Why does this Java code throw an exception?

Question

I am new with java. I having some exception while running my code:

import java.util.Random;

public class Example {
    public static void main(String[] args) {
        Random r = new Random();
        int[] num= new int[5];      

        for (int i= 0; 1<num.length; i++)
        {
            num[i]= r.nextInt(100)+1;
            System.out.println(num[i]);
        }
    }
}

It gives me the following exception:

Exception in thread "main" java.lang.ArrayIndexOutOfBoundsException: 5 at Example.main(Example.java:13)

Why am I getting this exception?

Answer 1

Firstly, you should always copy and paste the exception text, to save us trying to guess which type of exception it is or where it's happening

You'll be getting an ArrayOutOfBoundsIndexException , because the loop is endless yet the index always increments.

for (int i= 0; 1<num.length; i++)

One is always less than the length of num so it loops for ever, incrementing i each time until i is larger than the array size. At which point you'll try to do this

num[i] ...

And i will be out of bounds, throwing the exception.

Answer 2

import java.util.Random;

public class test {
public static void main(String[] args) {

    Random r = new Random();
    int[] num = new int[5];

    for (int i = 0; i < num.length; i++) {
        num[i] = r.nextInt(100) + 1;
        System.out.println(num[i]);
    }

}
}

This is how a loop works:

for (initialization; termination; increment) {
    statement(s)
} 

When the termination expression evaluates to false, the loop terminates. In your case the loop never terminates. That is why you are getting an ArrayOutOfBoundsIndexException .