什么样的列表<E>Collectors.toList() 会返回吗? [英] What kind of List<E> does Collectors.toList() return?

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

那么，这里使用的 List 的具体类型(子类)是什么?有什么保证吗?

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

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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