为什么此Java 8方法参考进行编译? [英] Why does this Java 8 method reference compile?

问题描述

我目前正在更深入地研究 Java 8 功能，例如Lambda和方法引用.玩一会儿，使我进入了以下示例:

I'm currently diving deeper into Java 8 features like Lambdas and method references. Playing around a bit brought me to the following example:

public class ConsumerTest {

  private static final String[] NAMES = {"Tony", "Bruce", "Steve", "Thor"};

   public static void main(String[] args) {
      Arrays.asList(NAMES).forEach(Objects::requireNonNull);
   }
}

我的问题是:

为什么main方法内的行会编译?

Why does the line inside the main method compile ?

如果我正确理解这件事，则引用方法的签名必须与功能接口的SAM签名相对应.在这种情况下，消费者需要以下签名:

If I understood the thing correctly, the referenced method's signature has to correspond to the functional interface's SAM signature. In this case, the Consumer requires the following signature:

void accept(T t);

但是， requireNonNull 方法返回 T 而不是void:

However, the requireNonNull method returns T instead of void:

public static <T> T requireNonNull(T obj)

推荐答案

Java语言规范版本8在

The Java Language Specification version 8 says in 15.13.2:

如果T是功能接口类型(

A method reference expression is compatible in an assignment context, invocation context, or casting context with a target type T if T is a functional interface type (§9.8) and the expression is congruent with the function type of the ground target type derived from T.

[..]

如果满足以下两个条件，则方法引用表达式与函数类型一致:

A method reference expression is congruent with a function type if both of the following are true:

• 函数类型标识与引用相对应的单个编译时声明.
• 以下条件之一为真:
  • 函数类型的结果为空.
  • 函数类型的结果为R，而应用捕获转换的结果为(
  • The function type identifies a single compile-time declaration corresponding to the reference.
  • One of the following is true:
    • The result of the function type is void.
    • The result of the function type is R, and the result of applying capture conversion (§5.1.10) to the return type of the invocation type (§15.12.2.6) of the chosen compile-time declaration is R' (where R is the target type that may be used to infer R'), and neither R nor R' is void, and R' is compatible with R in an assignment context.

    (重点是我的)

    因此，函数类型的结果为空的事实足以使其匹配.

    So the fact the result of the function type is void, is enough for it to allow it to match.

    JLS 15.12.2.5还特别提到了在匹配方法时使用void的情况.对于lambda表达式，存在 void-compatible 块的概念(

    JLS 15.12.2.5 also specifically mentions the use of void when matching methods. For lambda expression there is the notion of a void-compatible block (15.27.2) which is referenced in 15.12.2.1, but there is no equivalent definition for method references.

    我无法找到更具体的解释(但是JLS很难破解，所以也许我缺少一些相关的部分)，但是我认为这与允许您加入这一事实有关自己将非void方法作为语句调用(无赋值， return 等).

    I haven't been able to find a more specific explanation (but the JLS is a tough nut to crack so maybe I am missing some relevant sections), but I assume this has to do with the fact that you are also allowed to invoke non-void methods as a statement on its own (without assignment, return etc)).

    JLS 15.13.3方法引用的运行时评估还说:

    出于确定编译时结果的目的，如果调用方法的结果为空，则方法调用表达式为表达式语句，如果调用方法的结果为非无效，则返回语句的表达式.

    For the purpose of determining the compile-time result, the method invocation expression is an expression statement if the invocation method's result is void, and the Expression of a return statement if the invocation method's result is non-void.

    当方法引用的编译时声明为签名多态时，此确定的效果是:

    • 方法调用的参数类型是相应参数的类型.
    • 方法调用是void还是具有Object的返回类型，这取决于包围方法调用的调用方法是void还是具有返回类型.

    因此，生成的方法调用将无效以匹配功能类型.

    So the generated method invocation will be void to match the functional type.

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