Swift 的 Optional 如何比较包装类型的关系呢？

[sl0000]

var a: Int? = nil
print("\(type(of: a))")
print("\(a is String?)")

var b: Int? = 1
print("\(type(of: b))")
print("\(b is String?)")

/*
output::
Optional<Int>
true
Optional<Int>
false
 */

下面是 GPT 的答案

Imgur

[sl0000]

[img][/img]

[nobodyknows]

```swift
func isTypeEqual<T1, T2>(_ v: T1, _ t: T2.Type) -> Bool {
return ObjectIdentifier(T1.self) == ObjectIdentifier(T2.self)
}

func test() {
let a: Int? = nil
print("\(isTypeEqual(a, Optional<Int>.self))")
print("\(isTypeEqual(a, Optional<String>.self))")
}
```

[sl0000]

@nobodyknows 感谢，可能是我描述不够清晰，我其实是要做 optional<subclass>与 optional<class>的继承关系比较，而不是==

protocol OptionalProtocol {
static func wrappedType() -> Any.Type
func wrappedType() -> Any.Type
}
extension Optional: OptionalProtocol {
static func wrappedType() -> Any.Type {
return Wrapped.self
}
func wrappedType() -> Any.Type {
return Wrapped.self
}
}


class KeyValueCoding {
init() {
defaultKeyValue()
}

func defaultKeyValue() {
// let userDefault = UserDefaults.standard
let classPath = String(describing: self)
let mirror = Mirror(reflecting: self)
for child in mirror.children {
guard let label = child.label else { continue }
let fullLabel = classPath + "." + label
let type = type(of: child.value)
let value = child.value
print("\(fullLabel) \(type)")
if let wrappedType = (type as? OptionalProtocol.Type)?.wrappedType() {
print("optional \(wrappedType)")
} else {
print("not optional")
}
}
}
}

[agagega]

也许可以利用子类对象可以 upcast 到父类对象这一点？