recursive lambda as a local variable

[Remi Forax]

On 09/11/2013 07:32 PM, Zhong Yu wrote:
> State of the Lambda seems to suggest that this can be done for local variables:
>
> http://cr.openjdk.java.net/~briangoetz/lambda/lambda-state-4.html
>
>> When a lambda expression appears in any other context, such as a return expression, there is no way for it to refer to itself. The proper approach in such cases is to name the object with a variable declaration and replace the original expression with a variable reference.

This can be done, everything can be done with a computer,
but it means initializing a lambda that capture a reference to itself.
So you may observe a state where the lambda is not fully initialized,
in between.

And you can also use a method reference instead of a lambda
   static void run() {
     run();
   }
   Runnable r = Foo:run;

To summarize,this construct solves a corner case that is painful to 
implement,
and hard to get right so it was rule out by the EG.

Rémi

  On Wed, Sep 11, 2013 at 12:16 PM, Brian Goetz <brian.goetz at oracle.com> 
wrote:
>> Its not so much "intentional" as a consequence of language rules that
>> predate lambda.
>>
>> In the first example, r is a field; in the second, it is a local.  What the
>> first lambda really is shorthand for is:
>>
>>      Runnable r = () -> this.r.run();
>>
>> So this lambda captures 'this', whereas the second lambda captures 'r'.  The
>> latter runs into definite-assignment issues, but fields are not subject to
>> assignment and flow analysis.
>>
>> Of course, when you use 'this' from the initializer of a field, you are on
>> thin ice.
>>
>>
>> On 9/11/2013 12:37 PM, Zhong Yu wrote:
>>> This compiles: (b106)
>>>
>>>       class Test
>>>       {
>>>           Runnable r = ()->r.run();
>>>       }
>>>
>>> but this doesn't:
>>>
>>>       void test()
>>>       {
>>>           Runnable r = ()->r.run();
>>>       }
>>>
>>> is that intentional or a bug?
>>>
>>> Zhong Yu
>>>