lambda - 为什么允许为 BiConsumer 分配只接受单个参数的函数？

Question

举个例子:

public class MyConsumer {
    public void accept(int i) {}
    public static void biAccept(MyConsumer mc, int i) {}
}

public class BiConsumerDemo {

    public void accept(int i) { }
    public static void biAccept(MyConsumer mc, int i) { }

    private void testBiConsume() {
        BiConsumer<MyConsumer, Integer> accumulator = (x, y) -> {}; // no problem, method accepts 2 parameters
        accumulator = MyConsumer::accept; // accepts only one parameter and yet is treated as BiConsumer
        accumulator = MyConsumer::biAccept; // needed to be static
        accumulator = BiConsumerDemo::accept; // compilation error, only accepts single parameter
    }
}

为什么变量accumulator是一个BiConsumer，需要一个函数接受2个参数，可以用MyConsumer::accept进行赋值code> 当该方法只接受单个参数时？

Java这种语言设计背后的原理是什么？如果有一个术语，它是什么？

Answer 1

MyConsumer::accept 是对 MyConsumer 类的实例方法的方法引用，该类具有 int 类型的单个参数。调用该方法的 MyConsumer 实例被视为方法引用的隐式参数。

因此:

accumulator = MyConsumer::accept;

相当于:

accumulator = (MyConsumer m,Integer i) -> m.accept(i);

given a targeted function type with n parameters, a set of potentially applicable methods is identified:

• If the method reference expression has the form ReferenceType :: [TypeArguments] Identifier, the potentially applicable methods are the member methods of the type to search that have an appropriate name (given by Identifier), accessibility, arity (n or n-1), and type argument arity (derived from [TypeArguments]), as specified in §15.12.2.1.

  Two different arities, n and n-1, are considered, to account for the possibility that this form refers to either a static method or an instance method.

(来自 15.13.1. Compile-Time Declaration of a Method Reference )

在您的例子中，目标函数类型是 BiConsumer，它有 2 个参数。因此，MyConsumer 类中任何与名称 accept 匹配且具有 2 或 1 个参数的方法都会被考虑。

2个参数的静态方法和1个参数的实例方法都可以匹配目标函数类型。