[英]Using generics in Comparable
I am trying to implement generics in Java using Comparable<T>
interface. 我正在尝试使用Comparable<T>
接口在Java中实现泛型。
public static <T> T[] sort(T[] a) {
//need to compare 2 elements of a
}
Let's say, I want to override the compareTo
method for the above type T
in the Comparable
interface. 假设我想在Comparable
接口中覆盖上述类型T
的compareTo
方法。 Ie I need to compare two elements of my type T
, how will I do it? 即我需要比较我的T
型的两个元素,我将如何做? I don't know what my T
type will be. 我不知道我的T
型会是什么。
You need to set a type constraint on your method. 您需要在方法上设置类型约束。
public static <T extends Comparable<? super T>> T[] sort (T[] a)
{
//need to compare 2 elements of a
}
This forces the type T
to have the compareTo(T other)
method. 这会强制类型T
具有compareTo(T other)
方法。 This means you can do the following in your method: 这意味着您可以在方法中执行以下操作:
if (a[i].compareTo(a[j]) > 0) }
}
Old question but... 老问题但......
As jjnguy responded, you need to use: 正如jjnguy回应,你需要使用:
public static <T extends Comparable<? super T>> T[] sort(T[] a) {
...
}
Consider the following: 考虑以下:
public class A implements Comparable<A> {}
public class B extends A {}
The class B
implicitly implements Comparable<A>
, not Comparable<B>
, hence your sort method could not be used on an array of B
's if used Comparable<T>
instead of Comparable<? super T>
B
类隐式实现了Comparable<A>
,而不是Comparable<B>
,因此如果使用Comparable<T>
而不是Comparable<? super T>
那么你的sort方法不能用在B
的数组上Comparable<? super T>
Comparable<? super T>
. Comparable<? super T>
。 To be more explicit: 更明确一点:
public static <T extends Comparable<T>> T[] brokenSort(T[] a) {
...
}
would work just fine in the following case: 在以下情况下可以正常工作:
A[] data = new A[3];
...
data = brokenSort(A);
because in this case the type parameter T
would be bound to A
. 因为在这种情况下,类型参数T
将绑定到A
The following would produce a compiler error: 以下将产生编译器错误:
B[] data = new B[3];
...
data = brokenSort(B);
because T
cannot be bound to B
since B
does not implement Comparable<B>
. 因为T
不能绑定到B
因为B
没有实现Comparable<B>
。
尝试使用<T extends Comparable<T>>
然后compareTo
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