java display input in range [closed]

Question

Hi what was wrong with the following program? As i want it to display user input integer value in a range of 1-10, 11-20,21-30 ... 191-200 ?

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");
      }
    }
}

Contingency

You are passing int value to displayStatistics. However the method signature says, the input will be integer array. – Leek

The simple answer to the "what's wrong?" question is that your code isn't even attempting to do what you say it needs to do. – Paralipomena

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");
      }
    }

Remember that you are passing an integer to the function , not the complete array

Unconsidered

Answer 2

You must get compiler error from the above code. Change the method

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");
    }
}

Similarly you can add else if for another range check.

Lachman