[英]Generic <T extends Comparable<T>, V extends T> is V required
Came across this example in Java Complete Reference under Generics method.在 Java Complete Reference 中的泛型方法下遇到了这个例子。
static <T extends Comparable<T>, V extends T> boolean isIn(T x, V[] y) {
for(int i=0; i < y.length; i++)
if(x.equals(y[i])) return true;
return false;
}
Here V extends T
is confusing.这里V extends T
令人困惑。 Can we only have <T extends Comparable<T>>
to get the job done.我们能否只让<T extends Comparable<T>>
来完成工作。 Because T extends some type which means the type and its sub type.因为 T 扩展了一些类型,这意味着类型及其子类型。 Then why we need V extends T?.那为什么我们需要 V 扩展 T?。 Is there any special case for using <T extends Comparable<T>, V extends T>
?使用<T extends Comparable<T>, V extends T>
有什么特殊情况吗?
Yes, in this case, since the V
is only used in one place, in the type of a parameter as V[]
, and since array types are covariant, the V
is unnecessary and the method can be written with this signature:是的,在这种情况下,由于V
仅在一个地方使用,在V[]
参数类型中,并且由于数组类型是协变的,因此V
是不必要的,并且可以使用此签名编写该方法:
static <T extends Comparable<T>> boolean isIn(T x, T[] y)
The two method signatures would accept the same sets of arguments (assuming the caller does not specify an explicit type signature. You can see this as follows:这两个方法签名将接受相同的参数集(假设调用者没有指定显式类型签名。您可以看到如下:
V[]
can be passed to the second signature as T[]
, since V extends T
, so V[] extends T[]
(array types in Java are covariant).任何可以作为V[]
传递给第一个签名的参数都可以作为T[]
传递给第二个签名,因为V extends T
,所以V[] extends T[]
(Java 中的数组类型是协变的)。T[]
can be passed to the first signature as V[]
, because the compiler can always infer V
as T
, since T
is within V
's bounds.任何可以作为T[]
传递给第二个签名的参数都可以作为V[]
传递给第一个签名,因为编译器总是可以将V
推断为T
,因为T
在V
的范围内。 Such an inference will not cause problems anywhere else since V
is not used anywhere else.这样的推断不会在其他任何地方引起问题,因为V
没有在其他任何地方使用。
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