由于Java泛型的实现,你不能有这样的代码:

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

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

如何在保持类型安全的同时实现这一点?

我在Java论坛上看到一个解决方案是这样的:

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;
}

但我真的不明白这是怎么回事。


当前回答

试试这个。

private int m = 0;
private int n = 0;
private Element<T>[][] elements = null;

public MatrixData(int m, int n)
{
    this.m = m;
    this.n = n;

    this.elements = new Element[m][n];
    for (int i = 0; i < m; i++)
    {
        for (int j = 0; j < n; j++)
        {
            this.elements[i][j] = new Element<T>();
        }
    }
}

其他回答

我必须反过来问一个问题:你的发电机组是“检查”还是“未检查”? 这是什么意思?

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. -> 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]; } } Unchecked: weak typing. No type checking is actually done on any of the objects passed as argument. -> 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; } } 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]; } ... }

所有这些都是由于Java中泛型的一个已知且刻意的弱点:它是使用擦除来实现的,因此“泛型”类不知道它们在运行时是用什么类型参数创建的,因此不能提供类型安全,除非实现了一些显式机制(类型检查)。

在java中不允许创建通用数组,但是你可以这样做

class Stack<T> {
private final T[] array;
public Stack(int capacity) {
    array = (T[]) new Object[capacity];
 }
}

试试这个。

private int m = 0;
private int n = 0;
private Element<T>[][] elements = null;

public MatrixData(int m, int n)
{
    this.m = m;
    this.n = n;

    this.elements = new Element[m][n];
    for (int i = 0; i < m; i++)
    {
        for (int j = 0; j < n; j++)
        {
            this.elements[i][j] = new Element<T>();
        }
    }
}

数组不支持泛型(因为它是另一种类型的数据),但如果你不需要强制转换,你可以在创建它的时候使用未确定的泛型,顺便说一句,它比使用反射要好:

List<?>[] chars = new List[3];

现在我们得到了合法的泛型数组,即使没有Unchecked类型警告,

我想知道这段代码是否会创建一个有效的泛型数组?

public T [] createArray(int desiredSize){
    ArrayList<T> builder = new ArrayList<T>();
    for(int x=0;x<desiredSize;x++){
        builder.add(null);
    }
    return builder.toArray(zeroArray());
}

//zeroArray should, in theory, create a zero-sized array of T
//when it is not given any parameters.

private T [] zeroArray(T... i){
    return i;
}

编辑:也许另一种创建这样一个数组的方法,如果你所需要的大小是已知的和小的,将是简单地提供所需的“null”的数量到zeroArray命令?

虽然这显然不如使用createArray代码通用。