Why can not use protocol `Encodable` as a type in the func
Solution 1.
https://github.com/satishVekariya/SVCodable
Try this code, which extend encodable
extension Encodable {
func toJSONData() -> Data? { try? JSONEncoder().encode(self) }
}
Solution 2.
To avoid polluting Apple-provided protocols with extensions
protocol MyEncodable: Encodable {
func toJSONData() -> Data?
}
extension MyEncodable {
func toJSONData() -> Data?{ try? JSONEncoder().encode(self) }
}
Use
var dataSource2: Encodable?
dataSource2 = TestClass2()
let data = dataSource2?.toJSONData()
There are a number of approaches to solving this problem.
@SPatel solution of extending Encodable
is one possibility. However, I personally try to avoid polluting Apple-provided protocols with extensions.
If I am reading between the lines, it appears what you are wanting is to pass any construct that conforms to Encodable
to a function/method in some other struct/class.
Let's take an example of what I think you are trying to achieve:
struct Transform {
static func toJson(encodable: Encodable) throws -> Data {
return try JSONEncoder().encode(encodable)
}
}
However, Xcode will complain:
Protocol type 'Encodable' cannot conform to 'Encodable' because only concrete types can conform to protocols
A Swift-ier solution is to use a constrained generic on the function:
struct Transform {
static func toJson<EncodableType: Encodable>(encodable: EncodableType) throws -> Data {
return try JSONEncoder().encode(encodable)
}
}
Now the compiler can infer the type that conforms to Encodable
, and we can call the function as intended:
let dataSource = TestClass2()
let jsonData = try? Transform.toJson(encodable: dataSource)
You can't pass a protocol but you can use generics to require a class that conforms to one:
func printJSON<T: Encodable>(_ data: T) {
if let json = try? JSONEncoder().encode(data) {
if let str = String(data: json, encoding: .utf8) {
print(str)
}
}
}
// Now this should work
var dataSource2 = TestClass2()
printJSON(dataSource2!)