迅捷-具有继承的泛型类 [英] swift - Generic classes with inheritance

问题描述

当我尝试执行以下代码时，出现以下错误

When I try to execute the code below, I get the following error

错误:无法将类型"X"的值转换为指定的类型 'X'

error: cannot convert value of type 'X' to specified type 'X'

不迅速支持泛型继承吗?有解决方法吗?

Doesn't swift support inheritance with generics? Is there a workaround for this?

class Parent{ }

class Child:Parent{ }

class X<T>{
    var name: String?
}

var test:X<Parent> = X<Child>() //Compiler Error

推荐答案

在Swift中，泛型是不变，例如无论A和B之间的继承关系如何，任何X<A>都不会分配给X<B>.

In Swift, generics are invariant, e.g. any X<A> will never be assignable to X<B>, regardless of the inheritence relationship between A and B.

尽管如此，该规则还是有一些例外，涉及数组和可选项(以及其他一些类型):

Nevertheless, there are some exceptions to this rule, regarding Arrays and Optionals (and mabye some other types):

var array2:[Parent] = [Child]()
// same as:
var array1:Array<Parent> = Array<Child>()

var opt1:Parent? = Child()
// same as:
var opt2:Optional<Parent> = Optional<Child>(Child())

这些将编译(自Swift 3起)-但是这些特殊情况由编译器的一些硬编码规则处理.

These will compile (since Swift 3) - but these a special cases treated by some hard-coded rules of the the compiler.

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