为什么添加“.map(a->a)”允许编译这个？

Question

这与我对“流减少不兼容类型”的回答有关。我不知道为什么我的建议有效，霍尔格正确地敦促我这一点。但即使是他似乎也没有一个明确的解释为什么它能起作用。所以，让我们把它作为自己的问题来问：

以下代码不能在javac中编译（下面指向ideone的链接是sun-jdk-1.8.0_51，请参阅http://ideone.com/faq):

public <T> Object with(Stream<Predicate<? super T>> predicates) {
  return predicates.reduce(Predicate::or);
}

这样做是正确的：或者将这个流中的两个谓词结合在一起，就像写：

Predicate<? super T> a = null;
Predicate<? super T> b = null;
a.or(b);  // Compiler error!

public <T> Object with(Stream<Predicate<? super T>> predicates) {
  return predicates.map(a -> a).reduce(Predicate::or);
                // ^----------^ Added
}

IdeOne演示

尽管事实上我想尝试这一点，但我不太清楚为什么这会起作用。我的解释是，.map(a->a)的作用类似于“强制转换”，它给类型推断算法更大的灵活性来选择允许应用reduce的类型。但我不确定到底是什么类型。

请注意，这并不等同于使用.map(function.identity())，因为它被限制为返回输入类型。IdeOne演示

方法的返回类型可以变得更具体一些；我在上面省略了它，这样返回类型上的讨厌泛型就不会妨碍：

public <T> Optional<? extends Predicate<? super T>> with(
    Stream<Predicate<? super T>> predicates) {
  return predicates.map(a -> a).reduce(Predicate::or);
}

这是使用-xdverboseresolution=all编译的输出。不完全确定这是否是我可以发布的调试类型推断的最相关的输出；如果有更好的建议，请告知：

Interesting.java:5: Note: resolving method <init> in type Object to candidate 0
class Interesting {
^
  phase: BASIC
  with actuals: no arguments
  with type-args: no arguments
  candidates:
      #0 applicable method found: Object()

Interesting.java:7: Note: resolving method map in type Stream to candidate 0
    return predicates.map(a -> a).reduce(Predicate::or);
                     ^
  phase: BASIC
  with actuals: <none>
  with type-args: no arguments
  candidates:
      #0 applicable method found: <R>map(Function<? super T#1,? extends R>)
        (partially instantiated to: (Function<? super Predicate<? super T#2>,? extends Object>)Stream<Object>)
  where R,T#1,T#2 are type-variables:
    R extends Object declared in method <R>map(Function<? super T#1,? extends R>)
    T#1 extends Object declared in interface Stream
    T#2 extends Object declared in method <T#2>with(Stream<Predicate<? super T#2>>)

Interesting.java:7: Note: Deferred instantiation of method <R>map(Function<? super T#1,? extends R>)
    return predicates.map(a -> a).reduce(Predicate::or);
                         ^
  instantiated signature: (Function<? super Predicate<? super T#2>,? extends Predicate<CAP#1>>)Stream<Predicate<CAP#1>>
  target-type: <none>
  where R,T#1,T#2 are type-variables:
    R extends Object declared in method <R>map(Function<? super T#1,? extends R>)
    T#1 extends Object declared in interface Stream
    T#2 extends Object declared in method <T#2>with(Stream<Predicate<? super T#2>>)
  where CAP#1 is a fresh type-variable:
    CAP#1 extends Object super: T#2 from capture of ? super T#2

Interesting.java:7: Note: resolving method reduce in type Stream to candidate 1
    return predicates.map(a -> a).reduce(Predicate::or);
                                 ^
  phase: BASIC
  with actuals: <none>
  with type-args: no arguments
  candidates:
      #0 not applicable method found: <U>reduce(U,BiFunction<U,? super T,U>,BinaryOperator<U>)
        (cannot infer type-variable(s) U
          (actual and formal argument lists differ in length))
      #1 applicable method found: reduce(BinaryOperator<T>)
      #2 not applicable method found: reduce(T,BinaryOperator<T>)
        (actual and formal argument lists differ in length)
  where U,T are type-variables:
    U extends Object declared in method <U>reduce(U,BiFunction<U,? super T,U>,BinaryOperator<U>)
    T extends Object declared in interface Stream

Interesting.java:7: Note: resolving method metafactory in type LambdaMetafactory to candidate 0
    return predicates.map(a -> a).reduce(Predicate::or);
                          ^
  phase: BASIC
  with actuals: Lookup,String,MethodType,MethodType,MethodHandle,MethodType
  with type-args: no arguments
  candidates:
      #0 applicable method found: metafactory(Lookup,String,MethodType,MethodType,MethodHandle,MethodType)

Interesting.java:7: Note: resolving method metafactory in type LambdaMetafactory to candidate 0
    return predicates.map(a -> a).reduce(Predicate::or);
                                         ^
  phase: BASIC
  with actuals: Lookup,String,MethodType,MethodType,MethodHandle,MethodType
  with type-args: no arguments
  candidates:
      #0 applicable method found: metafactory(Lookup,String,MethodType,MethodType,MethodHandle,MethodType)

Answer 1

除非我在FunctionalInterface推断是如何发生的方面遗漏了什么，否则很明显，您不能在流上调用reduce<？超级谓词>，因为它没有足够的类型来推断为BinaryOperator。

方法引用隐藏了故事的一个非常重要的部分，第二个参数。

return predicates.map(a->a).reduce((predicate, other) -> predicate.or(other));

如果移除对映射的调用，编译器就没有机会适当地键入流以满足第二个捕获要求。使用map，编译器可以决定满足捕获所需的类型，但是如果没有泛型的具体绑定，两个捕获只能用一个对象流来满足，这很可能是通过map（）得到的结果。

public <T> Optional<? extends Predicate<? super T>> with(Stream<Predicate<Object>> predicates)
public <T> Optional<? extends Predicate<? super T>> with(Stream<Predicate<T>> predicates)