在Java类型参数中，&lt ;?扩展E>仅意味着严格的亚型？或者E也足够了？ [英] In Java type arguments, does <? extends E> mean strictly subtypes only? or would E also suffice?

问题描述

在Java类型参数中，仅仅意味着严格的子类型？或者E也足够了？

解决方案

是， super 和 extends 分别给出了包含的上下限。

以下是 Angelika Langer的泛型常见问题解答：

什么是有界通配符？

通配符的上界看起来像？类型类型类型的所有类型的族，表示类型 >被包含。 类型被称为上界。

带有下限的通配符看起来像？超类型，代表超类型为类型，类型类型被包含。 类型被称为下界。




In Java type arguments, does mean strictly subtypes only? or would E also suffice?

解决方案

Yes, super and extends gives inclusive lower and upper bounds respectively.

Here's a quote from Angelika Langer's Generics FAQ:

What is a bounded wildcard?

A wildcard with an upper bound looks like ? extends Type and stands for the family of all types that are subtypes of Type , type Type being included. Type is called the upper bound.

A wildcard with a lower bound looks like ? super Type and stands for the family of all types that are supertypes of Type , type Type being included. Type is called the lower bound.

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