如何迭代通配符泛型? [英] How to iterate over a wildcard generic?

问题描述

如何遍历通配符泛型?基本上我想内联以下方法:

How can I iterate over a wildcard generic? Basically I would like to inline the following method:

private <T extends Fact> void iterateFacts(FactManager<T> factManager) {
    for (T fact : factManager) {
        factManager.doSomething(fact);
    }
}

如果此代码在如图所示的单独方法中，它会起作用，因为泛型方法上下文允许定义可以迭代的通配符类型(此处为 T).如果尝试内联此方法，则方法上下文将消失，并且无法再迭代通配符类型.即使在 Eclipse 中自动执行此操作也会失败并显示以下(不可编译的)代码:

If this code is in a separate method as shown, it works because the generic method context allows to define a wildcard type (here T) over which one can iterate. If one tries to inline this method, the method context is gone and one cannot iterate over a wildcard type anymore. Even doing this automatically in Eclipse fails with the following (uncompilable) code:

...
for (FactManager<?> factManager : factManagers) {
    ...
    for ( fact : factManager) {
        factManager.doSomething(fact);
    }
    ...
}
...

我的问题很简单:有没有办法放置一些可以迭代的通配符类型，或者这是泛型的限制(意味着不可能这样做)?

My question is simply: Is there a way to put some wildcard type one can iterate over, or is this a limitation of generics (meaning it is impossible to do so)?

推荐答案

类型参数只能定义在

• 类型(即类/接口)，
• 方法和
• 构造函数.

您需要一个本地块的类型参数，这是不可能的.

You would need a type parameter for a local block, which is not possible.

是的，我有时也会错过这样的事情.

Yeah, I missed something like this sometimes, too.

但是在这里非内联方法并没有真正的问题 - 如果它出现性能瓶颈而内联会有所帮助，Hotspot 会再次内联它(不关心类型).

But there is not really a problem with having the method non-inlined here - if it presents a performance bottleneck where inlining would help, Hotspot will inline it again (not caring about the type).

此外，拥有一个单独的方法可以给它一个描述性的名称.

Additionally, having a separate method allows giving it a descriptive name.

只是一个想法，如果你经常需要这个:

Just an idea, if you need this often:

interface DoWithFM {
   void <T> run(FactManager<T> t);
}

...
for (FactManager<?> factManager : factManagers) {
    ...
    new DoWithFM() { public <T> run(FactManager<T> factManager) {
        for (T fact : factManager) {
            factManager.doSomething(fact);
        }
    }.run(factManager);
    ...
}
...

这篇关于如何迭代通配符泛型?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！