Swift中能否为CustomPublisher类定义Setter实现直接赋值？

我实现了一个类似@Published的不可完成版CurrentValueSubject（CustomPublisher类），现在想不用调用.send(value)，而是通过定义Setter，实现直接用customPublisher = newValue的方式完成赋值，请问这可行吗？

原调用方式：

customPublisher.send(newValue)

目标调用方式：

customPublisher = newValue

原自定义类代码：

final class CustomPublisher<Output: Any>: Publisher {
    typealias Failure = Never

    var value: Output
    private var wrapper: CurrentValueSubject<Output, Failure>

    init(_ value: Output) {
        self.value = value
        wrapper = .init(value)
    }

    func send(_ value: Output) {
        self.value = value
        wrapper.send(value)
    }

    func send(subscription: Subscription) {
        wrapper.send(subscription: subscription)
    }

    func receive<S>(subscriber: S) where S: Subscriber, Failure == S.Failure, Output == S.Input {
        wrapper.receive(subscriber: subscriber)
    }
}

直接通过customPublisher = newValue的语法触发原实例的send逻辑不可行，因为CustomPublisher是类（引用类型），这种赋值操作会直接替换整个实例的引用，而非修改现有实例的内部状态。不过可以通过以下几种方式实现类似的简洁赋值效果：

方式一：利用value属性的Setter触发发送 #

修改value属性的Setter，在赋值时自动调用send方法，只需给value赋值就能完成发送，语法接近目标需求：

final class CustomPublisher<Output: Any>: Publisher {
    typealias Failure = Never

    var value: Output {
        didSet {
            wrapper.send(value)
        }
    }
    private var wrapper: CurrentValueSubject<Output, Failure>

    init(_ value: Output) {
        self.value = value
        wrapper = .init(value)
    }

    func send(_ value: Output) {
        self.value = value
    }

    func send(subscription: Subscription) {
        wrapper.send(subscription: subscription)
    }

    func receive<S>(subscriber: S) where S: Subscriber, Failure == S.Failure, Output == S.Input {
        wrapper.receive(subscriber: subscriber)
    }
}

调用方式：

customPublisher.value = newValue

方式二：改为结构体（值类型）实现直接赋值 #

将CustomPublisher改为结构体，利用值类型的赋值特性，在赋值时触发内部状态更新并发送事件：

struct CustomPublisher<Output: Any>: Publisher {
    typealias Failure = Never

    private var wrapper: CurrentValueSubject<Output, Failure>

    var value: Output {
        get { wrapper.value }
        set { wrapper.send(newValue) }
    }

    init(_ value: Output) {
        wrapper = .init(value)
    }

    func send(_ value: Output) {
        wrapper.send(value)
    }

    func send(subscription: Subscription) {
        wrapper.send(subscription: subscription)
    }

    func receive<S>(subscriber: S) where S: Subscriber, Failure == S.Failure, Output == S.Input {
        wrapper.receive(subscriber: subscriber)
    }

    // 重载赋值运算符，确保赋值时更新内部状态
    static func = (lhs: inout CustomPublisher, rhs: Output) {
        lhs.value = rhs
    }
}

调用方式：

var customPublisher = CustomPublisher(initialValue)
customPublisher = newValue

方式三：实现为属性包装器（最接近@Published的用法） #

如果场景和@Published类似，希望给属性添加发布能力，可以将CustomPublisher改为属性包装器：

@propertyWrapper
struct CustomPublisher<Output: Any>: Publisher {
    typealias Failure = Never

    private var wrapper: CurrentValueSubject<Output, Failure>

    var wrappedValue: Output {
        get { wrapper.value }
        set { wrapper.send(newValue) }
    }

    var projectedValue: some Publisher {
        wrapper
    }

    init(wrappedValue value: Output) {
        wrapper = .init(value)
    }

    func send(subscription: Subscription) {
        wrapper.send(subscription: subscription)
    }

    func receive<S>(subscriber: S) where S: Subscriber, Failure == S.Failure, Output == S.Input {
        wrapper.receive(subscriber: subscriber)
    }
}

使用方式：

class ViewModel {
    @CustomPublisher var name: String = "Initial"
}

let vm = ViewModel()
vm.name = "New Value" // 自动触发发送

内容的提问来源于stack exchange，提问作者Petter Braka