java:使用Raw类型作为方法参数会擦除参数成员中所有已参数化的类型信息 [英] java: Use of Raw type as method parameter errases all parametrized type information in parameter members
问题描述
请向我解释为什么如果我在test()方法中使用原始类型A,则类型列表上的get()方法将返回一个Object而不是一个B.:
Please explain me why if I use the raw type A in the method test() , the get() method on my typed list returns an Object and not a B.:
public class test
{
public class B{}
public class C{}
public class A<T extends C>
{
private List<B> aBList;
public List<B> mGetBList()
{
return aBList;
}
}
public test(A pA) // Use of raw type - this is bad, I know!
{
B lB = pA.mGetBList().get(0); // Compile error: Type mismatch:
// cannot convert from Object to test.B
}
}
如果我声明
public test(A<?> pA)
get()方法返回预期的B.
the get() method returns a B as expected.
推荐答案
+1以获得有趣的测试用例.
+1 for interesting test case.
看起来像擦除会擦除所有内容,因此在这种情况下,您最终会得到结果.
Looks like erasure erases everything, so in this case, you end up with.
public List mGetBList()
擦除List
会导致public Object get( int )
,这当然不能分配给B
.
And erasure of List
will result in public Object get( int )
, which, of course, cannot be assigned to B
.
如果方法签名中对原始类型的需求不大,请使用您提供的通用形式,否则将对象强制转换为B
If there is no strong need for raw type in method signature, use generic form that you have provided, otherwise cast the object to B
B lB = (B) pA.mGetBList().get(0);
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