[swift-users] try? works on non-method-call?

[Sikhapol Saijit]

Hi Nick,

The code was simplified for demonstration.
You’re right that in a simple case like that I wouldn’t need to use `as?`.

In my actual case it was 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
    // …
}

Sure, this could be written as:

if let jsonObject = try? parse(JSON: json), 
let dict = jsonObject as? [String: Any] {
    // …
}

But I just wonder why the first case behave as it does and whether it should.

Sam

> On Jul 21, 2016, at 3:03 PM, Nicholas Outram <nicholas.outram at icloud.com> wrote:
> 
> Are we saying this function 
> 
> func couldFailButWillNot() throws -> Any {
>   return 42
> }
> 
> has an implicit Int return type?
> 
> If so, there is actually no need for as? (which will result in an Optional - which I thought that was the behaviour in Swift 2 as well?)
> 
> 
> 
>> On 21 Jul 2016, at 08:59, Nicholas Outram via swift-users <swift-users at swift.org <mailto:swift-users at swift.org>> wrote:
>> 
>> The issue is related to the as? 
>> 
>> Try this:
>> 
>> if let a = try? couldFailButWillNot()  {
>>    print(a)
>> }
>> 
>> 
>> Nick
>> 
>>> On 21 Jul 2016, at 07:13, Sikhapol Saijit via swift-users <swift-users at swift.org <mailto:swift-users at swift.org>> wrote:
>>> 
>>> Hi Swift Community,
>>> 
>>> 
>>> 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 2.2 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 non-method-call and return an optional of the type that follows?
>>> 
>>> 2. Should we design `try?` to have higher precedence than `as?`. 
>>> My intuition tells me that 
>>> `let a = try? couldFailButWillNot() as? Int`
>>> and 
>>> `let a = (try? couldFailButWillNot()) as? Int` 
>>> should be equivalent.
>>> 
>>> 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 <https://developer.apple.com/swift/blog/?id=12>))
>>> For me “Optional(nil)” (aka “Optional.Some(Optional.None))”) doesn’t make any sense at all. 
>>> 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 Optional(42).
>>> 
>>> 
>>> Thank you
>>> Sikhapol Saijit (Sam)
>>> iOS Developer, Taskworld, Bangkok
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>>> swift-users at swift.org <mailto:swift-users at swift.org>
>>> https://lists.swift.org/mailman/listinfo/swift-users <https://lists.swift.org/mailman/listinfo/swift-users>
>> 
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