C#•3y ago

Why won't the second try-catch block propagate an overflowexception?

When anything over 2.147b is entered for the first try-catch block, it successfully automatically throws (and then catches) an overflowexception. It does not need

if(int.Parse(input) < 0)

if(int.Parse(input) < 0)

But when anything over 2.147b is entered into the console for the second try-catch block, it does not automatically throw the overflowexception. Instead, it throws a formatexception. The second block needs

if(int.Parse(input) < 0)

if(int.Parse(input) < 0)

to throw overflow. WHY? Code below:

int? n = null;
string input;

while (n == null)
{

    Console.WriteLine("Please enter a positive integer");
    try
    {

        input = Console.ReadLine();
        if (int.Parse(input) < 0)
        {
            Console.Write("Number must be a positive integer. ");
            continue;
        }
        else if (!int.TryParse(input, out _))
        {
            throw new FormatException();

        }
        else if (int.TryParse(input, out _))
        {
            n = int.Parse(input);
        }


    }
    catch (OverflowException o)
    {
        Console.Write("OverflowException. ");
    }
    catch (FormatException f)
    {
        Console.Write("FormatException. ");
    }

}
int[] arr = new int[(int)n];

try
{
    for (int i = 0; i < (int)n; i++)
    {
        Console.WriteLine("Enter number {0}", i + 1);
        input = Console.ReadLine();

        if (!int.TryParse(input, out _))
        {
            throw new FormatException("FormatException.");

        }

        else if (int.TryParse(input, out _))
        {
            arr[i] = int.Parse(input);
        }
    }

    Console.WriteLine("Sum is " + Sum(arr));

}
catch (OverflowException o)
{
    Console.WriteLine("OverflowException.");
        
}

catch (FormatException f)
{
    Console.WriteLine("FormatException.");
        
}

//=======================================


static int Sum(int[] numbers)
{
    int sum = 0;
    foreach (int number in numbers)
    {
        sum += number;
    }
    return sum;
}

int? n = null;
string input;

while (n == null)
{

    Console.WriteLine("Please enter a positive integer");
    try
    {

        input = Console.ReadLine();
        if (int.Parse(input) < 0)
        {
            Console.Write("Number must be a positive integer. ");
            continue;
        }
        else if (!int.TryParse(input, out _))
        {
            throw new FormatException();

        }
        else if (int.TryParse(input, out _))
        {
            n = int.Parse(input);
        }


    }
    catch (OverflowException o)
    {
        Console.Write("OverflowException. ");
    }
    catch (FormatException f)
    {
        Console.Write("FormatException. ");
    }

}
int[] arr = new int[(int)n];

try
{
    for (int i = 0; i < (int)n; i++)
    {
        Console.WriteLine("Enter number {0}", i + 1);
        input = Console.ReadLine();

        if (!int.TryParse(input, out _))
        {
            throw new FormatException("FormatException.");

        }

        else if (int.TryParse(input, out _))
        {
            arr[i] = int.Parse(input);
        }
    }

    Console.WriteLine("Sum is " + Sum(arr));

}
catch (OverflowException o)
{
    Console.WriteLine("OverflowException.");
        
}

catch (FormatException f)
{
    Console.WriteLine("FormatException.");
        
}

//=======================================


static int Sum(int[] numbers)
{
    int sum = 0;
    foreach (int number in numbers)
    {
        sum += number;
    }
    return sum;
}

2 Replies

Kouhai•3y ago

You mean why TryParse doesn't throw OverflowException?

surwrenOP•3y ago

yeah i kinda got the answer by accident in the other thread but i'm trying to consoldiate it sec

Gaming

Programming

Why won't the second try-catch block propagate an overflowexception?