java 获取组件类型的数组Class
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Obtaining the array Class of a component type
提问by Tom Castle
If I have an instance of Class, is there a way of obtaining a Classinstance for its array type? What I'm essentially asking for is the equivalent of a method getArrayTypewhich is the inverse of the getComponentType()method, such that:
如果我有一个 的实例Class，有没有办法获得Class它的数组类型的实例？我本质上要求的是与方法getArrayType相反的getComponentType()方法的等价物，例如：
array.getClass().getComponentType().getArrayType() == array.getClass()
回答by Bozho
One thing that comes to mind is:
想到的一件事是：
java.lang.reflect.Array.newInstance(componentType, 0).getClass();
But it creates an unnecessary instance.
但它创建了一个不必要的实例。
Btw, this appears to work:
顺便说一句，这似乎有效：
Class clazz = Class.forName("[L" + componentType.getName() + ";");
Here is test. It prints true:
这里是测试。它打印true：
Integer[] ar = new Integer[1];
Class componentType = ar.getClass().getComponentType();
Class clazz = Class.forName("[L" + componentType.getName() + ";");
System.out.println(clazz == ar.getClass());
The documentation of Class#getName()defines strictly the format of array class names:
的文档Class#getName()严格定义了数组类名的格式：
If this class object represents a class of arrays, then the internal form of the name consists of the name of the element type preceded by one or more '[' characters representing the depth of the array nesting.
如果此类对象表示一类数组，则名称的内部形式由元素类型的名称组成，前面有一个或多个代表数组嵌套深度的“[”字符。
The Class.forName(..)approach won't directly work for primitives though - for them you'd have to create a mapping between the name (int) and the array shorthand - (I)
不过，该Class.forName(..)方法不会直接适用于原语 - 对于它们，您必须在名称 ( int) 和数组速记之间创建映射- ( I)
回答by FroMage
Actually due to ClassLoader, primitives and multi-dimensional arrays, the answer is a little more complex:
实际上由于ClassLoader, 基元和多维数组，答案有点复杂：
public static Class<?> getArrayClass(Class<?> componentType) throws ClassNotFoundException{
ClassLoader classLoader = componentType.getClassLoader();
String name;
if(componentType.isArray()){
// just add a leading "["
name = "["+componentType.getName();
}else if(componentType == boolean.class){
name = "[Z";
}else if(componentType == byte.class){
name = "[B";
}else if(componentType == char.class){
name = "[C";
}else if(componentType == double.class){
name = "[D";
}else if(componentType == float.class){
name = "[F";
}else if(componentType == int.class){
name = "[I";
}else if(componentType == long.class){
name = "[J";
}else if(componentType == short.class){
name = "[S";
}else{
// must be an object non-array class
name = "[L"+componentType.getName()+";";
}
return classLoader != null ? classLoader.loadClass(name) : Class.forName(name);
}
回答by Peter Lawrey
You can do the following
您可以执行以下操作
array.getClass() ==
Array.newInstance(array.getClass().getComponentType(), 0).getClass()
Usually, you don't need to know the type, you just want to create the array.
通常，您不需要知道类型，您只想创建数组。
回答by Leigh Klotz
Another possible refactoring is to use a generic superclass and pass in two class objects to the constructor.
另一种可能的重构是使用通用超类并将两个类对象传递给构造函数。
protected AbstractMetaProperty(Class<T> valueClass, Class<T[]> valueArrayClass) {
this.valueClass = valueClass;
this.valueArrayClass = valueArrayClass;
}
Then in subclasses:
然后在子类中：
public IntegerClass() {
super(Integer.class, Integer[].class);
}
Then in the abstract class you can use valueClass.cast(x), valueArrayClass.isInstance(x)etc.
然后在抽象类，你可以使用valueClass.cast(x)，valueArrayClass.isInstance(x)等等。
