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.
        }

}

无论其.....

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，默认情况下每个索引为0。然后，数组a直接进入list()方法，无论如何都不应该被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

multiple调用randomArray，然后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，因为您始终只访问数组的单个成员，并直接调用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;
}

在制作n的随机整数倍的数组时，程序任意修改数组

1 个答案: