Question

鉴于以下列表，我将如何迭代地使用参数FactoryWithArgs添加0 - 100类型的元素，同时仍然将此代码保持在（基本上）一行中？

final List<Factory> list = new ArrayList<Factory>(Arrays.asList(
        new Factory0(), new Factory1(), new Factory2(), 
        new FactoryWithArgs(0), new FactoryWithArgs(1), ... ,
        new FactoryWithArgs(99), new FactoryWithArgs(100),
        new Factory4(), new Factory5()));

Answer 1

由于@Brian评论（“双支撑”初始化是否定的否定）和向下投票，这是另一个建议。

对于FactoryWithArgs课程，我会创建一个FactoryWithArgsMaker课程，旨在将FactoryWithArgs个对象设为List。

类似的东西：

public static class FactoryWithArgsMaker {
    public static List<FactoryWithArgs> makeNew(int start, int end) {
        List<FactoryWithArgs> list = new ArrayList();
        for (int i = 0; i <= end; i++) {
            list.add(new FactoryWithArgs(i));
        }
        return list;
    }
}

只要您的“工厂”的顺序在您的列表中无关紧要，并且使用Java 8 Stream，您可以在一行中初始化您的列表，如下所示：

List<Factory> list = Stream.concat(
    Arrays.asList(
        new Factory0(), new Factory1(), new Factory2(), 
        new Factory3(), new Factory4(), new Factory5()
    ).stream(),
    FactoryWithArgsMaker.makeNew(0, 100).stream()
).collect(Collectors.toList());

Stream.concat()获取从Arrays.asList()返回的List<FactoryWithArgs>和FactoryWithArgsMaker.makeNew()的内容，并将它们合并为一个流，然后我们将其转换回List（ .collect(Collectors.toList()）;

Answer 2

    final List<Factory> list = new ArrayList<Factory>();        
    list.add(new Factory0());
    list.add(new Factory1());
    list.add(new Factory2());
    for (int i = 0; i < 101; i++) {
        list.add(new FactoryWithArgs(i));
    }
    list.add(new Factory4());
    list.add(new Factory5());

如果您愿意，可以将其放入方法中。

Java迭代在Arrays.asList

2 个答案: