玩Swift,来自Java背景,为什么要选择Struct而不是Class?看起来它们是一样的东西,只不过Struct提供的功能更少。那为什么选择它呢?
当前回答
以下是一些值得考虑的其他原因:
struct有一个自动初始化式,你根本不需要在代码中维护它。 struct MorphProperty { var类型:MorphPropertyValueType var键:字符串 var值:AnyObject enum MorphPropertyValueType { case字符串,Int, Double } } var m = MorphProperty(类型:.Int,键:“什么”,值:“blah”)
要在类中得到这个,你必须添加初始化式,并维护初始化式…
Basic collection types like Array are structs. The more you use them in your own code, the more you will get used to passing by value as opposed to reference. For instance: func removeLast(var array:[String]) { array.removeLast() println(array) // [one, two] } var someArray = ["one", "two", "three"] removeLast(someArray) println(someArray) // [one, two, three] Apparently immutability vs. mutability is a huge topic, but a lot of smart folks think immutability -- structs in this case -- is preferable. Mutable vs immutable objects
其他回答
假设我们知道Struct是值类型,Class是引用类型。
如果你不知道值类型和引用类型是什么,那么看看按引用传递和按值传递之间的区别是什么?
根据mikeash的帖子:
... Let's look at some extreme, obvious examples first. Integers are obviously copyable. They should be value types. Network sockets can't be sensibly copied. They should be reference types. Points, as in x, y pairs, are copyable. They should be value types. A controller that represents a disk can't be sensibly copied. That should be a reference type. Some types can be copied but it may not be something you want to happen all the time. This suggests that they should be reference types. For example, a button on the screen can conceptually be copied. The copy will not be quite identical to the original. A click on the copy will not activate the original. The copy will not occupy the same location on the screen. If you pass the button around or put it into a new variable you'll probably want to refer to the original button, and you'd only want to make a copy when it's explicitly requested. That means that your button type should be a reference type. View and window controllers are a similar example. They might be copyable, conceivably, but it's almost never what you'd want to do. They should be reference types. What about model types? You might have a User type representing a user on your system, or a Crime type representing an action taken by a User. These are pretty copyable, so they should probably be value types. However, you probably want updates to a User's Crime made in one place in your program to be visible to other parts of the program. This suggests that your Users should be managed by some sort of user controller which would be a reference type. e.g struct User {} class UserController { var users: [User] func add(user: User) { ... } func remove(userNamed: String) { ... } func ... } Collections are an interesting case. These include things like arrays and dictionaries, as well as strings. Are they copyable? Obviously. Is copying something you want to happen easily and often? That's less clear. Most languages say "no" to this and make their collections reference types. This is true in Objective-C and Java and Python and JavaScript and almost every other language I can think of. (One major exception is C++ with STL collection types, but C++ is the raving lunatic of the language world which does everything strangely.) Swift said "yes," which means that types like Array and Dictionary and String are structs rather than classes. They get copied on assignment, and on passing them as parameters. This is an entirely sensible choice as long as the copy is cheap, which Swift tries very hard to accomplish. ...
我个人不会这样命名我的类。我通常将我的命名为UserManager而不是UserController,但想法是一样的
另外,当你必须重写一个函数的每个实例(即它们没有任何共享功能)时,不要使用类。
所以不是一个类的几个子类。使用几个符合协议的结构体。
使用结构体的另一种合理情况是,当你想对新旧模型进行delta/diff运算时。对于引用类型,你不能开箱即用。对于值类型,突变是不共享的。
结构vs类
[堆栈vs堆] [值vs参考类型]
结构更可取。但是缺省情况下,Struct并不能解决所有问题。通常你会听说值类型是在堆栈上分配的,但这并不总是正确的。只有局部变量被分配到堆栈上
//simple blocks
struct ValueType {}
class ReferenceType {}
struct StructWithRef {
let ref1 = ReferenceType()
}
class ClassWithRef {
let ref1 = ReferenceType()
}
func foo() {
//simple blocks
let valueType1 = ValueType()
let refType1 = ReferenceType()
//RetainCount
//StructWithRef
let structWithRef1 = StructWithRef()
let structWithRef1Copy = structWithRef1
print("original:", CFGetRetainCount(structWithRef1 as CFTypeRef)) //1
print("ref1:", CFGetRetainCount(structWithRef1.ref1)) //2 (originally 3)
//ClassWithRef
let classWithRef1 = ClassWithRef()
let classWithRef1Copy = classWithRef1
print("original:", CFGetRetainCount(classWithRef1)) //2 (originally 3)
print("ref1:", CFGetRetainCount(classWithRef1.ref1)) //1 (originally 2)
}
*你不应该使用/依赖retainCount,因为它没有提供有用的信息
检查堆栈或堆
在编译过程中SIL(Swift中间语言)可以优化你的代码
swiftc -emit-silgen -<optimization> <file_name>.swift
//e.g.
swiftc -emit-silgen -Onone file.swift
//emit-silgen -> emit-sil(is used in any case)
//-emit-silgen Emit raw SIL file(s)
//-emit-sil Emit canonical SIL file(s)
//optimization: O, Osize, Onone. It is the same as Swift Compiler - Code Generation -> Optimization Level
在那里你可以找到alloc_stack(在堆栈上的分配)和alloc_box(在堆上的分配)
(优化级别(SWIFT_OPTIMIZATION_LEVEL)]
对于类,您获得继承并通过引用传递,而结构则没有继承并通过值传递。
有很多关于Swift的WWDC会议,其中一个会议详细回答了这个问题。确保你看了这些,因为它会让你更快地跟上语言指南或iBook。
在Swift中,引入了一种新的编程模式,称为面向协议编程。
创建型模式:
在swift中,Struct是一种自动克隆的值类型。因此,我们可以免费获得实现原型模式所需的行为。
而类是引用类型,在赋值过程中不会自动克隆。为了实现原型模式,类必须采用NSCopying协议。
浅拷贝只复制指向那些对象的引用,而深拷贝复制对象的引用。
为每种引用类型实现深度复制已成为一项乏味的任务。如果类包含进一步的引用类型,我们必须为每个引用属性实现原型模式。然后我们需要通过实现NSCopying协议复制整个对象图。
class Contact{
var firstName:String
var lastName:String
var workAddress:Address // Reference type
}
class Address{
var street:String
...
}
通过使用结构体和枚举,我们使我们的代码更简单,因为我们不需要实现复制逻辑。
许多Cocoa api需要NSObject子类,这迫使你使用class。但除此之外,你可以使用以下苹果Swift博客中的案例来决定是使用struct / enum值类型还是类引用类型。
https://developer.apple.com/swift/blog/?id=10