[英]Generic collection & wildcard in java
在以下情況下我遇到問題,請參閱下面的內聯評論:
public void exampleMethod() {
//Intuitively I would expect this to mean that test is set containing objects
//that subclass AbstractGroup
Set<? extends AbstractGroup> test;
//Yet the compiler complains here and I do not understand why?
test.add(new AnyAbstractGroupSubGroup());
//I would guess that a method call such as this at runtime
test = new HashSet<SubGroupA>()
//would mean that only objects of subgroupA can be added to the collection, but then
//what is the point in using the wildcard in the first place?
}
//Intuitively I would expect this to mean that test is set containing objects
//that subclass AbstractGroup
Set<? extends AbstractGroup> test;
不,這意味着它是一組特定的? 它擴展了AbstractGroup。 你和編制者都沒有辦法知道那是什么? 是的,所以你無法向該套裝添加任何東西。
您可以將集合的值分配給AbstractGroup類型的變量,但不能相反。
相反,你需要這個:
Set<? super AbstractGroup> test;
Set<? extends AbstractGroup> test;
這意味着您的集合可以是擴展AbstractGroup
的任何對象的集合,但通常編譯器不允許您向該集合添加內容(因為它無法判斷您是否SubGroupB
添加到Set<SubGroupA>
等)。
test = new HashSet<SubGroupA>()
您的實際集只包含SubGroupA
類型的SubGroupA
及其子類。 但是,編譯器仍然不知道test
的內容是什么(見上文)。
that map to AbstractGroup
(which the compiler checks for). 通配符的要點是:您可以將任何集合分配給使用AbstractGroup
或子類參數化的變量,從而確保您可以將該映射中所有對象為AbstractGroup
(編譯器檢查)。
如果您想擁有一個可以包含任何AbstractGroup
對象的集合,則不要使用通配符。
//this would compile (under the assumption that SubGroupA extends AbstractGroup)
Set<? extends AbstractGroup> test = new HashSet<SubGroupA>();
//this wouldn't compile, since the compiler doesn't know the type of test (it might change at runtime etc.)
test.add(new SubGroupA());
//this wouldn't compile since AbstractGroup and SubGroupA are not the exact same type (hence the wildcard would be needed here)
Set<AbstractGroup> test = new HashSet<SubGroupA>();
//this would compile
Set<AbstractGroup> test = new HashSet<AbstractGroup>();
test.add(new SubGroupA());
你不需要這里的通配符在你的情況下,它就足夠了
Set<AbstractGroup> test;
然后,您可以將AbstractGroup的任何子類放入集合中。
此外,它看起來不像您在上面的代碼中初始化測試。
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