关于performSelector返回值

Question

大家都知道ios在swift2.0中支持" performSelector“我想知道怎样才能得到performSelector的返回值？

let returnValue:Unmanaged<AnyObject> = (self.dataSource?.performSelector("myselector:", withObject: self))!
let value : Int? = returnValue.takeRetainedValue() as? Int

在最后一行出现崩溃，如何在swift2.0中从"performSelector“中获取值？

Answer 1

我认为目前Swift中的performSelector只支持符合AnyObject的返回值，在你的例子中，你有Int，它不支持。在这种情况下，performSelector将返回nil，并且您的代码将显式地展开可选的，从而导致程序崩溃。

因此，解决此问题的一种方法是创建一个返回NSNumber的方法。

您可以在游乐场中尝试以下代码：

@objc class B: NSObject {
  func myselector(arg: A) -> NSNumber {
    return 0
  }
}

@objc class A: NSObject {
  var dataSource: B? = B()
  func a() -> Int {
    let returnValue = self.dataSource?.performSelector("myselector:", withObject: self)
    let value = returnValue?.takeRetainedValue() as! NSNumber
    return value.integerValue
  }
}

A().a()

Answer 2

使用takeUnretainedValue

let foo = Foo()
let value = foo.performSelector(Selector("bar")) //type of value is Unmanaged<AnyObject?>
let uvalue = value.takeUnretainedValue() //type of uvalue is now AnyObject?
let result = value as? String //type of result is now String

针对Swift 5进行了更新：

同样适用于NSSelectorFromString:假设您有一个名为aViewController的UIViewController。

let value = aViewController.perform(NSSelectorFromString("view")) //type of value is Unmanaged<AnyObject?>
let uvalue = value.takeUnretainedValue() //type of uvalue is now AnyObject?
let myView = value as? UIView //type of result is now UIView

Answer 3

您可以通过以下方式获取performSelector的NSInteger结果

let returnValue = dataSource?.perform(#selector(myselector:), withObject: self)!
let value = Int(bitPattern: returnValue.toOpaque())