anonymous method call

[Marc Petit-Huguenin]

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On 06/01/2013 12:32 PM, Remi Forax wrote:
> On 06/01/2013 08:10 PM, Marc Petit-Huguenin wrote: Thanks Brian and Rémi
> for the response to my previous question.
> 
> Now that we established my lack of knowledge about the lambda type
> inference, I have another question, but please note that it is not a
> suggestion for improvement, merely a question about the limits of what
> would be possible.
> 
> A functional interface has only one abstract method, so is there any reason
> it would not be possible to use an anonymous call to this method?
> 
> For example:
> 
> IntUnaryOperator operator = f -> f + 1;
> 
> and instead of
> 
> int i = operator.applyAsInt(1);
> 
> being able to do
> 
> int i = operator(1);
> 
> 
> Thanks.
> 
>> It's possible to implement that but it exhibits a nasty property given 
>> the way Java scope works.
> 
>> In Java, the scope that contains local variables and the one that 
>> contains method calls are separated, when the compiler see
> 
>> int i = operator(1);
> 
>> it knowns that this is a method call, and it will not check if there is a
>> local variable named 'operator'.
> 
>> So to support the new syntax and still be able to compile old code, the 
>> compiler has to first check if a method is available and then check if a
>> local variable exist if no method was found.
> 
>> This rule is awful because it means that if you add a method in a class 
>> (or worst in a super class) you may broke an already existing code
>> because you introduce a method that will be used instead of calling the
>> lambda.
> 
>> so this syntax was rejected by the Expert Group :)

It makes sense, thanks.

FYI, I thought of this when trying to adapt the code in
http://www.righto.com/2009/03/y-combinator-in-arc-and-java.html.  The result
is far simpler than the original code, but I was wondering if there was a way
to make it even better:

interface HigherOrder<T> extends Function<HigherOrder<T>, T> {}

Function<UnaryOperator<IntUnaryOperator>, IntUnaryOperator> y = r ->
((HigherOrder<IntUnaryOperator>)(f -> f.apply(f))).apply(f -> r.apply(x ->
f.apply(f).applyAsInt(x)));
IntUnaryOperator factorial = y.apply(f -> n -> n == 0 ? 1 : n * f.applyAsInt(n
- - 1));
assert factorial.applyAsInt(10) == 3628800;
IntUnaryOperator fibonacci = y.apply(f -> n -> n == 0 ? 0 : n == 1 ? 1 :
f.applyAsInt(n - 1) + f.applyAsInt(n - 2));
assert fibonacci.applyAsInt(30) == 832040;

- -- 
Marc Petit-Huguenin
Email: marc at petit-huguenin.org
Blog: http://blog.marc.petit-huguenin.org
Profile: http://www.linkedin.com/in/petithug
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