在接收数组的方法中分配两个数组[重复]

Question

此问题已经在这里有了答案：

Is Java “pass-by-reference” or “pass-by-value”?（86个答案）

已关闭1分钟前。

在接收数组的方法中分配两个数组，但是内容未更改，为什么

public class Sol {

    public static void main(String[] args) {
        int[] list = {1, 2, 3, 4, 5}; 
        reverse(list); 

        for (int i = 0; i < list.length; i++) 
            System.out.print(list[i] + " ");// here it prints after reversing 1 2 3 4 5 why?
}

public static void reverse(int[] list) {
     int[] array = new int[list.length];

     for (int i = 0; i < list.length; i++) 
         array[i] = list[list.length - 1 - i];
         list = array;// is it call by value  //here assignment why does not change the contents after exit from the method
     }
}

Answer 1

Java对于原始类型按值传递，对于对象按引用传递。请参阅下面的链接以获取更多详细信息。

http://docs.oracle.com/javase/specs/jls/se8/html/jls-8.html#jls-8.4.1

public static void main(String[] args) {
        int[] list = { 1, 2, 3, 4, 5 };

        list = reverse(list);

        for (int i = 0; i < list.length; i++)
            System.out.print(list[i] + " "); // here it prints after reversing 1 2 3 4 5 why? }
    }

    public static int[] reverse(int[] list) {

        int[] tempList = new int[list.length];

        for (int i = 0; i < list.length; i++)
            tempList[i] = list[list.length - 1 - i];

        return tempList;
    }

Answer 2

int[] list = {1, 2, 3, 4, 5};

**reverse(list);**

for (int i = 0; i < list.length; i++)

System.out.print(list[i] + " ");// here it prints after reversing 1 2 3 4 5 why? }

public static void reverse(int[] list) {

int[] new List = new int[list.length];

for (int i = 0; i < list.length; i++)

new List[i] = list[list.length - 1 - i];

list = new List;}

}

在此之后，您没有将反向数组分配给任何变量。这就是这个问题的原因。像这样更改并检查列表=反向（列表）;