… this won’t compile, because the compiler doesn’t know that T is guaranteed to provide two critical things: the ability to divide (a / operator), and the ability to create new T from integer literals (i.e. to create a T with value 3 to divide by). Otherwise, what if we passed an array of strings or an array of arrays in?
To do this, we need to “constrain” T so our function will only accept as arguments types that provide these features. One such protocol we can use to constrain T is IntegerType, which does guarantee these features (as well as some other ones like +, * etc):
func divideby3Map<T: IntegerType>(source: [T]) -> [T] {
return source.map { $0 / 3 }
}
divideby3Map(a) // returns [0,0,1,1,1]
let smallInts: [UInt8] = [3,6,9]
divideby3Map(smallInts) // returns [1,2,3]
