Hi all,
Yesterday I tried this code:
func couldFailButWillNot() throws -> Any {
return 42
}
if let a = try? couldFailButWillNot() as? Int {
print(a)
}
And was surprised that the output was Optional(42) on both Swift 2 and Swift 3.
I always have the impression that when a variable is resolved with if let it
will never be optional.
So, with a little investigation, I found out that it happens because as? has
higher precedence than try? and is evaluated first.
And the whole expression `try? couldFailButWillNot() as? Int` evaluated as
Optional(Optional(42)).
Also, I’m surprised that try? can be used with non-method-call.
This code: `print(try? 42)` will print Optional(42).
So, the questions are:
1. Is it intentional that try? can be used with a "non-method-call" and return
an optional of the type that follows?
2. Should we design try? to have higher precedence than as? or any operators at
all?
My intuition tells me that
let a = try? couldFailButWillNot() as? Int
should be equivalent to
let a = (try? couldFailButWillNot()) as? Int
3. Do you think that doubly-nested optional (or multi-level-nested optional) is
confusing and should be removed from Swift? (Yes, I’ve seen this blog post
Optionals Case Study: valuesForKeys
<https://developer.apple.com/swift/blog/?id=12>).
For me Optional(nil) (aka Optional.Some(Optional.None))) doesn’t make much
sense.
Maybe, one of the solution is to always have optional of optional merged into a
single level optional? Like Optional(Optional(Optional(42))) should be the
merged to and evaluated as Optional(42).
BTW, the code above is merely for a demonstration. The actual code was more of
something like this:
func parse(JSON: Data) throws -> Any {
// …
}
if let dict = try? parse(JSON: json) as? [String: Any] {
// assume dict is a valid [String: Any] dictionary
// …
}
I’m new to this mailing list so I’m not sure if this belongs here. I’m sorry in
advance if it doesn’t.
Thank you,
Sam_______________________________________________
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