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iOS Protocols in Swift Creating Flexible Objects Using Protocols Swift's Standard Library Protocols

Johnny Nguyen
Johnny Nguyen
3,875 Points

Why not using this?

So can I do this? It works. Instead of using Equatable protocol.

user.name == anotherUser.name

1 Answer

Otto linden
Otto linden
5,857 Points

Hello! I think it's because of .name us just a small part of user while user itself need to check a lot of things! 🙂