I have a protocol that has a static method with a default parameter. I want to change the default value in a class that implements the protocol. Essentially doing what is easily done with classes and super.
I only have a solution when the Protocol has no associated type.
The following code works, but as soon as you uncomment the associated type declaration, it doesn't compile.
protocol Protocol {
// associatedtype AssociatedType
}
extension Protocol {
func sayHello(name: String = "World") {
print("Hello, \(name)!")
}
}
class Class<T>: Protocol {
typealias AssociatedType = T
func sayHello(name: String = "Stack Overflow") {
// Uncommenting the Protocol.AssociatedType causes:
// Protocol can only be used as a generic constraint because it has associated type requirements
(self as Protocol).sayHello(name)
}
}
Class<()>().sayHello()
I do understand why it doesn't compile: Protocol
has no concrete type for AssociatedType
.
So maybe the question should read "Can I explicitly specialize a protocol?", to which I believe the answer is no.
I have a partial workaround. But even when it works, it sucks.
Especially when you consider that I'm writing a library where sayHello
is public, so the following workaround forces me to have a second protocol, which has to be public, but is useless.
Here's the workaround:
protocol Parent {}
protocol Protocol: Parent {
associatedtype AssociatedType
}
extension Parent {
func sayHello(name: String = "World") {
print("Hello, \(name)!")
}
}
class Class<T>: Protocol {
typealias AssociatedType = T
func sayHello(name: String = "Stack Overflow") {
(self as Parent).sayHello(name)
}
}
Class<()>().sayHello()
But this doesn't work for me, because my sayHello
uses the associated type. So it can't be extracted to another protocol.
Just to be sure I'm clear, here's what I'd like, only substituting the class for a protocol:
class Protocol<T> {
func sayHello(name: String = "World") {
print("Hello, \(name)!")
}
}
class Class<T>: Protocol<T> {
override func sayHello(name: String = "Stack Overflow") {
super.sayHello(name)
}
}
Class<()>().sayHello()
Inspired by Rob Napier's answer, here's what I went with; good old overloading for defaults:
This does fit my needs, but doesn't answer the question. So I'm not 100% satisfied.
I believe Rust gets this one right by allowing traits/protocols to be generic both using
X<T>
and associated types.