Swift中未使用的类型参数

Question

Is there a keyword to place-hold unused type parameters?

In this example, receiver does not use T of MyGen . In Java language, it can be written MyGen<?> v . I could not find a counterpart in swift language documents.

import Foundation
class MyGen<T: Printable> {
    var value: T
    init(v: T) {
        value = v
    }
}

func receiver(v: MyGen<WHAT_COMES_HERE>) {
    println(v);
}

let s = MyGen<NSString>(v: "hello")
receiver(s)

I know that giving receiver a type parameter solves the problem, but it is not welcome because upper bound Printable is repeated as many as functions and the code has redundant information.

// This works, but not welcome
func receiver<T: Printable>(v: MyGen<T>) {
    println(v);
}

Answer 1

In Swift, despite it not being used, you have to declare a type parameters. But, you don't need :Printable in here, You can use just T or whatever as you like.

func receiver<A>(v: MyGen<A>) {
    println(v)
}

Use : for the type parameter, only if receiver requires more specialized MyGet.T . For example:

class MyGen<T: Printable> {
    var value: T
    init(v: T) {
        value = v
    }
}

// `receiver` accepts `MyGen` only if its `T` is a sequence of `Int` 
func receiver<A: SequenceType where A.Generator.Element == Int>(v: MyGen<A>) {
    println(reduce(v.value, 0, +))
}

let s = MyGen<[Int]>(v: [1,2,3])
receiver(s) // ->  6