Parsing value with try? and guard returning different values [duplicate] - swift

Question

This question already has answers here:
Optional binding with try? and as? still produces an optional type
(2 answers)
Closed 5 years ago.
Language Used: Swift 3
Xcode Version: 8.3.2 (8E2002)
I have an extension on Data which parses the data into a JSONObject of type Any
extension Data {
func toJsonObject() -> Any? {
do {
return try JSONSerialization.jsonObject(with: self, options: [])
} catch {
print(error)
}
return nil
}
}
Now the weird thing is when I use guard the result object seems to be different when using toJsonObject() and try?
For example
guard let dictionary = data.toJsonObject() as? [String: Any] else {
return
}
dictionary is now of type [String: Any]
Whereas when I use this:
guard let dictionary = try? JSONSerialization.jsonObject(with: data, options: []) as? [String: Any] else {
return
}
dictionary is now of type [String: Any]?
Isn't the result of the second code block supposed to be [String: Any] instead of the optional [String: Any]?
Is this a mistake on Swift or am I doing something wrong?

Answer 1

This is a feature of Swift, I'd say. The use of try? means that this becomes a double optional, it's trying to decode straight to that casting of as? [String: Any]. Plug this in and check the type on this to see the double optional in action:
// Becomes a type of `[String : Any]??`
let dictionary = try? JSONSerialization.jsonObject(with: Data(), options: []) as? [String: Any]
I think the answer you want is to just add some parentheses to clarify your intent:
guard let dictionary = (try? JSONSerialization.jsonObject(with: Data(), options: [])) as? [String: Any] else {
return
}