java 有什么问题： LinkedList<String> stringList = new LinkedList<String>();
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What's wrong with: LinkedList<String> stringList = new LinkedList<String>();
提问by Eric Wilson
When I try:
当我尝试：
LinkedList<String> stringList = new LinkedList<String>();
I get the following compilation error:
我收到以下编译错误：
type LinkedList does not take parameters
What am I missing? Can't you do this?
我错过了什么？你不能这样做吗？
回答by newacct
Check to make sure you don't have a compiled class named LinkedListin the same directory. (Especially since "linked list" is a common term, and it is something that people often try to implement as beginners.) This is important if you import your classes using something like import java.util.*;, because the *imports on-demand, so if there is a class with the same name in the package already, then that class is used and the java.util.LinkedListis not imported.
检查以确保您没有LinkedList在同一目录中命名的已编译类。（特别是因为“链表”是一个常用术语，人们经常在初学者时尝试实现它。）如果您使用类似的东西导入类，这一点很重要import java.util.*;，因为*按需导入，所以如果有已在包中具有相同名称的类，则使用该类并且java.util.LinkedList不导入该类。
回答by cletus
Are you possibly compiling against a JDK 1.4 or earlier? Or do you have your language setting in your build or IDE set to pre-5.0 (so no generics support)?
您是否可能针对 JDK 1.4 或更早版本进行编译？或者您是否将构建或 IDE 中的语言设置设置为 5.0 之前的版本（因此不支持泛型）？
By the way, the best way to do that is
顺便说一句，最好的方法是
List<String> stringList = new LinkedList<String>();
Use the interface rather than the implementation wherever possible.
尽可能使用接口而不是实现。
That being said, assuming you're compiling against a JDK 5.0+, have your language settings set to Java 5+ and that is a java.util.LinkedList then your code is perfectly valid.
话虽如此，假设您正在针对 JDK 5.0+ 进行编译，将您的语言设置设置为 Java 5+，并且这是一个 java.util.LinkedList，那么您的代码是完全有效的。
回答by saheer
Don't take the class name as class LinkedListinstead you can take class LinkedListDemoand rest of the declaration LinkedList<String> t = new LinkedList<String>();should be there and it will compile perfectly.
不要取类名，因为class LinkedList您可以取而代之class LinkedListDemo，其余的声明LinkedList<String> t = new LinkedList<String>();应该在那里，它会完美编译。
回答by the hero100
I had a same problem and I figured out that I mistakenly used: import java.awt.List; and got the following error message: "Type List doesn't take paramaters"
我遇到了同样的问题，我发现我错误地使用了： import java.awt.List; 并收到以下错误消息：“类型列表不接受参数”
Use this instead
改用这个
import java.util.List;
导入 java.util.List;
回答by TARUN NAGIA
You have used import java.util.*;
你用过 import java.util.*;
You will not face any issue if you use import java.util.LinkedList;
如果您使用，您将不会遇到任何问题 import java.util.LinkedList;
