java - 在制作一个n的随机整数倍数的数组时，程序任意修改一个数组

Question

    public class Tester {
        public static int randomInt(int low, int high){//this method finds a random integer in between the range of low and high.
            return (int)(Math.random()*(high-low)+low);
        }
        public static int[] randomArray(int[] a){//this method enters those random integers into an array of 100 length.
            for(int i = 0;i<a.length; i++){
                a[i] = randomInt(0, 100);
            }
            return a;
        }
        public static int[] multiple(int[] a, int n){//this method replaces the input array with a new array consisting of (random) multiples of n
            int[] x = new int[100];
            for(int i = 0; i<a.length; i++){
                x[i] = randomArray(a)[i]*n;
            }
            return x;
        }
        public static void list(int[] ints){//this method lists(prints) each indexes in the array in a i(index):ints[i](the random value) format.
            for(int i = 0; i<ints.length; i++){
                System.out.println(i+":"+ints[i]);
            }
        }
        public static void main(String[] args) {
            int[] a = new int[100];
            list(a);// this will yield 1~99:0 since the indexes in the array weren't assigned any values.
        }

}

然而......

public static void main(String[] args) {
        int[] a = new int[100];
        int[] b = multiple(a, 2);
        list(a);//this would instead give me a list of random numbers(0~99:random value) that are not even multiples of 2.

这没有意义，因为 multiple() 方法不应该修改 int[] a 的值。数组 a 的长度为 100，每个索引中的默认值为 0。然后，数组 a 直接进入 list() 方法，它不应该被 multiple() 改变。 multiple() 给出的新数组存储在数组b 中。为什么会这样？

Answer 1

多次 调用randomArray，后者依次修改作为参数传递的数组:

a[i] = randomInt(0, 100);

解决此问题的一种方法是完全删除此方法，因为无论如何您总是访问数组的单个成员，而只需直接调用 randomInt:

public static int[] multiple(int[] a, int n){
    int[] x = new int[100];
    for(int i = 0; i < a.length; i++) {
        x[i] = randomInt(0, 100) * n; // Here
    }
    return x;
}