Can i create an instance of an protocol in swift?
Like in java an instance of an interface?
Java:
public interface test {
void test();
}
new test() {
@Override
public void test() {
//...
}
}
Swift:
protocol ITransmitter {
func onExecuteSuccess(data:String)
}
//instance???
You can not create an instance of protocol.
For example
protocol ITransmitter {
func onExecuteSuccess(data:String)
}
var protocolInstance : ITransmitter = ITransmitter() // << Not allowed. This is an error
But however you can refer an object in your code using Protocol as the sole type. Let us say you have a class that conforms to this protocol, but in your code your requirement is only to be able to call the protocol method on it and you don't care any other methods the instance of the class supports.
For example-
class A{
func foo(){
}
}
extension A : ITransmitter{
func onExecuteSuccess(data:String){
//Do stuff here
}
}
//This function wants to run the ITransmitter objects, so it uses only protocol //type for its argument. The transmitter can be of any class/struct, but has to //conform to ITransmitter protocol
func runTransmittor(transmitter : ITransmitter){
//some other statements here..
transmitter. onExecuteSuccess(data :SomeData){
}
}
I created inner class which conforms to protocol and assigned object of that class
class Outer {
private final class Inner: ITransmitter {
unowned let parent: Outer
init(parent: Outer) {
self.parent = parent
}
func onExecuteSuccess(data:String){
}
}
var inner: Inner! = nil
init() {
inner = Inner(parent: self)
}
}
