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.r...@oracle.com> wrote:
On Mar 15, 2019, at 11:39 AM, Alex Buckley <alex.buck...@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
