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为什么是迭代器<Entry<Integer, Double> 不能转换为迭代器<Entry<? extends Integer, ? extends Double> &gt;

[英]Why Iterator<Entry<Integer, Double> cannot be converted to Iterator<Entry<? extends Integer, ? extends Double>>

I have the sample code below attempting to construct an immutable map out of an iterable:我有下面的示例代码,试图从一个可迭代对象中构建一个不可变的映射:

ImmutableList<Entry<Integer, Double>> list = ImmutableList.of(
    new AbstractMap.SimpleEntry<Integer, Double>(4, 2d),
    new AbstractMap.SimpleEntry<Integer, Double>(16, 4d));
ImmutableMap.<Integer, Double>builder().putAll(list); // No error
ImmutableMap.<Integer, Double>builder().putAll(() -> list.stream().iterator()); // Error - cannot convert type...

I am getting the error Bad return type in lambda expression: Iterator<Entry<Integer, Double>> cannot be converted to Iterator<Entry<? extends Integer, ? extends Double>>Bad return type in lambda expression: Iterator<Entry<Integer, Double>> cannot be converted to Iterator<Entry<? extends Integer, ? extends Double>>收到错误Bad return type in lambda expression: Iterator<Entry<Integer, Double>> cannot be converted to Iterator<Entry<? extends Integer, ? extends Double>>错误Bad return type in lambda expression: Iterator<Entry<Integer, Double>> cannot be converted to Iterator<Entry<? extends Integer, ? extends Double>> Bad return type in lambda expression: Iterator<Entry<Integer, Double>> cannot be converted to Iterator<Entry<? extends Integer, ? extends Double>> Bad return type in lambda expression: Iterator<Entry<Integer, Double>> cannot be converted to Iterator<Entry<? extends Integer, ? extends Double>> . Bad return type in lambda expression: Iterator<Entry<Integer, Double>> cannot be converted to Iterator<Entry<? extends Integer, ? extends Double>> From my understanding this should be valid because Integer is a valid upper bound of Integer, and Double is a valid upper bound of Double.根据我的理解,这应该是有效的,因为 Integer 是 Integer 的有效上限,而 Double 是 Double 的有效上限。 What is the problem here and how can I fix specifically the iterator code?这里有什么问题,我该如何修复迭代器代码?

I need to fix the iterator code in particular since I am trying to transform another collection into a map like below.我需要特别修复迭代器代码,因为我正在尝试将另一个集合转换为如下所示的地图。 I am using guava on android so do not have access to the default guava collect() methods (though I think I can implement them manually if there is really no other way to do this).我在 android 上使用 guava,所以无法访问默认的 guava collect() 方法(尽管我认为我可以手动实现它们,如果真的没有其他方法可以做到这一点)。

ImmutableList<Integer> numberList = ImmutableList.of(2,4,8,12,16,24,32);
ImmutableMap.<Integer, Double>builder().putAll(() -> 
   numberList.stream().map(num -> 
       (Entry<Integer, Double>) new AbstractMap.SimpleEntry<Integer, Double>(num, Math.sqrt(num))).iterator());

Oh boy, this one is ugly.哦,男孩,这个丑陋的。

You're right that Entry<? extends Integer, ? extends Double>你是对的那个Entry<? extends Integer, ? extends Double> Entry<? extends Integer, ? extends Double> Entry<? extends Integer, ? extends Double> is a "subtype" of Entry<Integer, Double> (I think it's not technically called a subtype, but is effectively one). Entry<? extends Integer, ? extends Double>Entry<Integer, Double>的“子类型”(我认为它在技术上不被称为子类型,但实际上是一个子类型)。 But, as you may know, Iterator<Subtype> is not a subtype of Iterator<Supertype> .但是,您可能知道, Iterator<Subtype>不是Iterator<Supertype>的子类型。 Instead you have to do Iterator<? extends Subtype>相反,您必须执行Iterator<? extends Subtype> Iterator<? extends Subtype> . Iterator<? extends Subtype>

So you'd need to add just one more ? extends所以你只需要再添加一个? extends ? extends in there:在那里? extends

Iterator<? extends Entry<? extends Integer, ? extends Double>>
         ^^^^^^^^^

I'm not sure offhand what the best way to do that is, though.不过,我不确定最好的方法是什么。

Integer is not a valid upper bound of Integer, you can't extend class A from class A. For example yout can't write this: Public class Actor extends Actor{ } So to solve your problem make Number instead Integer like this Iterator<Entry<? Integer 不是 Integer 的有效上限,您不能从 A 类扩展 A 类。例如,您不能这样写: Public class Actor extends Actor{ } 所以要解决您的问题,请使用 Number 代替 Integer 像这样 Iterator<条目<? extends Number, ?extends Number>>扩展数, ?扩展数>>

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