Java泛型和模板(Java Generics and templates)

 我有以下Java类定义： 
 
import java.util.*;

public class Test {

static public void copyTo(Iterator<? extends Number> it, List<? extends Number> out) {
    while(it.hasNext())
        out.add(it.next());
}
public static void main(String[] args) {
    List<Integer> in = new ArrayList<Integer>();
    for (int i = 1; i <= 3; i++) {
        in.add(i);
    }
    Iterator<Integer> it = in.iterator();
    List<Number> out = new ArrayList<Number>();
    copyTo(it, out);
    System.out.println(out.size());
}
 
 } 
 
 就是这样，我在Java使用wildcards定义了方法copyTo 。 我定义List<Number> out但是Iterator<Integer> it 。 我的想法是我可以将迭代Iterator<? extends Number>定义为Iterator<? extends Number> Iterator<? extends Number> ，然后键入match。 然而事实并非如此： 
 
Test.java:13: error: no suitable method found for add(Number)
            out.add(it.next());
               ^
    method List.add(int,CAP#1) is not applicable
      (actual and formal argument lists differ in length)
    method List.add(CAP#1) is not applicable
      (actual argument Number cannot be converted to CAP#1 by method invocation conversion)
    method Collection.add(CAP#1) is not applicable
      (actual argument Number cannot be converted to CAP#1 by method invocation conversion)
  where CAP#1 is a fresh type-variable:
    CAP#1 extends Number from capture of ? extends Number
1 error
 
 所以我继续前进，我为copyTo方法定义了另一个定义： 
 
static public void copyTo(Iterator<? super Integer> it, List<? super Integer> out) {
        while(it.hasNext())
            out.add(it.next());
    }
 
 它也不起作用。 在这种情况下使用wildcards的正确说法是什么？ 

I have the following Java class definition:
 
import java.util.*;

public class Test {

static public void copyTo(Iterator<? extends Number> it, List<? extends Number> out) {
    while(it.hasNext())
        out.add(it.next());
}
public static void main(String[] args) {
    List<Integer> in = new ArrayList<Integer>();
    for (int i = 1; i <= 3; i++) {
        in.add(i);
    }
    Iterator<Integer> it = in.iterator();
    List<Number> out = new ArrayList<Number>();
    copyTo(it, out);
    System.out.println(out.size());
}
 
}
 
That's it, I define the method copyTo using wildcards in Java. I define List<Number> out but Iterator<Integer> it. My thinking is I can define the iterator as Iterator<? extends Number> and that would type match. However that's not the case:
 
Test.java:13: error: no suitable method found for add(Number)
            out.add(it.next());
               ^
    method List.add(int,CAP#1) is not applicable
      (actual and formal argument lists differ in length)
    method List.add(CAP#1) is not applicable
      (actual argument Number cannot be converted to CAP#1 by method invocation conversion)
    method Collection.add(CAP#1) is not applicable
      (actual argument Number cannot be converted to CAP#1 by method invocation conversion)
  where CAP#1 is a fresh type-variable:
    CAP#1 extends Number from capture of ? extends Number
1 error
 
So I went ahead and I defined yet another definition for the copyTo method:
 
static public void copyTo(Iterator<? super Integer> it, List<? super Integer> out) {
        while(it.hasNext())
            out.add(it.next());
    }
 
It doesn't work either. What would be the correct say of using wildcards in this case?

原文：https://stackoverflow.com/questions/33586166