On Java lambda equality and/or instantiation [duplicate]

Question

Why does the snippet below print true on second pass through? Should it not be a new instance?

import java.util.function.Supplier;

public class Foo {
    public static void main(String[] args) throws Exception {
        Supplier<Long> old = () -> System.nanoTime();

        for (int i = 0; i < 3; i++) {
            /* false true true
            Supplier<Long> foo = System::nanoTime;*/

            Supplier<Long> foo = () -> System.nanoTime();

            /* false false false
            Supplier<Long> foo = new Supplier<Long>() {
                @Override
                public Long get() {
                    return System.nanoTime();
                }
            };
            //*/

            System.out.printf("%s %s %s%n", foo == old, foo, old);

            old = foo;
        }
    }
}

false Foo$$Lambda$2/122883338@1ddc4ec2 Foo$$Lambda$1/1534030866@133314b
true Foo$$Lambda$2/122883338@1ddc4ec2 Foo$$Lambda$2/122883338@1ddc4ec2
true Foo$$Lambda$2/122883338@1ddc4ec2 Foo$$Lambda$2/122883338@1ddc4ec2

Answer 1

Check out this article about how lambdas are implemented.

Essentially the compiler turned your two System.nanoTime() into the following static methods on your class:

static Long lambda$1() {
    return System.nanoTime();
}

static Long lambda$2() {
    return System.nanoTime();
}

And then created a constant reference to each target-typed to Supplier<Long> using the LambdaMetaFactory. Frankly, I'm disappointed that the Java compiler didn't realize that the lambda body was identical and only create one instance. If the Java compiler were sufficiently smart, every line should have printed true!