Exceptions (Java)

Question

Ok I have been trying this exercise because I am not good at exceptions.So here is the context of the exercise:Write a class with main method(the following code is give by the task):

public class task{

public static void main(
String[] args)
{
int[] array = new int[10];
// initialise array
int result =task.min(array); 
// Where the class task
// contains the min method
}

so I am asked to make sure that the method works even if the array contains only one element or none at all.So with the code give above I have to use exceptions to handle all the errors that could pop up.

This is what I did:

public static void main(String[] args){
int[] array = new int[10];
        array[0]=5;
        array[1]=7;

        int result =Exercise1.min(array); 
        if(array.length<=0){
            throw new IllegalArgumentException("empty array");
        }
        else if(array.length<10 && array.length>0){
            throw new IllegalArgumentException("I got just these numbers");
        }
            System.out.println(result);


            }
    public static int min(int[] array) {
        int min = array[0]; 
          for(int i=1;i<array.length;i++){  
                 if(array[i]< min){  
                 min = array[i];  
                    }  
          }
        return min;
    }
}

However no matter what I write the output is always 0 and I don't understand why.In case my approach is entirely wrong please offer some advice.

Your exceptions should be inside `min` method and you have to handle them in `main` method. Apart of that, your code should work as expected. Probably you have problems executing the code and you get the result of the last valid execution, where it printed 0. — Luiggi Mendoza, Aug 01 '14 at 16:47
`if(array[i]< min){min = array[i]; }` this makes min always get set to the lowest value, since `array[2-9] = 0`, min returns 0 — Kevin L, Aug 01 '14 at 16:48
your array will always have length 10 if it's declared as int[10]. it doesn't resize. — Nathan Hughes, Aug 01 '14 at 16:49

score 3 · Accepted Answer · answered Aug 01 '14 at 16:56

int[] array = new int[10];

This line of code sets 10 places of the array "array", all equal to "0".

In order for min to not return 0 you need to set all the values of the array or change the size to 2.

Setting them could easily be done with this function:

for(int i = 0; i < array.length; i++){
    array[i] = i + 1;
}

This will set all the values of the array so you can remove this code:

array[0]=5;
array[1]=7;

Setting these will fix your min value always being 0.

It will always be 1 if you use my suggested loop.

score 1 · Answer 2 · answered Aug 01 '14 at 16:50

1

When you create an array of int they are all set to zero, hence that is the minimum value in your array.

answered Aug 01 '14 at 16:50

user3757014

222
1
3

score 1 · Answer 3 · answered Aug 01 '14 at 17:03

If you can change your input:

ArrayList<Integer> array = new ArrayList<Integer>();
array.add(5);
[...]

Or you can change your condition, assuming YOU CAN'T HAVE the 0 value in your array.

for(int i=1;i<array.length && array[i]!=0;i++){  
  if(array[i]< min){  
    min = array[i];  
  }  
}

This will solve your current problem so you can focus on your Exception practise

Exceptions (Java)

3 Answers3