如何在Java中创建一个通用的阵列？ [英] How to create a generic array in Java?

问题描述

由于Java泛型的实现，你不能有code是这样的：

Due to the implementation of Java generics, you can't have code like this:

public class GenSet<E> {
    private E a[];

    public GenSet() {
        a = new E[INITIAL_ARRAY_LENGTH]; // error: generic array creation
    }
}

我如何能实现这一点的同时保持类型安全？

How can I implement this while maintaining type safety?

我看到是这样的关于Java论坛的解决方案：

I saw a solution on the Java forums that goes like this:

import java.lang.reflect.Array;

class Stack<T> {
    public Stack(Class<T> clazz, int capacity) {
        array = (T[])Array.newInstance(clazz, capacity);
    }

    private final T[] array;
}

但我真的不明白是怎么回事。谁能帮助？

But I really don't get what's going on. Can anyone help?

推荐答案

我要问的回报一个问题：在你的柴油发电机组检查或取消选中？
这是什么意思？

I have to ask a question in return: is your GenSet "checked" or "unchecked"? What does that mean?

• 经过：强类型的。 柴油发电机组知道明确它所包含的对象类型（即它的构造函数被显式调用了类＆LT; E＆GT; 参数，当它们被传递，并且是类型电子，见<一不参数的方法将抛出一个异常href=\"http://docs.oracle.com/javase/7/docs/api/java/util/Collections.html#checkedCollection%28java.util.Collection,%20java.lang.Class%29\"><$c$c>Collections.checkedCollection.

• Checked: strong typing. GenSet knows explicitly what type of objects it contains (i.e. its constructor was explicitly called with a Class<E> argument, and methods will throw an exception when they are passed arguments that are not of type E. See Collections.checkedCollection.

- >在这种情况下，你应该写：

-> in that case, you should write:

public class GenSet<E> {

    private E[] a;

    public GenSet(Class<E> c, int s) {
        // Use Array native method to create array
        // of a type only known at run time
        @SuppressWarnings("unchecked")
        final E[] a = (E[]) Array.newInstance(c, s);
        this.a = a;
    }

    E get(int i) {
        return a[i];
    }
}

未选中：弱类型的。没有类型检查任何作为参数传递的对象实际上完成的。

Unchecked: weak typing. No type checking is actually done on any of the objects passed as argument.

- >在这种情况下，你应该写

-> in that case, you should write

public class GenSet<E> {

    private Object[] a;

    public GenSet(int s) {
        a = new Object[s];
    }

    E get(int i) {
        @SuppressWarnings("unchecked")
        final E e = (E) a[i];
        return e;
    }
}

请注意，数组的组件类型应该是的删除的类型参数的：

Note that the component type of the array should be the erasure of the type parameter:

public class GenSet<E extends Foo> { // E has an upper bound of Foo

    private Foo[] a; // E erases to Foo, so use Foo[]

    public GenSet(int s) {
        a = new Foo[s];
    }

    ...
}

这一切都从已知的，和蓄意的，在Java泛型的弱点的结果：它使用擦除实现的，所以通用类不知道什么类型的说法，他们在运行时使用创建的，因此不能提供类型安全，除非一些明确的机制（类型检查）被实现。

All of this results from a known, and deliberate, weakness of generics in Java: it was implemented using erasure, so "generic" classes don't know what type argument they were created with at run time, and therefore can not provide type-safety unless some explicit mechanism (type-checking) is implemented.

这篇关于如何在Java中创建一个通用的阵列？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！