Consider the following example: let originalURL: NSURL = NSURL(string: "http://apple.com/iphone";)! let newURL = NSURL(string: "http://apple.com\(originalURL.path)/help")
What's the newURL? Nil, because it was being inited with http://apple.comOptional(/iphone)/help Since the path property is optional. Which is not something you figure out until runtime, wondering why the URL is nil. This is very annoying when you run into this issue repeatedly on several occasions because you forget to unwrap the value. *This* is IMHO an undesired and unexpected behavior. If interpolating optionals did become a warning, you'd immediately know about a potential bug: you don't check if path != nil. BTW if you still want the original behavior of printing Optional(my string value), you can still write "http://apple.com\(originalURL.path.debugDescription)/help" which invokes debugDescription on the Optional, not the String (since there is no "?"), and you get the original value anyway. And this could be the Fix-It for it as well to maintain original behavior. Krystof > On May 19, 2016, at 9:28 AM, Dan Appel <[email protected]> wrote: > > You know what's worse than seeing "Optional(my string value)" in a label? > Seeing "nil". When the optional is there, it is made clear to the developer > that the string they are showing can be nil. However, if you hide that from > the users you are less likely to unwrap that optional and hence more likely > to show the user "nil". This behavior really goes against some of the core > ideas of Swift - you want your code to be expressive but not abstract away > potentially useful information. > > On Thu, May 19, 2016 at 12:24 AM David Waite <[email protected] > <mailto:[email protected]>> wrote: > Making string interpolation reject just optional (at compile time) when it > doesn’t reject any other type sounds tricky to express. > > What if instead Optional just didn’t decorate the wrapped value, outputting > either the inner value or “nil” in these cases? > > The debugDescription could remain "Optional(data)" style. > > -DW > >> On May 19, 2016, at 12:52 AM, Valentin via swift-evolution >> <[email protected] <mailto:[email protected]>> wrote: >> >> From what I understand of this thread, the argument here is that directly >> using an optional in a string interpolation is almost never what you really >> want to do (except mainly for debugging purposes) but you wouldn't see this >> mistake until much later at runtime. >> And I feel like one of Swift goals is to enable us, imperfect human >> creatures, to detect as many problems or mistakes as possible long before >> runtime. >> >> On 19 mai 2016, at 00:56, Dan Appel via swift-evolution >> <[email protected] <mailto:[email protected]>> wrote: >> >>> -1. >>> >>> Optional(foo) better depicts the actual type (it's an options string, after >>> all). If you're not happy with it, just use the nil coalescing operator >>> such as "\(foo ?? "")". This is from the same series of proposals as >>> implicit casting - there are reasons it's done the way it is. >>> On Wed, May 18, 2016 at 3:49 PM Jacob Bandes-Storch via swift-evolution >>> <[email protected] <mailto:[email protected]>> wrote: >>> +1, personally I have taken to using `x+"str"+y` instead of >>> `"\(x)str\(y)"`, if x/y are strings, so I can get a compile-time error if I >>> do this accidentally. >>> >>> But I do see the appeal of being able to print("the data: \(data)") for >>> simple use cases. Didn't someone earlier propose some modifiers/labels like >>> "\(describing: x)" ? >>> >>> >>> On Wed, May 18, 2016 at 11:50 AM, Krystof Vasa via swift-evolution >>> <[email protected] <mailto:[email protected]>> wrote: >>> The string interpolation is one of the strong sides of Swift, but also one >>> of its weaknesses. >>> >>> It has happened to me more than once that I've used the interpolation with >>> an optional by mistake and the result is then far from the expected result. >>> >>> This happened mostly before Swift 2.0's guard expression, but has happened >>> since as well. >>> >>> The user will seldomly want to really get the output "Optional(something)", >>> but is almost always expecting just "something". I believe this should be >>> addressed by a warning to force the user to check the expression to prevent >>> unwanted results. If you indeed want the output of an optional, it's almost >>> always better to use the ?? operator and supply a null value placeholder, >>> e.g. "\(myOptional ?? "<<none>>")", or use myOptional.debugDescription - >>> which is a valid expression that will always return a non-optional value to >>> force the current behavior. >>> >>> Krystof >>> >>> _______________________________________________ >>> swift-evolution mailing list >>> [email protected] <mailto:[email protected]> >>> https://lists.swift.org/mailman/listinfo/swift-evolution >>> <https://lists.swift.org/mailman/listinfo/swift-evolution> >>> >>> _______________________________________________ >>> swift-evolution mailing list >>> [email protected] <mailto:[email protected]> >>> https://lists.swift.org/mailman/listinfo/swift-evolution >>> <https://lists.swift.org/mailman/listinfo/swift-evolution> >>> -- >>> Dan Appel >>> _______________________________________________ >>> swift-evolution mailing list >>> [email protected] <mailto:[email protected]> >>> https://lists.swift.org/mailman/listinfo/swift-evolution >>> <https://lists.swift.org/mailman/listinfo/swift-evolution> >> _______________________________________________ >> swift-evolution mailing list >> [email protected] <mailto:[email protected]> >> https://lists.swift.org/mailman/listinfo/swift-evolution >> <https://lists.swift.org/mailman/listinfo/swift-evolution> > > -- > Dan Appel
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