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Home/ Questions/Q 8735893
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Editorial Team
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Editorial Team
Asked: June 13, 20262026-06-13T10:10:29+00:00 2026-06-13T10:10:29+00:00

This code is simply throw exception, because short sNum is assigned big range value

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This code is simply throw exception, because short sNum is assigned big range value of int num, and conversion fail. Any way.
I want to loop request until valid rang of short is entered.

    static void Main()
    {
        int num = 40000;
        short sNum = 0;
        try
        {
            sNum = Convert.ToInt16(num);

        }
        catch (OverflowException ex)
        {
            // Request for input until no exception thrown.
            Console.WriteLine(ex.Message);
            sNum = Convert.ToInt16(Console.ReadLine());
        }

        Console.WriteLine("output is {0}",sNum);
                        Console.ReadLine();
    }

Thank you.

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  1. Editorial Team
    Editorial Team
    2026-06-13T10:10:30+00:00Added an answer on June 13, 2026 at 10:10 am

    The reason is you are throwing an exception when the conversion fails inside your catch block. The catch block technically is outside of the try block, so it will not get caught by the same catch as you seem to think. This is not really behaving as a loop as you appear to hope it will.

    Exceptions are not generally considered the best method for normal (non-exceptional) events in your code. The TryParse method and a loop would be much better in this case.

    static void Main()
    {
        string input = //get your user input;
        short sNum = 0;
    
        while(!short.TryParse(input,out sNum))
        {
            Console.WriteLine("Input invalid, please try again");
            input = //get your user input;
        }
    
        Console.WriteLine("output is {0}",sNum);
        Console.ReadLine();
    }
    
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