Question

你能解决我对Josh Blocks Effective Java异构容器所做的工作吗？我想在stringClass中存储String.class的想法，但它不起作用。你能解释一下原因并解决它吗？

package heterogeneous;

import java.util.HashMap;
import java.util.Map;

//Typesafe heterogeneous container pattern - implementation
public class TypedValue {
    private Map<Class<?>, Object> typedValues = new HashMap<Class<?>, Object>();

    public <T> TypedValue(Class<T> type, T instance) {
        put(type, instance);
    }
    public <T> T put(Class<T> type, T instance) {
        if (type == null)
            throw new NullPointerException("Type is null");

        @SuppressWarnings("unchecked")
        T oldValue = (T) typedValues.put(type, instance); 
        return oldValue; 
    }
    public <T> T get(Class<T> type) {
        return type.cast(typedValues.get(type));
    }

    // Typesafe heterogeneous container pattern - client
    public static void main(String[] args) {
        TypedValue j = new TypedValue(String.class, "Java");
        TypedValue i = new TypedValue(Integer.class, 0xcafebabe);
        TypedValue t = new TypedValue(Class.class, TypedValue.class);

        String favoriteString = j.get(String.class);
        int favoriteInteger = i.get(Integer.class);
        Class<?> favoriteClass = t.get(Class.class);

        // .-=== Fix and explain what was wrong ==-. //
        Class<?> stringClass = String.class;

        //Type mismatch: cannot convert from capture#1-of ? to String 
        favoriteString = j.get(stringClass); 
        // ^-=== Fix and explain what was wrong ==-^ //

        System.out.printf( "%s %x %s%n", 
                           favoriteString, 
                           favoriteInteger, 
                           favoriteClass.getName() );
    }
}

Answer 1

在这里使用通配符

Class<?> stringClass = String.class;

您明确表示您不了解（关心）基础类型。它可以是任何东西。但是你的方法

favoriteString = j.get(stringClass);

取决于该类型以确定其返回类型。由于它未知，因此返回类型也将推断为?。您不能将某种未知类型的值分配给String类型的变量。

将其更改为

Class<String> stringClass = String.class;

Typesafe异构容器拼图

1 个答案: