Java Reference
In-Depth Information
As the comments indicate,
num
is effectively final and can, therefore, be used inside
myLambda
. This is why the
println( )
statement outputs the number 18. When
func( )
is
called with the argument 8, the value of
v
inside the lambda is set by adding
num
(which is
10) to the value passed to
n
(which is 8). Thus,
func( )
returns 18. This works because
num
is not modified after it is initialized. However, if
num
were to be modified, either inside
the lambda or outside of it,
num
would lose its effectively
final
status. This would cause
an error, and the program would not compile.
It is important to emphasize that a lambda expression can use and modify an instance
variable from its invoking class. It just can't use a local variable of its enclosing scope un-
less that variable is effectively final.
