为什么这个 Java 8 方法参考编译？

Question

我目前正在深入研究 Java 8 功能，例如 Lambda 和方法引用。玩了一下让我想到了以下示例：

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 方法里面的那一行可以编译？

如果我没理解错的话，引用方法的签名必须对应于功能接口的 SAM 签名。在这种情况下，消费者需要以下签名：

void accept(T t);

然而，requireNonNull 方法 returns T 而不是 void:

public static <T> T requireNonNull(T obj)

Answer 1

Java 语言规范版本 8 在 15.13.2 中说：

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.

[..]

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

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.

（强调我的）

因此函数类型的结果为 void 这一事实足以让它匹配。

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，但没有对方法引用的等效定义。

我没能找到更具体的解释（但 JLS 是一个棘手的问题，所以也许我遗漏了一些相关部分），但我认为这与你是还允许调用非 void 方法作为其自身的语句（无赋值，return 等））。

JLS 15.13.3 Run-Time Evaluation of Method References 还说：

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.

The effect of this determination when the compile-time declaration of the method reference is signature polymorphic is that:

The types of the parameters for the method invocation are the types of the corresponding arguments.

The method invocation is either void or has a return type of Object, depending on whether the invocation method which encloses the method invocation is void or has a return type.

因此生成的方法调用将无效以匹配功能类型。

Answer 2

除了它在@Mark Rotteveel 的确切答案中所说的，它编译是因为在 Java 中你可以忽略任何方法调用的结果，例如在下面的例子中：

Map<String, String> map = new HashMap<>();

map.put("1", "a");

map.put("1", "A"); // Who cares about the returned value "a"?

由于您没有在 forEach() 消费者块中返回任何内容，因此根据规范它是有效的。

为什么这个 Java 8 方法参考编译？

Why does this Java 8 method reference compile?

java

java-8

method-reference