什么样的列表<E>Collectors.toList() 会返回吗?

问题描述

我正在阅读Lambda 现状:图书馆版，并且我对一个声明感到惊讶:

I am reading State of the Lambda: Libraries Edition, and am being surprised by one statement:

在 Streams 部分下，有以下内容:

Under the section Streams, there is the following:

List<Shape> blue = shapes.stream()
                         .filter(s -> s.getColor() == BLUE)
                         .collect(Collectors.toList());

文档没有说明 shapes 实际上是什么，我不知道它是否重要.

The document does not state what shapes actually is, and I do not know if it even matters.

令我困惑的是:这段代码返回什么样的具体List?

What confuses me is the following: What kind of concrete List does this block of code return?

• 它将变量分配给一个 List，这完全没问题.
• stream() 和 filter() 决定使用哪种列表.
• Collectors.toList() 均未指定List 的具体类型.

• It assigns the variable to a List<Shape>, which is completely fine.
• stream() nor filter() decide what kind of list to use.
• Collectors.toList() neither specifies the concrete type of List.

那么，这里使用了 List 的什么具体类型(子类)?有什么保证吗?

So, what concrete type (subclass) of List is being used here? Are there any guarantees?

推荐答案

如果您查看 Collectors#toList()，它指出 - 没有对类型、可变性、可序列化或线程的保证- 返回列表的安全性".如果您希望返回特定的实现，您可以使用 Collectors#toCollection(Supplier).

If you look at the documentation of Collectors#toList(), it states that - "There are no guarantees on the type, mutability, serializability, or thread-safety of the List returned". If you want a particular implementation to be returned, you can use Collectors#toCollection(Supplier) instead.

Supplier<List<Shape>> supplier = () -> new LinkedList<Shape>();

List<Shape> blue = shapes.stream()
            .filter(s -> s.getColor() == BLUE)
            .collect(Collectors.toCollection(supplier));

从 lambda 中，你可以返回任何你想要的 List 实现.

And from the lambda, you can return whatever implementation you want of List<Shape>.

更新:

或者，你甚至可以使用方法引用:

Or, you can even use method reference:

List<Shape> blue = shapes.stream()
            .filter(s -> s.getColor() == BLUE)
            .collect(Collectors.toCollection(LinkedList::new));

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