lambda expression not expected here

[Brian Goetz]

You've mostly got it.  A lambda expression needs a type, since it is compatible with many types.  It can get its target type from an assignment context, a method invocation context, a cast context, etc.  

You are right the error message could be better.  

On Jun 1, 2013, at 10:29 AM, 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.
> 
> - -- 
> 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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