ArgumentException Class : Exception System « Development « VB.Net

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VB.Net » Development » Exception SystemScreenshots 
ArgumentException Class
 
Public Class App
    Public Shared Sub Main()
        Try 
            DivideByTwo(7)
        Catch Ex As ArgumentException
            Console.WriteLine(Ex)
        End Try
    End Sub

    Private Shared Function DivideByTwo(ByVal num As IntegerAs Integer
        If ((num And 1)  1Then
            Throw New ArgumentException("Number must be even""num")
        End If
        Return (num / 2)
    End Function
End Class

   
  
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1.Use Argument ExceptionUse Argument Exception
2.Parse Number ExceptionParse Number Exception
3.OverflowException Class
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