如何在Swift中使用没有类型参数的泛型类? [英] How to use a generic class without the type argument in Swift?

问题描述

我想将通用对象封装在另一个类中，而无需设置通用类型参数.我创建了一个基类Animal<T>，并从中定义了其他子类.示例:

I want to encapsulate a generic object in another class without setting the generic type argument. I created a base Animal<T> class and defined other subclasses from it. Example:

public class Animal<T: YummyObject> {
    // Code
}

public class Dog: Animal<Bark> {
    // Code
}

public class Cat: Animal<Meow> {
    // Code
}

并在UITableView扩展名下面定义了一个Animal属性，不带类型参数:

and defined an Animal property, without the type argument, in the UITableView extension bellow:

extension UITableView {
    private static var animal: Animal!

    func addAnimal(animal: Animal) {
        UITableView.animal = animal
    }
}

但是这样做时出现以下编译错误:

but I get the following compile error when doing so:

对泛型类型Animal的引用需要<...>中的参数.

Reference to generic type Animal requires arguments in <...>.

这在 Java 中似乎可以正常工作.如何在 Swift 中完成同一件事?

This seems to work fine in Java. How can I accomplish the same thing in Swift as well?

推荐答案

Swift尚不支持 通配符样式的泛型 就像Java一样(即Animal<?>).因此，常见的模式是定义类型擦除的超类，协议(或包装器)以启用这种用法.例如:

Swift doesn’t yet support wildcard-style generics like Java does (i.e., Animal<?>). As such, a common pattern is to define a type-erased superclass, protocol (or wrapper) to enable such usage instead. For instance:

public class AnyAnimal {
    /* non-generic methods */
}

，然后将其用作您的超类:

and then use it as your superclass:

public class Animal<T: YummyObject>: AnyAnimal {
    ...
}

最后，在您的非通用代码中改为使用AnyAnimal:

Finally, use AnyAnimal in your non-generic code instead:

private static var animal: AnyAnimal!

Swift标准库中的示例.有关实际示例，请参见 KeyPath ， AnyKeyPath 类层次结构.它们遵循我上面概述的相同模式. 集合框架提供了进一步的类型擦除示例，但使用了包装器.

Examples in the Swift Standard Library. For a practical example, see the KeyPath, PartialKeyPath, and AnyKeyPath classes hierarchy. They follow the same pattern I outlined above. The Collections framework provides even further type-erasing examples, but using wrappers instead.

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