One of my longstanding frustrations with generic types and protocols has been
how hard it is to work with them when their type is unspecified.
Often I find myself wishing that I could write a function that takes a generic
type or protocol as a parameter, but doesn’t care what its generic type is.
For example, if I have a type:
struct Foo<T> {
let name: String
let value: T
}
or:
protocol Foo {
associatedtype T
var name: String { get }
var value: T { get }
}
And I want to write a function that only cares about Foo.name, I’d like to be
able to:
func sayHi(to foo: Foo) {
print("hi \(foo.name)")
}
But instead I get the error, “Reference to generic type Foo requires arguments
in <…>”
Also, when you want to have a polymorphic array of generic types, you can’t:
let foos: [Foo] = [Foo(name: "Int", value: 2), Foo(name: "Double", value: 2.0)]
And if you remove the explicit type coercion, you just get [Any]
let foos = [Foo(name: "Int", value: 2), Foo(name: "Double", value: 2.0)]
I wish that could be inferred to be [Foo]. I’d like to propose being able to
use the non-generic interface of a type normally.
I.e. if you have a type Foo<T>, it is implicitly of type Foo as well. The type
Foo could be used like any other type.
It could be a parameter in a function, a variable, or even the generic type of
another type (like a Dictionary<String, Foo>)
The only restriction is that if you want to call or access, directly or
indirectly, a function or member that requires the generic type,
the generic type would have to be known at that point.
Foo<T> should be able to be implicitly casted to Foo wherever you want, and Foo
could be cast to Foo<T> conditionally.
Initializers would still obviously have to know the generic type, but given the
above example, you should be able to:
let names = foos.map { $0.name }
However, you could not do the following:
let foos = [Foo]()
Because the initializer would need to know the generic type in order to
allocate the memory.
Let me know what you think!
—
Logan Shire
iOS @ Lyft
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