Java中的通用方法参数的getClass（） [英] getClass() of a Generic Method Parameter in a Java

问题描述

以下Java方法无法编译：

The following Java method fails to compile:

<T extends Number> void foo(T t)
{
    Class<? extends T> klass = t.getClass();
}

收到的错误是：
类型不匹配：无法转换为 Class< capture＃3-of？将Number> 扩展为 Class<？扩展T>

Error received is: Type mismatch: cannot convert from Class<capture#3-of ? extends Number> to Class<? extends T>

有人可以解释为什么 Class<？扩展T> 无效，但 Class<？ extends Number> 没问题？

Can someone explain why Class<? extends T> is invalid, but Class<? extends Number> is fine?

Javadoc 说：

实际结果类型是类< ;? extends | X |> 其中| X |是擦除调用getClass的表达式的静态类型。例如，此代码片段中不需要强制转换：

The actual result type is Class<? extends |X|> where |X| is the erasure of the static type of the expression on which getClass is called. For example, no cast is required in this code fragment:

Number n = 0;
Class<? extends Number> c = n.getClass(); 

推荐答案

因为 T class'type不从 T 延伸。相反，它从 Number 延伸，正如您在< T extends Number> 中声明的那样。就那么简单。

Because the T class' type doesn't extend from T. Instead, it extends from Number, exactly as you have declared yourself in <T extends Number>. Simple as that.

更好的问题是：

为什么不进行以下编译？

Why doesn't the following compile?

<T extends Number> void foo(T t)
{
    Class<T> class1 = t.getClass();
}

答案就是 Object＃getClass（）返回 Class<？> 带有无界通配符？因为对象本身并不直接了解其在任意方法中所期望的泛型类型。

The answer to that is that the Object#getClass() returns Class<?> with an unbounded wildcard ? because the object itself is not directly aware about its generic type which is been expected in an arbitrary method.

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