[swift-evolution] [swift-evolution-announce] [Review] SE-0099: Restructuring Condition Clauses

[Christopher Kornher]

I should have left the entire context in my reply.


> On May 31, 2016, at 1:59 PM, Xiaodi Wu <xiaodi.wu at gmail.com> wrote:
> 
> On Tue, May 31, 2016 at 2:51 PM, Christopher Kornher via swift-evolution <swift-evolution at swift.org <mailto:swift-evolution at swift.org>> wrote:
> 
>> On May 31, 2016, at 1:47 PM, Xiaodi Wu <xiaodi.wu at gmail.com <mailto:xiaodi.wu at gmail.com>> wrote:
>> 
>> 
>> 
>> On Tue, May 31, 2016 at 2:45 PM, Christopher Kornher via swift-evolution <swift-evolution at swift.org <mailto:swift-evolution at swift.org>> wrote:
>>> 
>>> Not allowed:
>>> …
>>> let a = a
>>> let b = b where b > 10 && a > 5
>>> 
>>> Why would this not be allowed by your rule? You're making use of `b` in your where clause. As I demonstrated above, essentially any assertion can be rewritten to work around your rule. In general:
>> 
>> It is not allowed because  ‘a’ is defined in the line above. It must be defined in the ‘if let’ associated with the where in which it is mentioned.
>> 
>> That's a much more restrictive where clause than you proposed earlier. You'd not be able to write:
>> 
>> ```
>> let b = b where b > anyOtherVariable
>> ```
> 
> 
> The definition is not a formal one, but that was the intent.
> 
> ```
> let b = b where b > anyOtherVariable
> ```
> is legal as long as `anyOtherVariable` is not defined within the entire condition clause
> 
> 
> You can propose that rule, but it doesn't solve the issue. If, today, I've got
> 
> ```
> let x = 1
> let y: Int? = 2
> let z = 3
> 
> if let y = y where x < z {
>   // do stuff
> }
> ```
> 
> your rule simply forces
> 
> ```
> if let y = y where y == y && x < z {
>   // do stuff
> }
> ```
> 
> The point is, the semantic relationship between what comes before and after `where` exists in the mind of the human reader only.

I meant to add that all boolean expressions after the where must use one of the constants defined in the associated `if let` so the `&& x < z` would not be allowed.

I don’t understand the 'y == y’ in your example


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