I have an anonymous inner class and an equivalent lambda. Why are the variable initialization rules stricter for the lambda, and is there a solution cleaner than an anonymou
The chapter on lambda expression bodies states
Unlike code appearing in anonymous class declarations, the meaning of names and the
this
andsuper
keywords appearing in a lambda body, along with the accessibility of referenced declarations, are the same as in the surrounding context (except that lambda parameters introduce new names).The transparency of
this
(both explicit and implicit) in the body of a lambda expression - that is, treating it the same as in the surrounding context - allows more flexibility for implementations, and prevents the meaning of unqualified names in the body from being dependent on overload resolution.
They're more strict because of that.
The surrounding context, in this case, is an assignment to a field and the issue at hand is an access of a field, val
, a blank final
field, in the right hand side of the expression.
The Java Language Specification states
Each local variable (§14.4) and every blank
final
field (§4.12.4, §8.3.1.2) must have a definitely assigned value when any access of its value occurs.An access to its value consists of the simple name of the variable (or, for a field, the simple name of the field qualified by
this
) occurring anywhere in an expression except as the left-hand operand of the simple assignment operator=
(§15.26.1).For every access of a local variable or blank
final
fieldx
,x
must be definitely assigned before the access, or a compile-time error occurs.
It then goes on to say
Let
C
be a class, and letV
be a blankfinal
non-static
member field ofC
, declared inC
. Then:
V
is definitely unassigned (and moreover is not definitely assigned) before the leftmost instance initializer (§8.6) or instance variable initializer ofC
.
V
is [un]assigned before an instance initializer or instance variable initializer ofC
other than the leftmost iffV
is [un]assigned after the preceding instance initializer or instance variable initializer ofC
.
Your code basically looks like this
private final int val;
// leftmost instance variable initializer, val still unassigned
private final Callable<String> anonInnerGetValString = ...
// still unassigned after preceding variable initializer
private final Callable<String> lambdaGetValString = ...
The compiler therefore determines that val
in unassigned when it's accessed within the initialization expression for lambdaGetValString
.
The rules above apply to the use of a simple name, val
, not to a qualified expression, this.val
. You can use
final Callable<String> lambdaGetValString = () -> String.valueOf(this.val);
In my case, I had a Predicate
which was trying to access a private final
instance variable. I made the Predicate
final as well which fixed it.
Before - Compiler error, this.availableCities might not have been initialized
class Service {
private final List<String> availableCities;
Service(List<String> availableCities) {
this.availableCities = availableCities;
}
private Predicate<String> isCityAvailable = city -> this.availableCities.contains(city);
}
After - No more error
class Service {
private final List<String> availableCities;
private final Predicate<String> isCityAvailable;
Service(List<String> availableCities) {
this.availableCities = availableCities;
this.isCityAvailable = city -> this.availableCities.contains(city);
}
}
I think the "may not have been initialized" compiler error has some merit, since otherwise there's nothing stopping you from calling the Predicate
from within the constructor before the final instance variable is initialized:
Potentially Why the Compiler Enforces This
class Service {
private final List<String> availableCities;
Service(List<String> availableCities, String topCity) {
boolean isTopCityAvailable = isCityAvailable.test(topCity); // Error: this.availableCities is not initialized yet
this.availableCities = availableCities;
}
private Predicate<String> isCityAvailable = city -> this.availableCities.contains(city);
}
This won't compile:
public class Example
{
private final int x;
private final int y = 2 * x;
public Example() {
x = 10;
}
}
but this will:
public class Example
{
private final int x;
private final int y;
public Example() {
x = 10;
y = 2 * x;
}
}
and so will this:
public class Example
{
private final int x = 10;
private final int y = 2 * x;
}
So it's nothing to do with lambdas. A field that is initialized on the same line that it is declared on is evaluated before the constructor is executed. So at that point, the variable 'val' (or in this example 'x') has not been initialized.