using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Windows.Forms;
namespace Exceptions
{
class ExceptionTest
{
static void Main(string[] args)
{
bool flag = true;
while (flag)
{
Random randomGenerator = new Random(); //Creates a New Random object
try
{
Console.Write("Please Enter Your Minimum Number: ");
int minNumber = Convert.ToInt32(Console.ReadLine());
Console.Write("Please Enter Your Maximum Number: ");
int maxNumber = Convert.ToInt32(Console.ReadLine());
for (int number = 0; number <= 4; number++)
{
int randomNumber = randomGenerator.Next(minNumber, maxNumber);
Console.WriteLine("{0}", randomNumber);
}
if (minNumber > maxNumber)
{
throw new Exception("Your Minimum Number is bigger than your Maximum Number");
}
flag = false;
}//End Try Block
catch (FormatException e)
{
MessageBox.Show(e.Message, "Error Detected", MessageBoxButtons.OK, MessageBoxIcon.Error);
}
catch (Exception e)
{
MessageBox.Show(e.Message, "Error Detected", MessageBoxButtons.OK, MessageBoxIcon.Error);
}
}
}
}
}
Message Box Problem
Started by Alex_j, Nov 20 2009 04:55 AM
2 replies to this topic
#1
Posted 20 November 2009 - 04:55 AM
I'm trying to display a message box when my new exception has been caught but its just showing a generic message from the system (I think) and not my custom message, any ideas of what I'm doing wrong, I had it working but now Its not :(
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#2
Posted 20 November 2009 - 06:00 AM
Put your test if min number is greater than max number before the for loop.
if (minNumber > maxNumber)
{
throw new Exception("Your Minimum Number is bigger than your Maximum number");
}
for (int number = 0; number <= 4; number++)
{
int randomNumber = randomGenerator.Next(minNumber, maxNumber);
Console.WriteLine("{0}", randomNumber);
}
Edited by Jaan, 20 November 2009 - 09:07 AM.
Please use code tags when you are posting your codes!
#3
Posted 20 November 2009 - 06:41 AM
Thanks works fine now, thought it would have been a simple solution! :thumbup:


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