invokeAll（）不愿意接受Collection< Callable< T>> [英] invokeAll() is not willing to accept a Collection<Callable<T>>

问题描述

我无法理解为什么这段代码无法编译

I fail to understand why this code won't compile

ExecutorService executor = new ScheduledThreadPoolExecutor(threads);

class DocFeeder implements Callable<Boolean> {....} 
... 
List<DocFeeder> list = new LinkedList<DocFeeder>();
list.add(new DocFeeder(1));
...
executor.invokeAll(list);

错误信息是：

The method invokeAll(Collection<Callable<T>>) in the type ExecutorService is 
not applicable for the arguments (List<DocFeeder>)  

list 是 Collection DocFeeder ，它实现了 Callable< Boolean> - 发生了什么？！

list is a Collection of DocFeeder, which implements Callable<Boolean> - What is going on?!

推荐答案

只是为了扩展saua的答案......

Just to expand on saua's answer a little...

在Java 5中，方法是声明为：

In Java 5, the method was declared as:

invokeAll(Collection<Callable<T>> tasks) 

在Java 6中，该方法声明为：

In Java 6, the method is declared as:

invokeAll(Collection<? extends Callable<T>> tasks) 

通配符差别非常大重要 - 因为列表< DocFeeder> 集合<？扩展Callable< T>> 但它不 Collection< Callable< T>> 。考虑一下这种方法会发生什么：

The wildcarding difference is very important - because List<DocFeeder> is a Collection<? extends Callable<T>> but it's not a Collection<Callable<T>>. Consider what would happen with this method:

public void addSomething(Collection<Callable<Boolean>> collection)
{
    collection.add(new SomeCallable<Boolean>());
}

这是合法的 - 但如果你能打电话给 addSomething 带有列表< DocFeeder> ，因为它会尝试将非DocFeeder添加到列表中。

That's legal - but it's clearly bad if you can call addSomething with a List<DocFeeder> as it will try to add a non-DocFeeder to the list.

因此，如果您遇到Java 5，则需要从列表< Callable< Boolean>> >列表< DocFeeder> 。

So, if you are stuck with Java 5, you need to create a List<Callable<Boolean>> from your List<DocFeeder>.

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