为什么等效的 lambda 表达式和方法引用在捕获静态字段值时表现不同？

Question

我对 Java lambda 和方法引用行为有点困惑。例如，我们有这个代码：

import java.util.function.Consumer;

public class Main {

    private static StringBuilder sBuilder = new StringBuilder("1");

    public static void main(String[] args) {
        Consumer<String> consumer = s -> sBuilder.append(s);
        sBuilder = new StringBuilder("2");
        consumer.accept("3");
        System.out.println(sBuilder);
    }

}

输出：

这按预期工作，但如果我们替换

s -> sBuilder.append(s)

和

sBuilder::append

输出将是：

你知道如何解释这个吗？这不是一样的东西吗？谢谢

Answer 1

在 lambda 表达式中，sBuilder 字段被捕获，但未被计算。它只会在相应的函数接口方法被调用时被评估。那时，sBuilder 引用创建并分配给字段的新实例

sBuilder = new StringBuilder("2");

在方法引用中，立即计算 sBuilder 字段以生成 Consumer 实例。该值引用在静态初始化程序中创建的实例

private static StringBuilder sBuilder = new StringBuilder("1");

并且 Consumer 将对那个进行操作。你打印新的。

来自 Java 语言规范，关于 Run-Time Evaluation of Method References

The body of an invocation method depends on the form of the method reference expression, as follows:

If the form is ExpressionName :: [TypeArguments] Identifier or Primary :: [TypeArguments] Identifier, then the body of the invocation method has the effect of a method invocation expression for a compile-time declaration which is the compile-time declaration of the method reference expression. Run-time evaluation of the method invocation expression is as specified in §15.12.4.3, §15.12.4.4, and §15.12.4.5, where:

The invocation mode is derived from the compile-time declaration as specified in §15.12.3.

The target reference is the value of ExpressionName or Primary, as determined when the method reference expression was evaluated.

The arguments to the method invocation expression are the formal parameters of the invocation method.

为什么等效的 lambda 表达式和方法引用在捕获静态字段值时表现不同？

Why do equivalent lambda expression and method reference behave differently when capturing static field value?

java

lambda

java-8

method-reference