Question

以下程序出了什么问题？因为我希望它显示范围为1-10, 11-20,21-30 ... 191-200？

的用户输入整数值

public class Program
{
    /**
     * This is the main entry point for the application
     */
   public static void main(String args[]) 
{
  int a[] = new int[100];
  int i = 0; 
  Scanner in = new Scanner(System.in);
  while(i<100)
  {
  System.out.println("Enter a int");
  a[i] = in.nextInt();
  displayStatistics(a[i]);


  }

}

    public static void displayStatistics(integer[] a[i])
    {
        if(a[i]>=1 && a[i]<=100) 
      {
        i++;
        System.out.println();  ----> need to display in range 1-10, 11-20,21-30 ... 191-200
      } else {
      System.out.println("Integer not in range of 1-200");
      }
    }
}

Answer 1

public static void displayStatistics(int k)
    {
        if(k>=1 && k<=200) 
      {
        int low,high;
        if(k%10==0)
        {
            low=k-9;
            high=k;
        }
        else 
        {
            low=k-k%10+1;
            high=k-k%10+10;
        }
        System.out.println("value in range " +low+" -"+high); 

      } else {
      System.out.println("Integer not in range of 1-200");
      }
    }

请记住，您将整数传递给函数，而不是完整的数组

Answer 2

您必须从上面的代码中获得编译器错误。改变方法

public static void displayStatistics(int a) {
    if (a >= 1 && a <= 100) {
        System.out.println("Input[" + a + "] is within the range 1 to 100");
    } else {
        System.out.println("Integer not in range of 1-200");
    }
}

同样，您可以添加else if进行另一次范围检查。

java显示输入范围

2 个答案: