在Java Lambda相等和/或实例化上 [英] On Java lambda equality and/or instantiation

问题描述

下面的代码段为什么在第二遍打印时显示为真?应该不是新实例吗?

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

推荐答案

查看此

Check out this article about how lambdas are implemented.

本质上，编译器将您的两个 System.nanoTime()转换为类上的以下静态方法:

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();
}

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