Why can not use protocol `Encodable` as a type in

问题:

I'm trying to get data by encode model which conforms to Encodable protocol. But it's failed to invoke func encode like code below:

// MARK: - Demo2

class TestClass2: NSObject, Encodable {
    var x = 1
    var y = 2
}


var dataSource2: Encodable?

dataSource2 = TestClass2()

// error: `Cannot invoke 'encode' with an argument list of type '(Encodable)'`
let _ = try JSONEncoder().encode(dataSource2!)
//func encode<T>(_ value: T) throws -> Data where T : Encodable

But in another demo, it works well, why?

// MARK: - Demo1

protocol TestProtocol {
    func test()
}

class TestClass1: NSObject, TestProtocol {
    func test() {
        print("1")
    }

    var x = 1
    var y = 2
}


var dataSource1: TestProtocol?

dataSource1 = TestClass1()


func logItem(_ value: TestProtocol) {
    value.test()
}

logItem(dataSource1!)

回答1:

Try this code, which extend encodable

extension Encodable {
    func toJSONData() -> Data? {
        return try? JSONEncoder().encode(self)
    }
}

Use

var dataSource2: Encodable?
dataSource2 = TestClass2()
let data = dataSource2?.toJSONData()

回答2:

Your 2 examples are different.

JSONEncoder().encode() expects a concrete class which conforms to the procotol Encodable. The reference dataSource2 holds a protocol and not a concrete class.

logItem on the other hands, only takes a protocol as input, and NO concrete class which conforms to the protocol. This is the difference between your examples and why your second case is working and the first case does not.

With your current setup, it will not work. You need to pass in a concrete class to the JSONEncoder.

回答3:

As long as TestClass2 is Encodable you can use following code. encode should know what to encode. It refers to class properties to do that. Encodable itself doesn't contain any properties.

JSONEncoder().encode(dataSource2 as! TestClass2)