将通用ArrayList元素转换为基元类型（长）(Cast a generic ArrayList element into a primitive type (long))

将通用ArrayList元素转换为基元类型（长）(Cast a generic ArrayList element into a primitive type (long))

我想编写一个方法来处理一个泛型数组列表的一个元素，其中包含一个很长的数字。 我知道该数组只包含Long值，所以我不需要检查它。 这是我的方法：

searchSum(ArrayList <T> array, long n);

我需要做的对抗是：

array.get(index)==n

I would like to write a method which confronts an element of a generic ArrayList with a long number. I know that the array contains only Long values, so I don’t need to check this. This is my method:

searchSum(ArrayList <T> array, long n);

The confrontation I need to do is:

array.get(index)==n

最满意答案

您无法将列表转换为基元。 Java列表包含对象，并且基元不是对象。 但取决于你想做什么，我很确定有一个简单的方法来实现它。 不幸的是，从你的问题，我很难理解目标是什么。

如果您想查找列表是否包含您可以使用的元素：

array.indexOf(new Long(n)); 它会告诉你元素的第一次提示的索引。

如果你只是想看看它是否存在，你可以使用array.contains(new Long(n));

如果你想总结所有元素（因为你的方法被称为searchSum），你可以这样做：

searchSum(ArrayList <T> array, long n){ for (Long element: array){ if(Long.valueOf(n).equals(element)){ .... do something here... } } }

You cannot cast list to primitives. Java list contains Objects and the primitives are not objects. But depending on what you want to do I am pretty sure there is an easy way to achieve it. Unfortunately from your question I can hardly understand what is the goal.

If you want to find if the list contains an element you can use:

array.indexOf(new Long(n)); and it will tell you the index of the first ocurance of the element.

If you just want to see if it is there you can use array.contains(new Long(n));

If you want to sum all elements (since your method is called searchSum) you can do:

searchSum(ArrayList <T> array, long n){ for (Long element: array){ if(Long.valueOf(n).equals(element)){ .... do something here... } } }