结构中的快速类在分配期间是不是通过副本传递？

Posted 2023-02-25

技术标签:

【中文标题】结构中的快速类在分配期间是不是通过副本传递？【英文标题】：Are swift classes inside struct passed by copy during assignment?结构中的快速类在分配期间是否通过副本传递？ 【发布时间】：2017-04-24 01:19:34 【问题描述】：

如果我在 swift 中有一个带有类属性的结构，并且我复制了该结构对象，那么类属性是复制还是通过引用传递？

【问题讨论】：

【参考方案1】：

通过引用传递。你可以测试一下。声明：

class A
struct B  let a = A()

然后：

let b = B()
print("A = \(unsafeAddressOf(b.a))")//0x0000600000019450
let b_copy = b
print("A = \(unsafeAddressOf(b_copy.a))")//0x0000600000019450

【讨论】：

要使其与 swift 5 一起使用：print(Unmanaged.passUnretained(b.a).toOpaque()) 为什么不再调用init？【参考方案2】：

结构体的所有属性在复制结构体时被复制（就像您将旧结构体的每个属性分配 (=) 到新结构体的相应属性一样），无论输入。

当您说“类属性”时，我假设您的意思是引用类型的变量。（与类同名的类型表示指向该类对象的引用的引用类型。）复制引用类型的值（引用）会产生另一个指向同一对象的引用。请注意，“对象”在 Swift 中不是值——没有“对象类型”——相反，对象总是通过指向它们的引用来操作。

【讨论】：

如果对象是值类型对象（结构）并且其内容中包含少于 3 个单词（ivars），则可以在没有指向它们的引用的情况下表示对象。这些 ivars 的类型必须允许它们在堆栈上分配。在其他情况下，对象将使用相应的指针在堆上分配。来源developer.apple.com/videos/play/wwdc2016/416【参考方案3】：

我在 swift 5 中测试了上述实验：让我们看看结果：

class A 
    var id: Int
    init(id: Int) 
        self.id = id
    


struct B 
    var grade: Int
    var a: A

根据结果进行实验：

var a = A(id: 1)
var a_copy = a

var b1 = B(grade: 2, a: a)
var copy_b1 = b1

print(b1.a.id)
b1.a.id = 5
print(copy_b1.a.id)

print(b1.grade)
b1.grade = 3
print(copy_b1.grade)

输出：

1
5 // call by reference, same result
2
2 // call by value, no change in result

结论：

struct 在我们创建它的另一个对象时会复制。它复制其结构属性（按值调用）但引用相同的类属性实例（按引用调用）

通过地址做实验：

在类上做一个实验：

var a = A(id: 1)
var a_copy = a

withUnsafePointer(to: &a)  (address) in
    print("address of a (class) = \(address)")

withUnsafePointer(to: &a_copy)  (address) in
    print("address of a_copy (class) = \(address)")

withUnsafePointer(to: &a.id)  (address) in
    print("address of a.id (struct) = \(address)")

withUnsafePointer(to: &a_copy.id)  (address) in
    print("address of a_copy.id (struct) = \(address)")

输出

address of a (class) = 0x0000000114747f80
address of a_copy (class) = 0x0000000114747f88
address of a.id (struct) = 0x000060000285a390
address of a_copy.id (struct) = 0x000060000285a390

观察 1：

类的两个实例都指向其属性的相同位置。

让我们在 struct 上做实验：

print("\n\n\n")
withUnsafePointer(to: &b1)  (address) in
    print("address of b1 (struct) = \(address)")

withUnsafePointer(to: &b1.grade)  (address) in
    print("address of b1.grade (struct) = \(address)")

withUnsafePointer(to: &b1.a)  (address) in
    print("address of b1.a (class) = \(address)")

withUnsafePointer(to: &b1.a.id)  (address) in
    print("address of b1.a.id (class) = \(address)")

输出：

address of b1 (struct) = 0x0000000109382770
address of b1.grade (struct) = 0x0000000109382770
address of b1.a (class) = 0x0000000109382778
address of b1.a.id (class) = 0x0000600001e5cfd0

print("\n\n\n")
withUnsafePointer(to: &copy_b1)  (address) in
    print("address of copy_b1 (struct) = \(address)")

withUnsafePointer(to: &copy_b1.grade)  (address) in
    print("address of copy_b1.grade (struct) = \(address)")

withUnsafePointer(to: &copy_b1.a)  (address) in
    print("address of copy_b1.a (class) = \(address)")

withUnsafePointer(to: &copy_b1.a.id)  (address) in
    print("address of copy_b1.a.id (class) = \(address)")

输出：

address of copy_b1 (struct) = 0x0000000109382780
address of copy_b1.grade (struct) = 0x0000000109382780
address of copy_b1.a (class) = 0x0000000109382788
address of copy_b1.a.id (class) = 0x0000600001e5cfd0