Java8-显式类型如何匹配一个变体-而不是其他类型? [英] Java8 - How does explicit type matches one variant - not other type?

问题描述

我有一个简单的代码段，如下所示.我提到了此

I have a simple snippet as below. I referred this

List<Document> list = new LinkedList<Document>();
FindIterable<Document> itr = collection.find(findQuery)
                       .forEach((Document doc) -> list.add(doc));
return list;

它编译没有任何问题.

It compiles without any issues.

1. 我猜我们是在告诉编译器doc是Document类型. 为什么需要它?

1. I guess that we are telling compiler that doc is of type Document. Why is it needed?

但是，如果我执行以下操作，则会引发模棱两可的错误.我提到了此但无法准确地联系和理解.

But If I do the below, it throws ambiguous error. I referred this But couldn't relate and understand exactly.

collection.find(findQuery).forEach(list::add);

1. 有人可以解释为什么第二个语句不起作用吗?

有没有更好的方式来编写第一个[工作的] ?

Java版本:1.8.0_231

Java version: 1.8.0_231

导入语句:

import java.util.List;
import java.util.Optional;
import com.mongodb.client.FindIterable;
import org.bson.Document;

推荐答案

FindIterable继承了两个forEach方法:

• com.mongodb.client.MongoIterable.forEach(Block<? super TResult>)

java.lang.Iterable.forEach(Consumer<? super T>)

您可以用其中任何一个重写参数

You could rewrite your paramter with either

Consumer<Document> consumer = documents::add;
Block<Document> block = list::add;

任何一个都可以.这些也将起作用:

And either will work. These too will work:

.forEach((Consumer<Document>) doc -> list.add(doc))
.forEach((Consumer<Document>) list::add);

但是，当您调用forEach(list::add)或forEach(doc -> list.add(doc))时，编译器无法选择哪个重载将确定方法引用的目标类型(因为该表达式在该上下文中与两者兼容).

However, when you call forEach(list::add) or forEach(doc -> list.add(doc)), the compiler is unable to pick which overload will determine the method reference's target type (because the expression is compatible with both in that context).

现在，我不确定100％为什么.forEach((Document doc) -> list.add(doc))成功选择/链接Consumer<Document>的签名而不是Block<? super Document>的签名，我推测这与通用范围有关(但是我还在继续阅读).

Now, I'm not 100% sure why .forEach((Document doc) -> list.add(doc)) successfully selects/links the signature of Consumer<Document> instead of the one with Block<? super Document>, I'm surmizing it has to do with the generic bounds (but I'm still reading on this).

因为Block<? super Document>版本已被弃用，所以您的选择应该很容易.

The choice for you should be easy because the Block<? super Document> version is deprecated.

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