如何在 Java 中创建一个通用数组？

Question

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
    }
}

How can I implement this while maintaining type safety?

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;
}

But I really don't get what's going on.

转载于:https://stackoverflow.com/questions/529085/how-to-create-a-generic-array-in-java

Answer 1

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

• 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];
      }
  
      ...
  }
  

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.