Help with my code to find MinOfAll numbers

I am new to Programming world and so to Java,

I tried my hands on writing the following code to find minimum of all the numbers entered via command prompt,
but I am getting the error

Exception in thread "main" java.lang.ArrayIndexOutOfBound Exception :5 at MinOfAll.main

Kindly tell me where am I going wrong?

Code:

---

public class MinOfAll{ 
 
public static void main(String[] args){ 
 
int result = Integer.parseInt(args[0]); 
for (int i =0 ; i< args.length; i++) 
{ 
    int current=Integer.parseInt(args[i+1]); 
 if (current < result){ 
    result = current; 
 } 
       
} 
System.out.println("Min of all is : " + result); 
 
 
} 
} 
 
[color=red]I/P at console : java MinOfAll 2 3 4 5 6[/color]

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Why did you do that?

Code:

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int current=Integer.parseInt(args[i+1]);

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should be,

Code:

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int current=Integer.parseInt(args[i]);

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Because arrays in Java are zero based indexing.

I did that to start the comparison with next element in the array.

Even tough I edit it to,

Code:

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int current=Integer.parseInt(args[i]);

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the output error still persists.

How did you run your code? Did you pass relevant parameters?

This is what I do on console

java MinOfAll 2 3 4 5 6

So that's the error. Can you show your correct code. I don't think that you fix the correct place.

Code:

---

public class MinOfAll{

public static void main(String[] args){

int result = Integer.parseInt(args[0]);
for (int i =0 ; i< args.length; i++)
{
        int current=Integer.parseInt(args[i]);
 if (current <= result){
        result = current;
 }
         
}
System.out.println("Min of all is : " + result);


}
}

---

Right, so it' working. There cannot be any error at all. You say that you get the same error? It cannot happen.

My bad I had kept the same code at 2 places ad was running the unpdated one.

Thanks again!!

You are welcome. :)

If you are satisfied with the solution, please mark the thread solved.