[swift-evolution] Brainstorming: Optional sugar inferred map

[Jacob Bandes-Storch]

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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