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

[Zhao Xin]

I think the best way is

do {

    let a = try couldFailButWillNot()

    if let b = a as? Int {

        print(b)

    }

} catch let error {

    print(error)

}

It is longer, but much clearer.

Below code will work, but not as clear. You don't know whether it is a or b
causes the failure. It is not a good practice to use try? if that is not on
purpose.

if let a = try? couldFailButWillNot(), let b = a as? Int {

    print(b)

}
or

if let a = (try? couldFailButWillNot()) as? Int {

    print(a)

}

Zhaoxin

On Thu, Jul 21, 2016 at 4:25 PM, Sikhapol Saijit via swift-users <
swift-users at swift.org> wrote:

> 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> 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> 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))
> 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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