[swift-evolution] Brainstorming: Optional sugar inferred map

[Jacob Bandes-Storch]

Another conundrum: are the optional-pattern arguments evaluated before any
other arguments? After?
On Fri, Jan 29, 2016 at 9:54 AM Paul Ossenbruggen <possen at gmail.com> wrote:

> Yeah I was thinking about that too, multiple optionals. It gets a bit
> confusing as to the scope.
>
> Wondering if we should pass it on as an optional or not allow it?
> Something else?
>
> Sent from my iPhone
>
> On Jan 29, 2016, at 9:26 AM, Jacob Bandes-Storch <jtbandes at gmail.com>
> wrote:
>
> Hi Paul,
> I think the two statements would behave separately (as if you had used map
> twice).
>
> If you want them to act together, you could do this:
>
> { manager.doSomething(data: data, count: $0); doSomethingElse($0) }(n?)
>
> Another aspect worth considering is whether this syntax should work with
> multiple optionals used in the same expression. (e.g. if they are all
> non-nil, the expression is evaluated.)
> On Fri, Jan 29, 2016 at 1:53 AM Paul Ossenbruggen <possen at gmail.com>
> wrote:
>
>> Jacob,
>>
>> While working on proposal, I am just trying to understand one thing with
>> this. In your second example using map, it is clear where the block of code
>> begins. How would you deal with multiple statements or ones that returned
>> no value:
>>
>>     manager.doSomething(data: data, count: n?); doSomethingElse( n?)
>>
>> the equivalent with map is
>>
>> n.map { manager.doSomething(data: data, count: $0); doSomethingElse($0)  }
>>
>> Thanks,
>> - Paul
>>
>> On Jan 28, 2016, at 8:41 PM, Jacob Bandes-Storch <jtbandes at gmail.com>
>> wrote:
>>
>> That'd be the point. If doSomething were not optional, when "n?" appears
>> in it, it becomes optional (it's basically optional chaining, but for
>> function calls and other expressions instead of just dot-notation).
>> On Thu, Jan 28, 2016 at 8:38 PM Craig Cruden <ccruden at novafore.com>
>> wrote:
>>
>>>
>>>     manager.doSomething(data: data, count: n?)
>>>
>>>
>>> What if the return value of doSomething is not an optional?  Expressions
>>> are easy — but there might be some conflicts with this one.
>>>
>>>
>>> On 2016-01-28, at 14:34:58, Jacob Bandes-Storch via swift-evolution <
>>> swift-evolution at swift.org> wrote:
>>>
>>> I've wanted this sort of thing a lot. It would also work for other
>>> functions, such as
>>>
>>>     manager.doSomething(data: data, count: n?)
>>>
>>> which is equivalent to
>>>
>>>     n.map { manager.doSomething(data: data, count: $0) }
>>>
>>> It might be hard to see exactly which operator/function applications
>>> such a thing applies to, if used in the context of a complex expression.
>>>
>>> Jacob
>>>
>>> On Wed, Jan 27, 2016 at 9:50 PM, Paul Ossenbruggen via swift-evolution <
>>> swift-evolution at swift.org> wrote:
>>>
>>>> Maybe something like this?
>>>>
>>>> let n : Int? = 5
>>>>
>>>> let r = n? + 5
>>>>
>>>>
>>>> On Jan 27, 2016, at 9:46 PM, Thorsten Seitz <tseitz42 at icloud.com>
>>>> wrote:
>>>>
>>>> Too much hidden magic IMO. This would mean loosing the benefits of
>>>> optionals, i.e. making explicit where optional cases occur and that
>>>> handling the missing case has to be considered.
>>>>
>>>> -Thorsten
>>>>
>>>> Am 28.01.2016 um 06:30 schrieb Craig Cruden via swift-evolution <
>>>> swift-evolution at swift.org>:
>>>>
>>>> Yes
>>>>
>>>> On 2016-01-28, at 12:28:40, Paul Ossenbruggen <possen at gmail.com> wrote:
>>>>
>>>> Trying to see if I got this. So the type of r would be Int? at the end
>>>> of this? And if n were nil then r would be nil? Otherwise it r is 10?
>>>>
>>>>
>>>>
>>>> On Jan 27, 2016, at 9:19 PM, Craig Cruden via swift-evolution <
>>>> swift-evolution at swift.org> wrote:
>>>>
>>>> Swift currently encourages a lot of conditional code - especially when
>>>> it comes to optionals.  In most cases when you are computing etc. on an
>>>> Optional you would expect that you would want an optional result and things
>>>> to be able to use optionals.
>>>>
>>>> In another language I generally just `map` one optional to another -
>>>> which may not be the most readable code to some not use to optionals.
>>>>
>>>> I was wondering if maybe an expression is not available that it would
>>>> rewrite the syntax to map from one to another value.
>>>>
>>>> So things like:
>>>>
>>>> let n : Int? = 5
>>>>
>>>> let r = n + 5
>>>>
>>>> would actually compile as
>>>>
>>>> let r = n.map {$0 + 5}
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>>
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