Re: Simple Swift question
Re: Simple Swift question
- Subject: Re: Simple Swift question
- From: Rick Mann <email@hidden>
- Date: Mon, 29 Jun 2015 15:53:45 -0700
> On Jun 29, 2015, at 15:43 , Greg Parker <email@hidden> wrote:
>
>> On Jun 29, 2015, at 3:18 PM, Rick Mann <email@hidden> wrote:
>>
>> How are you supposed to do simple things like this, succinctly?
>>
>> let url = NSURL(string: "<some url>")
>> let req = NSURLRequest(URL: url)
>> self.webView.loadRequest(req)
>>
>> This, obviously, doesn't work, since the results are optionals, and so need to be unwrapped. But it gets rapidly ridiculous with a bunch of nested if-lets (imagine you had more than just two operations to get from here to there). What's the right way to do this?
>
> If you "know" that the call cannot actually fail (for example, NSURL(string:) with a valid hardcoded URL) then you can simply force-unwrap. If you are wrong then the program will halt at runtime.
> let url = NSURL(string: "<some url>")!
Yep, part of the problem was I was expecting the two type to have the same construction semantics, which isn't the case.
> If you are using Swift 2.0, you can use the new `guard let` to get optional checking without the cascade of indentation.
> guard let url = NSURL(string: "<some url>")
> else { /* handle failure and either halt or return */ }
> // URL is now in scope and non-optional.
> let req = NSURLRequest(URL: url)
I am, but this may not work as well if there are additional steps after this one that should run anyway. Can I enclose the bit in braces to limit the scope of the break? Probably not, huh? e.g.:
{
guard let url = ... else { break }
<use url>
}
{
...other init...
}
Quincey Morris pointed out that I can add multiple let clauses with commas. I tried using &&, and even a where clause, but was being tripped up by NSURL being optional and NSURLRequest not. Here's a contrived example, though. Would this be gross?
let url = NSURL(string: "http://my.matterport.com/show?m=YJrquK9RCiE";)
if let url2 = NSURL(string: url!.absoluteString) where url != nil
{
...
}
>> Oh, I think I figured it out. NSURL(string:) is optional, but NSURLRequest(URL:) can't. Very unexpected, and the error message I was getting did not clue me in.
>
> You can often improve diagnostics by adding the explicit types that you expect each variable to have. That can help pinpoint the place where your understanding of the code and the compiler's understanding of the code don't match.
>
> For example, in your code above, writing `let url: NSURL = …` moves the error message from the NSURLRequest line to the NSURL line.
That sure seems to defeat the purpose of a highly publicized feature of the language. Or are you suggesting just doing it temporarily to figure out what the compiler really wants?
The compiler's message was, "Could not find an overload for 'init' that accepts the supplied arguments." Could it instead say "NSURLRequst.init(URL:) does not return Optional type"?
The confusion stems from it talking about what the init *accepts*, rather than what it returns.
--
Rick Mann
email@hidden
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