Lambda expression vs method reference implementati

Question

Given this:

class MyClass {
    static class A {
        public boolean property() {
            return Math.random() < 0.5;
        }
    }

    static List<A> filterLambda(List<A> list) {
        return list.stream().filter(a -> a.property()).collect(Collectors.toList());
    }

    static List<A> filterMethodCall(List<A> list) {
        return list.stream().filter(A::property).collect(Collectors.toList());
    }
}

• What are the differences in what the compiler does for each method?
• In case there's any, is there a difference in memory usage or runtime? (even if it's small, the question is just academic)

PD: I know the question is similar to this one but I think it's not been addressed correctly.

Answer 1

This is an extract from the Brian Goetz's doc linked by Brett Oken:

When the compiler encounters a lambda expression, it first lowers (desugars) the lambda body into a method whose argument list and return type match that of the lambda expression, possibly with some additional arguments (for values captured from the lexical scope, if any.) At the point at which the lambda expression would be captured, it generates an invokedynamic call site, which, when invoked, returns an instance of the functional interface to which the lambda is being converted. This call site is called the lambda factory for a given lambda. The dynamic arguments to the lambda factory are the values captured from the lexical scope. The bootstrap method of the lambda factory is a standardized method in the Java language runtime library, called the lambda metafactory. The static bootstrap arguments capture information known about the lambda at compile time (the functional interface to which it will be converted, a method handle for the desugared lambda body, information about whether the SAM type is serializable, etc.)

Method references are treated the same way as lambda expressions, except that most method references do not need to be desugared into a new method; we can simply load a constant method handle for the referenced method and pass that to the metafactory.

Examples extracted from same doc:

As an example, consider a lambda that captures a field minSize:

list.filter(e -> e.getSize() < minSize )

We desugar this as an instance method, and pass the receiver as the first captured argument:

list.forEach(INDY((MH(metaFactory), MH(invokeVirtual Predicate.apply),
                    MH(invokeVirtual B.lambda$1))( this ))));

private boolean lambda$1(Element e) {
    return e.getSize() < minSize; }

While

list.filter(String::isEmpty)

is translated as:

list.filter(indy(MH(metaFactory), MH(invokeVirtual Predicate.apply),
             MH(invokeVirtual String.isEmpty))()))

Answer 2

One case where we can find the difference lambda expression & method reference is using as part of Supplier interface. Lets say we have below static method

public static <T, E> T catchException(Supplier<T> resolver) {
    try {
        T result = resolver.get();
        return  result ;
    } catch (Exception e) {
        System.out.println("Exception");
        return null;
    }
}

When we call the method using lambda expression as below, code work fine since the lambda expression passed as part of Supplier & executed only when get() method is called where the exception is caught.

List<String> underlyers=null;
System.out.println(catchException(()->underlyers.size()));

When we call the method using method reference as below, we get NullPointerExecption before calling the method since the reference is executed before passing in to the supplier. The control does not reach get() method in this case.

List<String> underlyers=null;
System.out.println(catchException(underlyers::size));

Answer 3

Here is the java language spec on method references.

Here is the spec for lambda expressions.

Lambdas are quite a bit more complex from a rules perspective.

In both cases, however, the result is an invokedynamic call.

Brian Goetz has written a doc on how this works.