lambda expression not expected here

[Remi Forax]

On 06/01/2013 07:29 PM, Marc Petit-Huguenin wrote:
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> Hi,
>
> The compilation of this code fails with a "lambda expression not expected
> here" error:
>
> int i = (f -> f + 1).applyAsInt(1);
>
> I am not sure to understand why this code is invalid as the following code
> compiles fine:
>
> int i = ((IntUnaryOperator)(f -> f + 1)).applyAsInt(1);
>
> If it is because the compiler cannot infer the type, then perhaps the error
> text could be improved.
>
> Thanks.

Hi Marc,
no the compiler can not infer the type, there is not enough information.

In your example,

int i = (f -> f + 1).applyAsInt(1);

One lambda can be converted to different functional interfaces,
by example,
interface IntToIntFun { int fun(int value); }
interface Operator<T> { T apply(T value); }

so the following statements are both valid:
IntToIntFun fun = f -> f + 1;
Operator<Integer> operator = f -> f + 1;

There is no way for the compiler to infer the type because they are more 
than one possible choice.
Here, we pay the price that Java has no type for a function and uses the 
more flexible and more java-like concept of function interface.

cheers,
Rémi

>
> - -- 
> 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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