Comparator.reversed（）不使用lambda进行编译 [英] Comparator.reversed() does not compile using lambda

问题描述

我有一个包含一些User对象的列表，我正在尝试对列表进行排序，但只能使用方法引用，使用lambda表达式，编译器会给出错误：

I have a list with some User objects and i'm trying to sort the list, but only works using method reference, with lambda expression the compiler gives an error:

List<User> userList = Arrays.asList(u1, u2, u3);
userList.sort(Comparator.comparing(u -> u.getName())); // works
userList.sort(Comparator.comparing(User::getName).reversed()); // works
userList.sort(Comparator.comparing(u -> u.getName()).reversed()); // Compiler error

错误：

com\java8\collectionapi\CollectionTest.java:35: error: cannot find symbol
            userList.sort(Comparator.comparing(u -> u.getName()).reversed());
                                                     ^
symbol:   method getName()
location: variable u of type Object
1 error

推荐答案

这是编译器类型推理机制的一个弱点。为了推断lambda中 u 的类型，需要建立lambda的目标类型。这完成如下。 userList.sort（）期望参数类型为 Comparator< User> 。在第一行中， Comparator.comparing（）需要返回 Comparator< User> 。这意味着 Comparator.comparing（）需要一个函数，需要用户参数。因此，在第一行的lambda中， u 必须是用户类型，一切正常。

This is a weakness in the compiler's type inferencing mechanism. In order to infer the type of u in the lambda, the target type for the lambda needs to be established. This is accomplished as follows. userList.sort() is expecting an argument of type Comparator<User>. In the first line, Comparator.comparing() needs to return Comparator<User>. This implies that Comparator.comparing() needs a Function that takes a User argument. Thus in the lambda on the first line, u must be of type User and everything works.

在第二行和第三行中，由于存在对 reversed（）的调用，目标类型被中断。我不完全确定为什么;收件人和返回类型 reversed（）都是 Comparator< T> 所以它似乎应该是目标类型被传播回接收器，但事实并非如此。 （就像我说的那样，这是一个弱点。）

In the second and third lines, the target typing is disrupted by the presence of the call to reversed(). I'm not entirely sure why; both the receiver and the return type of reversed() are Comparator<T> so it seems like the target type should be propagated back to the receiver, but it isn't. (Like I said, it's a weakness.)

在第二行中，方法参考提供了填补这一空白的其他类型信息。第三行没有此信息，因此编译器推断 u 为对象（最后的推理回退） ），失败。

In the second line, the method reference provides additional type information that fills this gap. This information is absent from the third line, so the compiler infers u to be Object (the inference fallback of last resort), which fails.

显然，如果你可以使用方法参考，那么这样做就可以了。有时您不能使用方法引用，例如，如果要传递其他参数，则必须使用lambda表达式。在这种情况下，您将在lambda中提供显式参数类型：

Obviously if you can use a method reference, do that and it'll work. Sometimes you can't use a method reference, e.g., if you want to pass an additional parameter, so you have to use a lambda expression. In that case you'd provide an explicit parameter type in the lambda:

userList.sort(Comparator.comparing((User u) -> u.getName()).reversed());

在将来的版本中，可能会增强编译器以涵盖这种情况。

It might be possible for the compiler to be enhanced to cover this case in a future release.

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