第一种是类型安全的，因为列表为空，但仍然没有建议。在这里使用原始类型没有好处。第二个肯定是不是类型安全的，因为 theList $ c>可能是列表<整数> ，例如：



  import java.util。*; 

 public class Test {

 public static void main（String [] args）throws Exception {
 List< Integer>整数= new ArrayList<>（）; 
 integers.add（0）; 

列表< String> strings = new ArrayList（整数）; 
 // Bang！ 
 String x = strings.get（0）; 




 $ b $ p 
 $ b 
请注意构造函数本身是如何在没有异常的情况下被调用的 - 没有办法让它知道你真的试图构建什么样的列表，所以它不会执行任何转换。但是，如果您然后获取值，则隐式转换为 String ，您将得到 ClassCastException 。
 
I have some code like this:
@SuppressWarnings({"unchecked", "rawtypes"})
List<String> theList = new ArrayList();

Is this type-safe? I think it is safe because I don't assign the raw type to anything else. I can even demonstrate that it performs type checking when I call add:
theList.add(601); // compilation error

I have read "What is a raw type and why shouldn't we use it?" but I don't think it applies here because I only create the list with a raw type. After that, I assign it to a parameterized type, so what could go wrong?

Also, what about this?
@SuppressWarnings({"unchecked", "rawtypes"})
List<String> anotherList = new ArrayList(theList);




The first is type-safe because the list is empty, but still not advised. There's no benefit in using a raw type here. Better to design away from a warning than suppress it.

The second is definitely not type-safe, as theList may be a List<Integer> for example:
import java.util.*;

public class Test {

    public static void main(String[] args) throws Exception {
        List<Integer> integers = new ArrayList<>();
        integers.add(0);

        List<String> strings = new ArrayList(integers);
        // Bang!
        String x = strings.get(0);
    }
}

Note how the constructor itself is called without an exception - there's no way for it to know what kind of list you're really trying to construct, so it doesn't perform any casts. However, when you then fetch a value, that implicitly casts to String and you get a ClassCastException.