ArrayList problem (finding largest no)

[bugger]

I am working on an assignment and I have to find largest number from an ArrayList using Iterator. I wrote following code.

Code:

		ArrayList <Integer>alist = new ArrayList<Integer>();
		// populating
		alist.add(101);
		alist.add(777);
		alist.add(872);
		alist.add(102);
		alist.add(23);
		
		// finding the largest number in the ArrayList
		Iterator<Integer> it = alist.iterator();
		int max = 0;
		while(it.hasNext())
		{
			if( it.next() > max)
				max =  it.next();
		}
		System.out.println(max);

It compiles fine but the output is not as required.

Output:

Code:

102

Output should be 872. I am not sure what is happening.
Please look into this.

Thanks.

[revathi17]

Code:

import java.util.ArrayList;
import java.util.Iterator;

public class LargestNumber {
	public static void main(String[] args){
		ArrayList<Integer> aList = new ArrayList<Integer>();
		aList.add(101);
		aList.add(872);
		aList.add(777);
		aList.add(102);
		aList.add(23);
		
		Iterator<Integer> it = aList.iterator();
		int max = 0;
		while(it.hasNext()){
			int temp = it.next();
			if(temp > max)
				max = temp;
		}	
		System.out.println("Max value:"+max);
	}

}

I think what happens is, only if you call it.next() it goes to the next element and continues iteration. This was something new for me too, i just tried giving a temp variable and it worked!!

-R

[gulapala]

Bugger,

When you call it->next, it goes to the next iterator element. So in your code the statement is being called twice due to which, it goes to 3 element, if it satisfies the if condition. So, you should call it only once.

[bugger]

Thanks for the feedback. Yes, its works.
A small thing can make such a big difference