Why can't the var keyword in Java be assigned a lambda expression?

Question

It is allowed to assign var in Java 10 with a string like:

var foo = \"boo\";

While it is not allowed to assign it with a lambda e

Answer 1

As several people have already mentioned, what type should var infer and why should it?

The statement:

var predicateVar = apple -> apple.getColor().equals("red");

is ambiguous and there is no valid reason why the compiler should pick Function<Apple, Boolean> over Predicate<Apple> or vice versa assuming the apple identifier in the lambda represents an Apple isntance.

Another reason is that a lambda in its own doesn't have a speakable type hence there is no way for the compiler to infer it.

Also, "if this was possible" imagine the overhead as the compiler would have to go through all the functional interfaces and determine which functional interface is the most appropriate each time you assign a lambda to a var variable.

Answer 2

This has nothing to do with var. It has to do with whether a lambda has a standalone type. The way var works is that it computes the standalone type of the initializer on the RHS, and infers that.

Since their introduction in Java 8, lambda expressions and method references have no standalone type -- they require a target type, which must be a functional interface.

If you try:

Object o = (String s) -> s.length();

you also get a type error, because the compiler has no idea what functional interface you intend to convert the lambda to.

Asking for inference with var just makes it harder, but since the easier question can't be answered, the harder one cannot either.

Note that you could provide a target type by other means (such as a cast) and then it would work:

var x = (Predicate<String>) s -> s.isEmpty();

because now the RHS has a standalone type. But you are better off providing the target type by giving x a manifest type.

Answer 3

To answer this we have to go into details and understand what a lambda is and how it works.

First we should understand what a lambda is:

A lambda expression always implements a functional interface, so that when you have to supply a functional interface like Runnable, instead of having to create a whole new class that implements the interface, you can just use the lambda syntax to create a method that the functional interface requires. Keep in mind though that the lambda still has the type of the functional interface that it is implementing.

With that in mind, lets take this a step further:

This works great as in the case of Runnable, I can just create a new thread like this new Thread(()->{//put code to run here}); instead of creating a whole new object to implement the functional interface. This works since the compiler knows that Thread() takes an object of type Runnable, so it knows what type the lambda expression has to be.

However, in a case of assigning a lambda to a local variable, the compiler has no clue what functional interface this lambda is implementing so it can't infer what type var should be. Since maybe it's implementing a functional interface the user created or maybe it's the runnable interface, there is just no way to know.

This is why lambdas do not work with the var keyword.

Answer 4

Because that is a non-feature:

This treatment would be restricted to local variables with initializers, indexes in the enhanced for-loop, and locals declared in a traditional for-loop; it would not be available for method formals, constructor formals, method return types, fields, catch formals, or any other kind of variable declaration.

http://openjdk.java.net/jeps/286

Answer 5

To everyone who is saying this is impossible, undesired, or unwanted, I just want to point out that Scala can infer the lambda's type by specifying only the argument type:

val predicateVar = (apple: Apple) => apple.getColor().equals("red")

And in Haskell, because getColor would be a standalone function not attached to an object, and because it does full Hindley-Milner inference, you don't need to specify even the argument type:

predicateVar = \apple -> getColor apple == "red"

This is extraordinarily handy, because it's not the simple types that are annoying for programmers to explicitly specify, it's the more complex ones.

In other words, it's not a feature in Java 10. It's a limitation of their implementation and previous design choices.

Answer 6

In a nutshell, the types of a var and lambda expression both need inference, but in opposite way. The type of a var is inferred by the initializer:

var a = new Apple();

The type of a lambda expression is set by the context. The type expected by the context is called the target type, and is usually inferred by the declaration e.g.

// Variable assignment
Function<Integer, Integer> l = (n) -> 2 * n;
// Method argument 
List<Integer> map(List<Integer> list, Function<Integer, Integer> fn){
    //...
}
map(List.of(1, 2, 3), (n) -> 2 * n);
// Method return 
Function<Integer, Integer> foo(boolean flag){
    //...
    return (n) -> 2 * n;
}

So when a var and lambda expression are used together, the type of the former needs to be inferred by the latter while the type of the latter needs to be inferred by the former.

var a = (n) -> 2 * n;

The root of this dilemma is Java cannot decide the type of a lambda expression uniquely, which is further caused by Java's nominal instead of structural type system. That is, two types with identical structures but different names are not deemed as the same, e.g.

class A{
    public int count;
    int value(){
        return count;
    }
}

class B{
    public int count;
    int value(){
        return count;
    }
}

Function<Integer, Boolean>
Predicate<Integer>