如何在 Java 中使用泛型类型获取类
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How to get class with generics types in Java
提问by ZZ Coder
I am trying to make a method call like this,
我正在尝试进行这样的方法调用，
public class GenericsTest<T> {
public static <T> Map<String, T> createMap(Class<? extends Map<String, T>> clazz) {
return null;
}
public static void main(String[] argv) {
Map<String, Integer> result = createMap(TreeMap.class);
}
}
But I am getting this error,
但我收到这个错误，
<T>createMap(java.lang.Class<? extends java.util.Map<java.lang.String,T>>) in test.GenericsTest<T> cannot be applied to (java.lang.Class<java.util.TreeMap>)
How to fix this problem?
如何解决这个问题？
采纳答案by pickypg
Map<String, Integer> instance = new TreeMap<String, Integer>();
@SuppressWarnings("unchecked")
Map<String, Integer> map =
createMap((Class<? extends Map<String, Integer>>)instance.getClass());
map.put("x", 1);
System.out.println("THIS IS x: " + map.get("x"));
This will appropriately print out 1. The implementation of the method is most likely
这将适当地打印出 1. 方法的实现是最有可能的
try
{
return clazz.newInstance();
}
catch (Exception e)
{
throw new RuntimeException(e);
}
A better implementation of their API would be for them to ask you for the type, T, and for them to give back a Mapof their choosing instead of asking you for all of the details. Otherwise, as long as they are not filling in the Mapwith any data, you can instantiate a Mapwith the generic type argument yourself like so:
他们的 API 的更好实现是让他们询问您的类型，T并让他们返回Map他们选择的a ，而不是询问您所有的细节。否则，只要它们没有填充Map任何数据，您就可以Map像这样自己使用泛型类型参数实例化 a ：
public static <T> Map<String, T> getMap()
{
return new TreeMap<String, T>();
}
You can then access that without a warning:
然后您可以在没有警告的情况下访问它：
// note the lack of type arguments, which are inferred
Map<String, Integer> instance = getMap();
// alternatively, you could do it more explicitly:
// Map<String, Integer> instance = ClassName.<Integer>getMap();
There's really no reason for them to ask you for the Classtype of your Mapexcept to give you back an exact match to the implementation (e.g., if you stick in a HashMap, then you will get back a HashMap, and if you stick in a TreeMap, then you will get back a TreeMap). However, I suspect that the TreeMapwill lose any Comparatorthat it was constructed with, and since that is an immutable (final) field of TreeMap, then you cannot fix that; that means that the Mapis not the same in that case, nor is it likely to be what you want.
他们真的没有理由问你你的Class类型，Map除了给你一个完全匹配的实现（例如，如果你坚持 a HashMap，那么你会得到 a HashMap，如果你坚持 a TreeMap，那么你会回来的TreeMap）。但是，我怀疑TreeMap将丢失Comparator它构建的任何内容，并且由于这是一个不可变的 ( final) 字段TreeMap，因此您无法修复它；这意味着Map在那种情况下是不一样的，也不可能是你想要的。
If they are filling in the Mapwith data, then it makes even less sense. You could always pass in an instance of a Mapto fill, or have them return a Mapthat you can simply wrap (e.g., new TreeMap<String, Integer>(instance);), and they should know which Mapoffers the most utility to the data.
如果他们正在填写Map数据，那么就更没有意义了。你总是可以传入一个 a 的实例Map来填充，或者让他们返回一个Map你可以简单地包装的（例如，new TreeMap<String, Integer>(instance);），他们应该知道哪个Map为数据提供了最多的效用。
