如何在类层次结构中应用Swift泛型/协议?
让我们从我要解决的问题开始。我正在将XML文档解析为模型对象的层次结构。所有模型对象都有一个具有一组公共属性的公共基类。然后,每个特定的模型类都有一些附加属性 以下是几个模型类的简化示例:如何在类层次结构中应用Swift泛型/协议?,swift,oop,generics,protocols,Swift,Oop,Generics,Protocols,让我们从我要解决的问题开始。我正在将XML文档解析为模型对象的层次结构。所有模型对象都有一个具有一组公共属性的公共基类。然后,每个特定的模型类都有一些附加属性 以下是几个模型类的简化示例: class Base { var id: String? var name: String? var children = [Base]() } class General: Base { var thing: String? } class Specific: Gener
class Base {
var id: String?
var name: String?
var children = [Base]()
}
class General: Base {
var thing: String?
}
class Specific: General {
var boring: String?
}
class Other: Base {
var something: String?
var another: String?
}
我遇到的问题是实现一种干净的方法来编写XML解析器类来处理这个模型层次结构。我正在尝试编写与模型层次结构匹配的解析器层次结构。以下是我的尝试:
protocol ObjectParser {
associatedtype ObjectType
func createObject() -> ObjectType
func parseAttributes(element: XMLElement, object: ObjectType)
func parseElement(_ element: XMLElement) -> ObjectType
}
class BaseParser: ObjectParser {
typealias ObjectType = Base
var shouldParseChildren: Bool {
return true
}
func createObject() -> Base {
return Base()
}
func parseAttributes(element: XMLElement, object: Base) {
object.id = element.attribute(forName: "id")?.stringValue
object.name = element.attribute(forName: "name")?.stringValue
}
func parseChildren(_ element: XMLElement, parent: Base) {
if let children = element.children {
for child in children {
if let elem = child as? XMLElement, let name = elem.name {
var parser: BaseParser? = nil
switch name {
case "general":
parser = GeneralParser()
case "specific":
parser = SpecificParser()
case "other":
parser = OtherParser()
default:
break
}
if let parser = parser {
let res = parser.parseElement(elem)
parent.children.append(res)
}
}
}
}
}
func parseElement(_ element: XMLElement) -> Base {
let res = createObject()
parseAttributes(element: element, object: res)
if shouldParseChildren {
parseChildren(element, parent: res)
}
return res
}
}
class GeneralParser: BaseParser {
typealias ObjectType = General
override func createObject() -> General {
return General()
}
func parseAttributes(element: XMLElement, object: General) {
super.parseAttributes(element: element, object: object)
object.thing = element.attribute(forName: "thing")?.stringValue
}
}
class SpecificParser: GeneralParser {
typealias ObjectType = Specific
override func createObject() -> Specific {
return Specific()
}
func parseAttributes(element: XMLElement, object: Specific) {
super.parseAttributes(element: element, object: object)
object.boring = element.attribute(forName: "boring")?.stringValue
}
}
还有OtherParser
,它与GeneralParser
相同,只是将General
替换为Other
。当然,在我的层次结构中还有更多的模型对象和相关的解析器
这个版本的代码几乎可以正常工作。您会注意到,GeneralParser
和SpecificParser
类中的parseAttributes
方法没有override
。我认为这是由于对象
参数的类型不同所致。这样做的结果是,不会从BaseParser
的parseElement
方法调用特定于解析器的parseAttributes
方法。我通过将所有parseAttributes
签名更新为:
func parseAttributes(element: XMLElement, object: Base)
然后,在非基本解析器中,我必须使用强制转换(并添加override
,如GeneralParser
中的以下内容:
override func parseAttributes(element: XMLElement, object: Base) {
super.parseAttributes(element: element, object: object)
let general = object as! General
general.thing = element.attribute(forName: "thing")?.stringValue
}
最后,问题是:
如何消除在parseAttributes
方法层次结构中强制转换的需要,并利用协议的关联类型?更一般地说,这是解决此问题的正确方法吗?是否有更“快速”的方法来解决此问题
以下是一些基于此简化对象模型的合成XML(如果需要):
<other id="top-level" name="Hi">
<general thing="whatever">
<specific boring="yes"/>
<specific boring="probably"/>
<other id="mid-level">
<specific/>
</other>
</general>
</other>
以下是我解决此问题的方法:
func createObject(from element: XMLElement) -> Base {
switch element.name {
case "base":
let base = Base()
initialize(base: base, from: element)
return base
case "general":
let general = General()
initialize(general: general, from: element)
return general
case "specific":
let specific = Specific()
initialize(specific: specific, from: element)
return specific
case "other":
let other = Other()
initialize(other: other, from: element)
return other
default:
fatalError()
}
}
func initialize(base: Base, from element: XMLElement) {
base.id = element.attribute(forName: "id")?.stringValue
base.name = element.attribute(forName: "name")?.stringValue
base.children = element.children.map { createObject(from: $0) }
}
func initialize(general: General, from element: XMLElement) {
general.thing = element.attribute(forName: "thing")?.stringValue
initialize(base: general, from: element)
}
func initialize(specific: Specific, from element: XMLElement) {
specific.boring = element.attribute(forName: "boring")?.stringValue
initialize(general: specific, from: element)
}
func initialize(other: Other, from element: XMLElement) {
other.something = element.attribute(forName: "something")?.stringValue
other.another = element.attribute(forName: "another")?.stringValue
initialize(base: other, from: element)
}
我真的不认为需要解析器类的镜像继承层次结构。我最初尝试将initialize
函数作为扩展中的构造函数,但您不能覆盖扩展方法。当然,您可以将它们作为类本身的init
方法,但我假设您希望保留XML与型号代码分开的特定代码
--加成--
我仍然很想知道是否有一个更普遍的解决方案
处理调用重载(非重写)的总体问题
Swift中基类中的方法(如parseAttributes)
你可以用与其他语言相同的方式来做。你可以投射对象(如果需要的话),然后调用方法。在这方面,斯威夫特没有什么神奇或特别之处
class Foo {
func bar(with: Int) {
print("bar with int called")
}
}
class SubFoo: Foo {
func bar(with: String) {
print("bar with string called")
}
}
let foo: Foo = SubFoo()
foo.bar(with: 12) // can't access bar(with: Double) here because foo is of type Foo
(foo as? SubFoo)?.bar(with: "hello") // (foo as? SubFoo)? will allow you to call the overload if foo is a SubFoo
let subFoo = SubFoo()
// can call either here
subFoo.bar(with: "hello")
subFoo.bar(with: 12)
下面是我解决这个问题的方法:
func createObject(from element: XMLElement) -> Base {
switch element.name {
case "base":
let base = Base()
initialize(base: base, from: element)
return base
case "general":
let general = General()
initialize(general: general, from: element)
return general
case "specific":
let specific = Specific()
initialize(specific: specific, from: element)
return specific
case "other":
let other = Other()
initialize(other: other, from: element)
return other
default:
fatalError()
}
}
func initialize(base: Base, from element: XMLElement) {
base.id = element.attribute(forName: "id")?.stringValue
base.name = element.attribute(forName: "name")?.stringValue
base.children = element.children.map { createObject(from: $0) }
}
func initialize(general: General, from element: XMLElement) {
general.thing = element.attribute(forName: "thing")?.stringValue
initialize(base: general, from: element)
}
func initialize(specific: Specific, from element: XMLElement) {
specific.boring = element.attribute(forName: "boring")?.stringValue
initialize(general: specific, from: element)
}
func initialize(other: Other, from element: XMLElement) {
other.something = element.attribute(forName: "something")?.stringValue
other.another = element.attribute(forName: "another")?.stringValue
initialize(base: other, from: element)
}
我真的不认为需要解析器类的镜像继承层次结构。我最初尝试将initialize
函数作为扩展中的构造函数,但您不能覆盖扩展方法。当然,您可以将它们作为类本身的init
方法,但我假设您希望保留XML与型号代码分开的特定代码
--加成--
我仍然很想知道是否有一个更普遍的解决方案
处理调用重载(非重写)的总体问题
Swift中基类中的方法(如parseAttributes)
你可以用与其他语言相同的方式来做。你可以投射对象(如果需要的话),然后调用方法。在这方面,斯威夫特没有什么神奇或特别之处
class Foo {
func bar(with: Int) {
print("bar with int called")
}
}
class SubFoo: Foo {
func bar(with: String) {
print("bar with string called")
}
}
let foo: Foo = SubFoo()
foo.bar(with: 12) // can't access bar(with: Double) here because foo is of type Foo
(foo as? SubFoo)?.bar(with: "hello") // (foo as? SubFoo)? will allow you to call the overload if foo is a SubFoo
let subFoo = SubFoo()
// can call either here
subFoo.bar(with: "hello")
subFoo.bar(with: 12)
给我们一个
XML
给玩具with@Alexander我在问题的末尾添加了一些XML,但您确实不需要它来提供答案,它们不是从哪里调用的?@courteouselk来自BaseParser
的parseElement
方法。您将解析器作为一个单独的层次结构来编写有什么原因吗?例如,我只需要一个必需的init(from:xmldecker)
在您的模型类中,让每个类从给定的XMLDecoder
实例中获取它们的属性值,这将只是一个简单的类型,可以为XML元素的给定属性名提供属性值。然后您只需将此解码器实例传递到模型层次结构中。为我们提供一个XML
with@Alexander我在问题的末尾添加了一些XML,但您确实不需要它来提供答案,它们不是从哪里调用的?@courteouselk来自BaseParser
的parseElement
方法。您将解析器作为一个单独的层次结构来编写有什么原因吗?例如,我只需要一个必需的init(from:xmldecker)
在您的模型类中,让每个类从给定的XMLDecoder
实例获取它们的属性值,这将只是一个简单的类型,可以为XML元素的给定属性名提供属性值。然后,您只需将此解码器实例传递到模型层次结构中。