swift - 当只有符合较小类型的变量时如何调用重载函数？

Question

在给定符合每个重载参数的公共(public)协议(protocol)的变量的情况下，Swift 是否有办法正确推断要调用哪个重载函数？

在下面的代码示例中，我希望能够将类型为 Command 的通用对象传递给许多可能能够处理它的其他对象。

但是，当一个对象向下转换为基本协议(protocol)时，类型信息就会“丢失”。这使我无法根据命令类型使用函数重载。

有没有办法“返回”该类型信息，以便 Swift 有足够的信息来正确调用适当的函数？

protocol Command {}
protocol ToDoCommand : Command {}
protocol UserCommand : Command {}

struct AddToDoCommand  : ToDoCommand {}
struct EditToDoCommand : ToDoCommand {}

struct AddUserCommand  : UserCommand {}
struct EditUserCommand : UserCommand {}

class ToDoManager {
  func performCommand(_ command:Command) {
    guard let todoCommand = command as? ToDoCommand else {
      return
    }

    // Perform some tasks that are common to all ToDoCommands...

    // This produces a compiler error because 'todoCommand' is of 
    // type ToDoCommand, which is not specific enough for Swift
    // to deduce which overloaded function to call. Can this
    // be overcome?

    performToDoCommand(todoCommand)
  }

  func performToDoCommand(_ command:AddToDoCommand) {
    print("Add ToDo")
  }

  func performToDoCommand(_ command:EditToDoCommand) {
    print("Edit ToDo")
  }
}

class UserManager {
  func performCommand(_ command:Command) {
    guard let userCommand = command as? UserCommand else {
      return
    }

    // Perform some tasks that are common to all UserCommands...

    // See note above...
    performUserCommand(userCommand)
  }

  func performUserCommand(_ command:AddUserCommand) {
    print("Add User")
  }

  func performUserCommand(_ command:EditUserCommand) {
    print("Edit User")
  }

}

let todoManager = ToDoManager()
let userManager = UserManager()

let command = AddUserCommand()
todoManager.performCommand(command)
userManager.performCommand(command)

Answer 1

有几种方法可以解决这个问题......

使用开关重新建立类型

Swift 需要在编译时知道它正在调用哪个重载函数。如果 Swift 在编译时不知道它拥有哪种类型的变量，这就不会发生。

要取回类型信息，可以使用开关重新建立类型:

func performCommand(_ command:Command) {
    guard let todoCommand = command as? ToDoCommand else {
        return
    }

    // Perform some tasks that are common to all ToDoCommands...

    switch todoCommand {
    case let command as AddToDoCommand:
        performCommand(command)
    case let command as EditToDoCommand:
        performCommand(command)
    default: break
    }
}

---

使用多态性

让 Swift 决定在运行时运行哪个 performToDoCommand() 命令的方法是使用多态性。

将实现func performToDoCommand() 的要求添加到ToDoCommand 协议(protocol)，然后为符合 的每个struct 实现它>ToDo 命令。调用正确的方法很简单...

protocol Command {}
protocol ToDoCommand : Command {
    func performToDoCommand()
}
protocol UserCommand : Command {}

struct AddToDoCommand  : ToDoCommand {
    func performToDoCommand() {
        print("Add ToDo")
    }
}

struct EditToDoCommand : ToDoCommand {
    func performToDoCommand() {
        print("Edit ToDo")
    }
}

struct AddUserCommand  : UserCommand {}
struct EditUserCommand : UserCommand {}

class ToDoManager {
    func performCommand(_ command:Command) {
        guard let todoCommand = command as? ToDoCommand else {
            return
        }

        todoCommand.performToDoCommand()
    }
}