I can't find the syntax, but I want to do something like this:
class MyClass {
let stringValue: String // filled in later
let integerValue: Int // filled in later
init(stringValue: String) {
self.stringValue = stringValue
self.integerValue = stringValue.hashValue
}
init(integerValue: Int) {
self.integerValue = integerValue
self.stringValue = String(integerValue)
}
}
extension MyClass {
// This is invalid syntax, but I think you can understand
// vvvvvvvvv I'm trying to give back an enum whose type is inferred
var enumValue<T: enum>: T? {
get {
// This is also invalid; I want to check the type of the enum's raw value
if T is String {
return T(rawValue: self.stringValue)
} else if T is Int {
return T(rawValue: self.integerValue)
} else {
return nil
}
}
}
}
The usage would be like:
enum MyEnum: String {
case foo
case bar
}
func baz(_ some: MyClass) {
if let myEnum: MyEnum = some.enumValue {
print(myEnum)
}
}
let some = MyClass(stringValue: "foo")
baz(some) // prints "foo"
Is this possible in Swift? That is, to have a generically-typed field or function whose type is constricted to enums and inferred based on usage, then use that to instantiate an enum value?
A possible solution would be a generic overloaded function:
extension MyClass {
func enumValue<T: RawRepresentable>() -> T? where T.RawValue == String {
return T(rawValue: stringValue)
}
func enumValue<T: RawRepresentable>() -> T? where T.RawValue == Int {
return T(rawValue: integerValue)
}
}
which is then called as
func baz(_ some: MyClass) {
if let myEnum: MyEnum = some.enumValue() {
print(myEnum)
}
}
Alternatively, pass the enum type as an argument:
extension MyClass {
func enumValue<T: RawRepresentable>(_ type: T.Type) -> T? where T.RawValue == String {
return T(rawValue: stringValue)
}
func enumValue<T: RawRepresentable>(_ type: T.Type) -> T? where T.RawValue == Int {
return T(rawValue: integerValue)
}
}
and call it like
func baz(_ some: MyClass) {
if let myEnum = some.enumValue(MyEnum.self) {
print(myEnum)
}
}