Switch expressions spec

[Alex Buckley]

OK, we intend at least one result expression to be required, so the spec 
is correct as is.

(I should have been clearer that my belief was about the intent of the 
spec, rather than about how I personally think completion should occur.)

Manoj didn't say what javac build he is testing with, but this is a 
substantial discrepancy between compiler and spec. I hope that Leonid 
Arbouzov (cc'd) can tell us what conformance tests exist in this area.

Alex

On 3/15/2019 12:09 PM, Brian Goetz wrote:
> At the same time, we also reaffirmed our choice to _not_ allow throw from one half of a conditional:
>
>      int x = foo ? 3 : throw new FooException()
>
> But John has this right — the high order bit is that every expression should have a defined normal completion, and a type, even if computing sub-expressions (or in this case, sub-statements) might throw.  And without at least one arm yielding a value, it would be impossible to infer the type of the expression.
>
>> On Mar 15, 2019, at 3:01 PM, John Rose <john.r.rose at oracle.com> wrote:
>>
>> On Mar 15, 2019, at 11:39 AM, Alex Buckley <alex.buckley at oracle.com> wrote:
>>>
>>> In a switch expression, I believe it should be legal for every `case`/`default` arm to complete abruptly _for a reason other than a break with value_.
>>
>> My reading of Gavin's draft is that he is doing something very
>> subtle there, which is to retain an existing feature in the language
>> that an expression always has a defined normal completion.
>>
>> We also don't have expressions of the form "throw e".  Allowing
>> a switch expression to complete without a value on *every* arm
>> raises the same question as "throw e" as an expression.  How do
>> you type "f(throw e)"?  If you can answer that, then you can also
>> have switch expressions that refuse to break with any values.
>>
>> BTW, if an expression has a defined normal completion, it also
>> has a possible type.  By possible type I mean at least one correct
>> typing (poly-expressions can have many).  So one obvious
>> result of Gavin's draft is that you derive possible types from
>> the arms of the switch expression that break with values.
>>
>> But the root requirement, I think, is to preserve the possible
>> normal normal of every expression.
>>
>> "What about some form of 1/0?"  That's a good question.
>> What about it?  It completes normally with a type of int.
>> Dynamically, the normal completion is never taken.
>> Gavin might call that a "notional normal completion"
>> (I like that word) provided to uphold the general principle
>> even where static analysis proves that the Turing machine
>> fails to return normally.
>>
>> — John
>