Java通用方法问题

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问题描述

考虑以下代码:

public <T> List<T> meth(List<?> type)
{
   System.out.println(type); // 1
   return new ArrayList<String>(); // 2
}

它不会在第2行进行编译，表示必须使用List.

It does not compile at line 2, saying that List is required.

现在，如果将其更改为:

Now, if it's changed to:

public <T> List<?> meth(List<T> type)
{
   System.out.println(type); // 1
   return new ArrayList<String>(); // 2
}

它确实可以编译.为什么?我认为用T声明通用类型和使用通配符之间的区别在于，当使用通配符时，不能将新元素添加到集合中.为什么<?>允许返回List的子类型?我在这里遗漏了什么，明确的规则是什么以及如何应用它?

It does compile. Why? I thought the difference between declaring a generic type with T and using the wildcard was that, when using the wildcard, one cannot add new elements to a collection. Why would <?> allow a subtype of List to be returned? I'm missing something here, what's the explicit rule and how it's being applied?

推荐答案

区别在于返回类型声明. List<String>不是List<T>的子类型，但是它是List<?>的子类型.

The difference is in the return type declaration. List<String> is not a subtype of List<T>, but it is a subtype of List<?>.

List<?>对其类型变量不做任何假设，因此以下语句是有效的:

List<?> makes no assumptions regarding its type variable, so the following statements are valid:

List<?> l0 = new ArrayList<String>();
List<?> l1 = new ArrayList<Object>();
List<? extends Number> ltemp = null;
List<?> l2 = ltemp;

List<T>假定将类型参数声明为List<String>或List<Object>时，将在客户端的上下文中解析该类型参数(例如，type use).在方法主体中，您也不能对此做任何假设.

List<T> assumes that the type argument will be resolved in the context of client (e.g. type use), when you declared it as List<String> or List<Object>. Within the method body, you cannot make any assumptions about it either.

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