这种原始类型分配类型安全吗？列表与LT; T> = new ArrayList（）; [英] Is this raw type assignment type-safe? List<T> = new ArrayList();

问题描述

我有这样的代码：

  @SuppressWarnings（{unchecked，rawtypes}）
列表< String> theList = new ArrayList（）; 
  

这种类型安全吗？我认为这是安全的，因为我不把原始类型分配给其他任何东西。当我调用 add 时，我甚至可以证明它执行类型检查：

  theList.add（601）; //编译错误
  

我已阅读 什么是原始类型，为什么我们不应该使用它？ ，但我不认为它适用于此，因为我只用 创建具有原始类型的列表。之后，我将它分配给一个参数化类型，那么会出现什么问题？

另外，这又如何？

  @SuppressWarnings（{unchecked，rawtypes}）
 List< String> anotherList = new ArrayList（theList）; 
  

解决方案

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 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);

解决方案
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.

                        
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