' This example demonstrates Math.IEEERemainder()

Class Example
   Public Shared Sub Demo(ByVal outputBlock As System.Windows.Controls.TextBlock)
      Dim doubleResult As Double
      Dim divisor As Double

      Dim str1 As [String] = "The IEEE remainder of {0:e}/{1:f} is {2:e}"
      divisor = 2.0
      outputBlock.Text += String.Format("{0}Divide two double-precision floating-point values:", vbCrLf) & vbCrLf
      doubleResult = Math.IEEERemainder([Double].MaxValue, divisor)
      outputBlock.Text &= "1) "
      outputBlock.Text += String.Format(str1, [Double].MaxValue, divisor, doubleResult) & vbCrLf

      divisor = 3.0
      doubleResult = Math.IEEERemainder([Double].MaxValue, divisor)
      outputBlock.Text &= "2) "
      outputBlock.Text += String.Format(str1, [Double].MaxValue, divisor, doubleResult) & vbCrLf
      outputBlock.Text &= "Note that two positive numbers can yield a negative remainder." & vbCrLf
   End Sub 
End Class 
'
'This example produces the following results:
'   Divide two double-precision floating-point values:
'   1) The IEEE remainder of 1.797693e+308/2.00 is 0.000000e+000
'   2) The IEEE remainder of 1.797693e+308/3.00 is -1.000000e+000
'   Note that two positive numbers can yield a negative remainder.

// This example demonstrates Math.IEEERemainder()
using System;

class Example
{
   public static void Demo(System.Windows.Controls.TextBlock outputBlock)
   {
      double doubleResult;
      double divisor;

      String str1 = "The IEEE remainder of {0:e}/{1:f} is {2:e}";
      divisor = 2.0;
      outputBlock.Text += "\nDivide two double-precision floating-point values:" + "\n";
      doubleResult = Math.IEEERemainder(Double.MaxValue, divisor);
      outputBlock.Text += "1) ";
      outputBlock.Text += String.Format(str1, Double.MaxValue, divisor, doubleResult) + "\n";

      divisor = 3.0;
      doubleResult = Math.IEEERemainder(Double.MaxValue, divisor);
      outputBlock.Text += "2) ";
      outputBlock.Text += String.Format(str1, Double.MaxValue, divisor, doubleResult) + "\n";
      outputBlock.Text += "Note that two positive numbers can yield a negative remainder." + "\n";
   }
}
/*
This example produces the following results:

   Divide two double-precision floating-point values:
   1) The IEEE remainder of 1.797693e+308/2.00 is 0.000000e+000
   2) The IEEE remainder of 1.797693e+308/3.00 is -1.000000e+000
   Note that two positive numbers can yield a negative remainder.

*/