JAVA: displaying largest number of an array bug

german · 06-28-2004, 07:29 PM

so... to circumvent this problem you use arrays where you can just add to your array pointer a fixed amount of memory * the number of elements to skip, and that gives you your get()... is that more or less correct?

aluser · 06-28-2004, 07:53 PM

that will help in (some|most) cases, but indexing will still be slower than in a real array and insertions/deletions into the middle of the list will not be great. Plus, get() would get slower and slower as the array grew since it would have to iterate over all the previous skip amounts. It could be the way ArrayList is done, I'm not sure.

The point I wanted to make is that since the OP is *only* iterating through the array, and not really indexing it except as a method of iteration, plus he wants to append to it an undefined number of times, a plain old linked list is the right structure.

german · 06-28-2004, 07:59 PM

right... yes I didn't mean to extend the tangent, but you know... I got interested in it...

anyway as it has been stated before, the right way to write this program would be something like this with error handling:

Code:

int max = 0;
boolean reading = true;
while(reading) {
  int n = Integer.parseInt(JOptionPane.showInputDialog("Number:"));
  if(n > max) max = n;
  if(n == 0) reading = false;
}

JOptionPane.showMessageDialog("Max was " + max);

Thanks for the introspective aluser, very interesting.

B.