transparent lambda

[Peter Levart]

Isn't it true that method 'four' only calls lambda 'two' once? Lambda
'two' returns from method 'four' (since lambda is transparent) before
the second call.
My question was about how can a program differentiate multiple
invocations of a method from the same call site? Does it maintain a
counter?

Regards, peter


2009/12/30, Neal Gafter <neal at gafter.com>:
> I'm trying to avoid a "solution" that allows this puzzler to be puzzling:
>
> *#int() two = null;
> synchronized int four() {
>   if (two == null) two = #() { return 2; };
>   return two() + two();
> }
> public static void main(String[] args) {
>   System.out.println(four());
> }
> *
>
> What does this do?
>
>    1. prints "4"
>    2. throws an exception
>    3. prints "2"
>    4. Sometimes prints "2", sometimes throws an exception
>    5. does not compile
>
> If you make statement lambdas transparent, the answer is choice 4: the
> method prints "2" on its first execution, and throws an exception (unmatched
> transfer) thereafter.
>
> Cheers,
> Neal
>

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