Question

这个代码尽可能地从更复杂的类结构中简化。在实际代码中，我在这里使用了Integer和Double类型的子类型。

我正在尝试使用带有类型参数的Java Generics。如果用户请求Number.class类型，我们希望将List<Integer>列表和List<Double>列表合并到一个列表中。

虽然代码有效，但我无法获得未经检查的强制转换警告（请参阅TODO标记）。警告是：

Type safety: Unchecked cast from List<Integer> to Collection<? extends T>

但是，如果我删除了强制转换，我会收到编译错误：

The method addAll(Collection<? extends T>) in the type List<T> is not applicable for the arguments (List<Integer>).

我的代码：

import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collection;
import java.util.List;

public class Generics1 {

    static final List<Integer> intList = new ArrayList<Integer>(Arrays.asList(
        1, 2, 3, 4));
    static final List<Double> dblList = new ArrayList<Double>(Arrays.asList(
        1.1, 2.2, 3.3));

    public static <T extends Number> List<T> getObjects(Class<T> type) {
        List<T> outList = new ArrayList<T>();
        if (type == Number.class) {
            // user asked for everything
            // TODO: unchecked cast warnings here should be fixed
            outList.addAll((Collection<? extends T>) intList);
            outList.addAll((Collection<? extends T>) dblList);
        } else {
            // user asked for subtype of number
            if (Integer.class.isAssignableFrom(type)) for (Integer i : intList)
                if (type.isInstance(i)) {
                    T obj = type.cast(i);
                    outList.add(obj);
                }
            if (Double.class.isAssignableFrom(type)) for (Double d : dblList)
                if (type.isInstance(d)) {
                    T obj = type.cast(d);
                    outList.add(obj);
                }
        }
        return outList;
    }

    public static void main(String[] args) {
        System.out.println("HI!");
        System.out.println("integers: " + getObjects(Integer.class));
        System.out.println("doubles: " + getObjects(Double.class));
        System.out.println("numbers: " + getObjects(Number.class));
    }
}

Answer 1

您可以将其添加到您的代码中：

@SuppressWarnings("unchecked")

这是另一篇SO帖子，解释了“含义”：What is SuppressWarnings ("unchecked") in Java?

这是另一个处理转换可能有用的链接的问题：How do I fix "The expression of type List needs unchecked conversion...'?

虽然有一些重新编码，但你可能会让警告完全消失而不需要被压制。

Answer 2

（先前的答案已删除）

以下是使用番石榴的方法：

@SuppressWarnings("unchecked")
public static <T> List<T> filterAndCollapse(final Class<T> type,
        Collection<?> a, Collection<?> b) {
    List combined = new ArrayList();
    Predicate<Object> filter = new Predicate<Object>() {

        public boolean apply(Object obj) {
            return type.isInstance(obj);
        }
    };
    combined.addAll(Collections2.filter(a, filter));
    combined.addAll(Collections2.filter(b, filter));
    return combined;
}
// ...
filter(Number.class, intList, dblList);

编辑：完全类型安全的比较方式。

public static <T> List<T> filterAndCollapse(final Class<T> type,
        Collection<?> a, Collection<?> b) {
    List<T> combined = new ArrayList<T>();
    Predicate<Object> filter = new Predicate<Object>() {

        public boolean apply(Object obj) {
            return type.isInstance(obj);
        }
    };
    Function<Object, T> transform = new Function<Object, T>() {

        public T apply(Object obj) {
            return type.cast(obj);
        }
    };
    combined.addAll(Collections2.transform(Collections2.filter(a, filter),
        transform));
    combined.addAll(Collections2.transform(Collections2.filter(b, filter),
        transform));
    return combined;
}

不幸的是，根据我的知识，没有办法用Guava过滤和转换一步。

Answer 3

    (Class<T> type)
    List<T> outList = new ArrayList<T>();

    if (type == Number.class) {
        // obviously, T==Number here, though the compiler doesn't know that
        // so we do the cast. compiler will still warn. since the cast makes 
        // perfect sense and is obviously correct, we are ok with it.   
        List<Number> numList = (List<Number>)outList;
        numList.addAll( intList);
        numList.addAll( dblList);
    } else {

更好的解决方案，简单

for list in lists
  for item in list 
     if item instance of type
        add item to result

使用Java Generics，类型参数和返回列表取消选中转发警告

3 个答案: