java List<Integer> 不能转换为 ArrayList<Integer>
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List<Integer> cannot be converted to ArrayList<Integer>
提问by beepretty
At the beginning of my code, there is:
在我的代码的开头，有：
List<List<Integer>> result = new ArrayList();
And then, (here is to reverse a sub-list):
然后，（这里是反转一个子列表）：
List<Integer> sub = new ArrayList();
re = reverse(result.get(j)); // error occurs here
There is this method:
有这个方法：
public ArrayList<Integer> reverse(ArrayList<Integer> list) {
List<Integer> newList = new ArrayList<>();
for(int i=list.size()-1; i>=0; i--) {
newList.add(list.get(i));}
return newList;
}
}
The error message is:
错误信息是：
List cannot be converted to ArrayList
列表无法转换为 ArrayList
Why?
为什么？
回答by Captain Man
Think of it like this, you have a Fruitcalled re(I use this name because it's the name of the variable you are using).
可以这样想，您有一个Fruit调用re（我使用这个名称是因为它是您正在使用的变量的名称）。
Fruit re;
You have a method reversewhose input type is Apple.
您有一个reverse输入类型为的方法Apple。
public Apple reverse(Apple a) {
// ...
}
We have a variable rethat we declared as Fruitwhich means we're saying it's always going to be some kind of Fruit, perhaps Apple, but maybe Orange-- or even Banana.
我们有一个re我们声明为的变量，Fruit这意味着我们说它总是会是某种Fruit，也许是Apple，但也许Orange- 甚至是Banana。
When you try to give the Fruitto the method taking Applethe compiler stops you, because it can't be certain that it's 100% an Apple. For example...
当您尝试给予Fruit服用方法Apple编译器停止你，因为它不能确定它是100％的Apple。例如...
Fruit re = new Orange();
reverse(re);
Yikes! We are putting a square peg into a round hole so to speak. reversetakes Apple, not Orange. Bad things could happen!
哎呀！可以这么说，我们正在将一个方钉插入一个圆孔中。reverse需要Apple，不是Orange。可能会发生不好的事情！
Side note:Why is it okay then to assign
Appleto something declared asFruitthen? (reversereturns anApple,Fruit f = reverse(re);is legal.) Because anAppleisaFruit. If it were declared as the more specificAppleand the return type were the more generalFruit, thenthere would be an issue here. (IfreversereturnedFruit,Apple a = reverse(re);would be illegal.)
旁注：为什么可以分配
Apple给那样声明的东西Fruit？（reverse返回 anApple,Fruit f = reverse(re);是合法的。）因为 anApple是aFruit。如果它被声明为更具体的Apple并且返回类型更通用Fruit，那么这里就会出现问题。（如果reverse返回Fruit，Apple a = reverse(re);将是非法的。）
If you didn't follow the metaphor,replace Fruitwith Listand Applewith ArrayListand read the above again. Listis Fruit, a general way to describe an abstract idea. ArrayListis Apple, a specific implementation of the abstract idea. (LinkedListcould be Orangetoo.)
如果你没有按照比喻，替换Fruit用List，并Apple与ArrayList以上再次读取。List是Fruit，描述抽象思想的一般方式。ArrayList是Apple，抽象思想的具体实现。（LinkedList也可能是Orange。）
In general you want to declare things as the most general thing you can to get the functionality you need. Making the below change should fix your problem.
通常，您希望将事物声明为获得所需功能的最一般事物。进行以下更改应该可以解决您的问题。
public List<Integer> reverse(List<Integer> list) {
We are taking some kind of Listof Integers and returning some kind of Listof Integers.
我们正在采取某种List的IntegerS和返回某种List的Integer秒。
回答by Reut Sharabani
You're using a specific implementation of the Listinterface as a return type for the reversefunction (namely ArrayList). This forces what you return to be an ArrayList, but you're defining it as List.
您正在使用List接口的特定实现作为reverse函数的返回类型（即ArrayList）。这会强制您返回的内容为ArrayList，但您将其定义为List。
To understand, try and change the construction part to use a LinkedList, which also implements List:
要理解，请尝试将构造部分更改为使用 a LinkedList，它还实现了List：
List<Integer> newList = new LinkedList<Integer>();
This construction is valid, but now we try and return a LinkedListfrom a function that returns an ArrayList. This makes no sense since they aren't related (But notice this: they both implement List).
这种构造是有效的，但现在我们尝试LinkedList从返回 a的函数中返回 a ArrayList。这是没有意义的，因为它们不相关（但请注意：它们都实现了List）。
So now you have two options:
所以现在你有两个选择：
Make the implementation more specific by forcing use of ArrayLists:
通过强制使用ArrayLists使实现更加具体：
ArrayList<Integer> newList = new ArrayList<>();
Or better, make the implementation more general by changing the function's return type (any Listcan be plugged in):
或者更好的是，通过更改函数的返回类型（List可以插入任何类型）使实现更通用：
public List<Integer> reverse(List<Integer> list){
回答by StansTec
It's because your result varible has a type of List, but your method want an ArrayList Try it with:
这是因为你的结果变量有一个 List 类型，但你的方法想要一个 ArrayList 试试看：
List<ArrayList<Integer> result = new ArrayList<>();
orchange your function in:
或更改您的功能：
public List<Integer> reverse(List<Integer> list){...
