>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
> <swift-evolution@swift.org> 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
>> <swift-evolution@swift.org> 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
>>> <swift-evolution@swift.org> 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
>>>
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>>
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>
> --
> Dan Appel
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