在 Objective-C 属性中转换为 Swift 4 的反射错误

Question

收到错误消息

无法转换类型为"不安全的可变指针？(又名"可选<不安全的可变指针>")到指定类型"不安全的可变指针<objc_property_t？>"(又名"不安全的可变指针<可选><不透明指针>>")

在这一行上

let properties : UnsafeMutablePointer <objc_property_t?> = class_copyPropertyList(self.classForCoder, &count)

完整代码在这里

var count = UInt32()
let properties : UnsafeMutablePointer <objc_property_t?> = class_copyPropertyList(self.classForCoder, &count)
var propertyNames = [String]()
let intCount = Int(count)
for i in 0..<intCount {
let property : objc_property_t = properties[i]!
guard let propertyName = NSString(utf8String: property_getName(property)) as? String else {
debugPrint("Couldn't unwrap property name for (property)")
break
}
propertyNames.append(propertyName)
}

Answer 1

您收到错误是因为class_copyPropertyList的返回类型不是UnsafeMutablePointer<objc_property_t?>。

您的行应为：

let properties : UnsafeMutablePointer <objc_property_t> = class_copyPropertyList(self.classForCoder, &count)

Answer 2

class_copyPropertyList()返回UnsafeMutablePointer<objc_property_t>?而不是UnsafeMutablePointer<objc_property_t?>。 通常最好避免显式类型注释，而只写

let properties = class_copyPropertyList(self.classForCoder, &count)

并让编译器推断类型。然后必须解开可选包，例如使用guard：

guard let properties = class_copyPropertyList(self.classForCoder, &count) else {
return // Handle error ...
}

SwiftString的创建也可以简化，从而导致

var count = UInt32()
guard let properties = class_copyPropertyList(self.classForCoder, &count) else {
return
}
var propertyNames = [String]()
for i in 0..<Int(count) {
let propertyName = String(cString: property_getName(properties[i]))
propertyNames.append(propertyName)
}

Answer 3

您可以删除类型注释，如下所示：

var count = UInt32()
let properties = class_copyPropertyList(self.classForCoder, &count)

现在还可以映射属性：

if let properties = class_copyPropertyList(self.classForCoder, &count) {
let range = 0..<Int(count)
let names = range.map {
String(cString: property_getName(properties[$0]))
}
}