数组—我在将java程序拆分为同一类中的不同方法时遇到问题,有人能解决吗?

ogq8wdun  于 2021-07-03  发布在  Java
关注(0)|答案(0)|浏览(172)

我的任务是拆分我的程序,它允许用户输入一个数字数组,在1到10之间的奇数之后检查奇数是否是数组中5个数字的一个因子。当我运行程序时,复选框会给出一个输出,但程序的其余部分不会,有人解释为什么吗?。有人能帮我一下吗?或者寄一份样品,看看要换什么样的衣服才能用?程序如下:

import java.util.Scanner;
public class CheckboxExample{  
    Scanner s = new Scanner(System.in);
    int num;
    public static void main(String args[])  {  
        CheckBox c = new CheckBox();
        new CheckboxExample();  // links to checkbox class
        int array[] = new int[10]; 
    }

    public static void readInputArray(Scanner s){
        System.out.println ("Please enter 10 random numbers"); // prompts the user to enter 10 numbers
        int array[] = new int[10]; 
        // declares variable num
        try{
            for (int i = 0; i < 10; i++) {
                array[i] = s.nextInt(); // array declaration
            }
        }catch (Exception e){
            System.out.println ("You have an error");
        }
    }

    public static void readOddValue(Scanner s, int num){
        System.out.println ("Please enter an odd number between 1 and 10");
        try{
            num = s.nextInt ();
            if (num % 2 == 0){
                do{
                    System.out.println ("\nYour number is even, enter an odd one");
                    num = s.nextInt ();
                }while (num % 2 == 0);
            } 

            if (num < 0 | num > 10){
                do{  
                    System.out.println ("Your number is outside of the range, try again");
                    num = s.nextInt ();  
                }while (num < 0 | num > 10);
            } 
        }catch (Exception e){
            System.out.println ("error");
        }
    }

    public static void calculateFactors(int[] array, int num){
        try{
            for (int i = 0; i < 10 ; i++){
                if (array[i] % num  == 0) {
                    System.out.println("Your number is a factor of " + array[i] );
                } 
            }
        }catch (Exception e){
            System.out.println ("error");
        }
    }  
}

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