java 如何调用泛型类型对象的方法？
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How do I call a method of a generic type object?
提问by Callum
The below code gives me the error:
下面的代码给了我错误：
SceneNode.java:17: cannot find symbol
symbol : method execute() location:
class java.lang.Object
operation.execute();
^ 1 error
Code:
代码：
import java.util.LinkedList;
import java.util.Iterator;
public class SceneNode<T>{
T operation;
public SceneNode() {
}
public SceneNode(T operation) {
this.operation = operation;
}
public void setOperation(T operation) {
this.operation = operation;
}
public void doOperation() {
operation.execute();
}
}
It's a cut down (for your readability) start of a simple scene graph. The node could be a model, transformation, switch, etc., so I made a variable called operationthat's type is defined by the Tclass variables. This way I can pass a Transformation/ Model/ Switchobject (that has an executemethod) and pass it like this:
这是一个简单场景图的缩减（为了您的可读性）开始。节点可以是模型、转换、开关等，所以我创建了一个名为operation该类型的T变量，该变量由类变量定义。这样我可以通过一个Transformation/ Model/Switch对象（即具有一个execute方法），并通过它是这样的：
SceneNode<Transformation> = new SceneNode<Transformation>(myTransformation);
I'm pretty sure having a base class of SceneNodeand subclassing for all the various types of nodes would be a better idea (I was trying out generics, only learnt about them recently). Why doesn't this work? I must be missing something fundamental about generics.
我很确定SceneNode为所有各种类型的节点建立一个基类和子类将是一个更好的主意（我正在尝试泛型，最近才了解它们）。为什么这不起作用？我一定遗漏了一些关于泛型的基本知识。
回答by Jon Skeet
It doesn't work because Tcould be anytype, and Java is statically typed. The compiler has no idea whether you'll try to create a SceneNode<String>- then what would executedo?
它不起作用，因为T可以是任何类型，而 Java 是静态类型的。编译器不知道您是否会尝试创建一个SceneNode<String>- 那么怎么execute办？
One option is to create an appropriate interface, e.g.
一种选择是创建一个适当的界面，例如
public interface Executable {
void execute();
}
and then to constrain Tin SceneNodeto implement Executable:
然后限制T在SceneNode实施Executable：
public class SceneNode<T extends Executable> {
...
}
(I find it a little bit odd that Thas to extendExecutablerather than implementit in the source code, but then Tcould end up being an interface itself, so I guess it makes sense.)
（我发现它T必须扩展Executable而不是在源代码中实现它有点奇怪，但T最终可能会成为一个接口本身，所以我想这是有道理的。）
Then it should work fine. Of course you couldmake Executablean abstract superclass instead - or even a (non-final) concrete class - if you wanted, but I would generally prefer to use an interface unless I had some reason not to.
那么它应该可以正常工作。当然，如果你愿意，你可以创建Executable一个抽象的超类——或者甚至是一个（非最终的）具体类——但我通常更喜欢使用接口，除非我有一些理由不这样做。
回答by dty
I'm guessing you come from a C++ background.
我猜你来自 C++ 背景。
The compiler has no idea what kind of a thing T might be because you haven't told it.
编译器不知道 T 可能是什么东西，因为你没有告诉它。
If you had an interface called, for example, Executablewhich defines your execute()method, then you would need to do:
例如，如果您调用了一个接口Executable来定义您的execute()方法，那么您需要执行以下操作：
public class SceneNode<T extends Executable> {
// ...
}
Now, the compiler will know that T is an Executable, and will give you access to all the methods on that interface.
现在，编译器会知道 T 是一个Executable，并且会让您访问该接口上的所有方法。
回答by Bozho
Java is statically typed language. You must know the type at compile-time in order to be able to invoke a method. Instead of a subclass you can have an interface Executablethat defines the execute()method. T(without any <T extends SomeClass>) has only the methods defined by java.lang.Object.
Java 是静态类型语言。您必须在编译时知道类型才能调用方法。您可以拥有一个Executable定义execute()方法的接口，而不是子类。T（没有任何<T extends SomeClass>）只有由 定义的方法java.lang.Object。
