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VistasOverload dictionary subscript two times and forward call
I'm trying to extend Dictionary and allow extracting values casted to a certain types and with a given default value. For this I added two overloads for the subscript function, one with a default value, one without:
extension Dictionary {
subscript<T>(_ key: Key, as type: T.Type, defaultValue: T?) -> T? {
// the actual function is more complex than this :)
return nil
}
subscript<T>(_ key: Key, as type: T.Type) -> T? {
// the following line errors out:
// Extraneous argument label 'defaultValue:' in subscript
return self[key, as: type, defaultValue: nil]
}
}
However when calling the three-argument subscript from the two-argument one I get the following error:
Extraneous argument label 'defaultValue:' in subscript
Is this a Swift limitation? Or am I missing something?
I'm using Xcode 10.2 beta 2.
P.S. I know there are other alternatives to this, like dedicated functions or nil coalescing, trying to understand what went wrong in this particular situation.
1 Respuestas
Responde la pregunta0
Subscripts have different rules than functions when it comes to argument labels. With functions, argument labels default to the parameter name – for example if you define:
func foo(x: Int) {}
you would call it as foo(x: 0).
However for subscripts, parameters don't have argument labels by default. Therefore if you define:
subscript(x: Int) -> X { ... }
you would call it as foo[0] rather than foo[x: 0].
Therefore in your example with the subscript:
subscript<T>(_ key: Key, as type: T.Type, defaultValue: T?) -> T? {
// the actual function is more complex than this :)
return nil
}
The defaultValue: parameter has no argument label, meaning that the subscript would have to be called as self[key, as: type, nil]. In order to add the argument label, you need to specify it twice:
subscript<T>(key: Key, as type: T.Type, defaultValue defaultValue: T?) -> T? {
// the actual function is more complex than this :)
return nil
}