在目标C中不可见的swift init [英] swift init not visible in objective-C

问题描述

我试图在Swift中创建init函数，并从Objective-C创建实例.问题是我在Project-Swift.h文件中看不到它，并且在初始化时找不到该函数.我有一个定义如下的函数:

I'm trying to create init functions in Swift and create instances from Objective-C. The problem is that I don't see it in Project-Swift.h file and I'm not able to find the function while initializing. I have a function defined as below:

public init(userId: Int!) {
    self.init(style: UITableViewStyle.Plain)
    self.userId = userId
}

我什至尝试放置@objc(initWithUserId:)，并且再次出现相同的错误.还有什么我想念的吗?如何使Objective-C代码可见构造函数?

I even tried putting @objc(initWithUserId:) and I keep getting the same error again. Is there anything else I'm missing? How do I get the constructor visible to Objective-C code?

我为此阅读了以下内容:

I read the below for this:

https://developer.apple.com/library/ios/documentation/swift/conceptual/swift_programming_language/Initialization.html

https://developer.apple. com/library/ios/documentation/swift/conceptual/buildingcocoaapps/interactingwithobjective-capis.html

如何在Swift中编写Init方法

如何在Swift协议中定义可选方法?

推荐答案

您看到的问题是Swift无法桥接可选值类型-Int是值类型，因此Int!不能被桥接.可选引用类型(即任何类)可以正确桥接，因为它们在Objective-C中始终可以是nil.您的两个选择是使参数成为非可选参数，在这种情况下，它将作为int或NSInteger桥接到ObjC:

The issue you're seeing is that Swift can't bridge optional value types -- Int is a value type, so Int! can't be bridged. Optional reference types (i.e., any class) bridge correctly, since they can always be nil in Objective-C. Your two options are to make the parameter non-optional, in which case it would be bridged to ObjC as an int or NSInteger:

// Swift
public init(userId: Int) {
    self.init(style: UITableViewStyle.Plain)
    self.userId = userId
}

// ObjC
MyClass *instance = [[MyClass alloc] initWithUserId: 10];

或使用可选的NSNumber?，因为可以将其桥接为可选值:

Or use an optional NSNumber?, since that can be bridged as an optional value:

// Swift
public init(userId: NSNumber?) {
    self.init(style: UITableViewStyle.Plain)
    self.userId = userId?.integerValue
}

// ObjC
MyClass *instance = [[MyClass alloc] initWithUserId: @10];    // note the @-literal

但是，请注意，您实际上并没有像对待可选参数那样对待参数-除非self.userId也是可选参数，否则您将自己设置为可能会导致运行时崩溃.

Note, however, you're not actually treating the parameter like an optional - unless self.userId is also an optional you're setting yourself up for potential runtime crashes this way.

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