Programming is, to a large extent, the science of dealing with data. In the previous chapter, you saw how to use .NET strings and regular expressions to manipulate text. In this chapter, we consider the rest of the core types in the Microsoft .NET Framework, including numbers, dates, and specialty types such as GUIDs and enumerations.

Some of the recipes in this chapter focus on fundamentals for dealing with data types. For example, you'll learn how to validate dates, format numeric values, access trigonometric functions, and convert binary and hexadecimal values. Other recipes present more advanced techniques that might require custom classes. Recipes 2.5, 2.6, 2.7 show you how to deal with complex numbers, vectors, and matrixes—all of which are types of data that have no core support in the .NET Framework. Fortunately, .NET makes it easy to write a custom class that encapsulates all the functionality you need.

Note |
This chapter includes lengthy examples that work with |

This chapter also presents recipes that demonstrate how to convert basic data types to binary and back, and how to pull off a few more enigmatic tricks, such as evaluating a mathematical expression contained in a string (recipe 2.9), determining the day of a week for a specific date (recipe 2.12), and converting the name of an enumeration into the corresponding enumerated value (recipe 2.20).

You need to perform a mathematical operation such as taking the sine, logarithm, absolute value, and so on.

Use the *System.Math* class, which provides a collection of helper utilities for this task.

Microsoft Visual Basic .NET provides operators for common tasks such as addition and subtraction (*+*, *-*), multiplication and division (***, */*), exponentiation (*^*), integer division (), and finding the remainder (*Mod*). For all other operations, you can use the methods exposed by the *Math* class. Here are three examples:

' Get a sin of an angle measured in radians. x = Math.Sin(y) ' Round a number to two decimal places. x = Math.Round(y, 2) ' Get the absolute value of a number. x = Math.Abs(y)

Useful *Math* class members include:

- Constants
*PI*and*E*(the natural logarithmic base). - Trigonometric functions
*Sin*,*Cos*,*Tan*,*Asin*,*Acos*,*Atan*, and hyperbolic*Sinh*,*Cosh*, and*Tanh*. - Logarithmic function
*Log*and*Log10*. - Power functions
*Exp*(returns e raised to a specified power),*Pow*(which raises a number to an indicated power), and*Sqrt*. - Comparative functions
*Max*and*Min*, which return the higher or lower of two numbers, respectively. - Boundary functions
*Floor*(returns the largest whole number less than or equal to the supplied number, effectively truncating it), and*Ceiling*(returns the smallest whole number greater than or equal to the specified number). - Sign-related functions
*Abs*(which removes the sign). *Round*, which rounds a number to a specified number of decimal places.

You want to convert a numeric type (*Decimal*, *Int32*, *Double*, and so on) into a formatted string.

Use the overloaded version of the *ToString* method that accepts a format string.

There are two types of numeric format strings: standard and custom. *Standard format strings* use a preset format according to the current culture and are identified by a single letter. You use that letter, in conjunction with a number that indicates the precision, when converting a number to a string. Here are a few common examples:

Dim Number As Decimal = 12345.6D ' The format specifier for currency is "C" Dim MoneyString As String = Number.ToString("C") ' MoneyString is now "12,345.60" ' The format specifier for scientific notation is "E" ' The code also specifies 2 decimal places. Dim ScientificString As String = Number.ToString("E2") ' ScientificString is now "1.23E+004" ' The format specifier for ordinary numbers is "N" Dim NormalString As String = Number.ToString("N") ' NormalString is now "12,345.60"

Table 2-1 presents a full list of standard numeric format specifiers.

String |
Name |
Description |
---|---|---|

C or c |
Currency |
The number is converted to a string that represents a currency amount (such as "$12,345.68"). |

D or d |
Decimal |
The number is padded with zeros on the left side according the precision. For example, D8 would format the number 12345 to "00012345". This is only supported for integer types. |

E or e |
Scientific |
The number is converted to a string in exponential form with one digit preceding the decimal point. The precision specifier indicates the desired number of digits after the decimal point (the default is six). For example, 12345.6789 will be formatted as "1.234568E+004" with the default precision. |

F or f |
Fixed-point |
The number is converted to a string with a number of decimal places equal to the precision specifier (the default is two). Rounding is performed as needed. For example, the number 12345.6789 would format to "12345.68". |

G or g |
General |
The number is converted to the most compact decimal form, using fixed or scientific notation. The precision specifier determines the number of significant digits in the resulting string. If the precision specifier is omitted, the number of significant digits is determined by the type of number being converted (5 for an |

N or n |
Number |
The number is converted to a string with comma separators between each group of three digits to the left of the decimal point. The precision specifier indicates the desired number of decimal places (the default is two). Rounding is performed as needed. For example, 12345.6789 formats to "12,345.68". |

P or p |
Percent |
The number is converted to a string that represents a percent. The converted number is multiplied by 100, and a percent sign (%) is appended. For example, 0.126 formats to "12.60 %". |

R or r |
Round-trip |
The round-trip format guarantees that a floating-point number can be converted to a string and back to the number (using the |

X or x |
Hexadecimal |
The number is converted to a string of hexadecimal digits, so 123456789 formats to "75bcd15". The case of the format specifier indicates whether to use uppercase or lowercase characters for alphabetic digits. The precision specifier indicates the minimum number of digits desired in the resulting string, and the number will be padded with zeros accordingly. This is only supported for integer types. |

You can also create your own *custom format strings* and pass them to the *ToString* method. This is rarely necessary, but you can consult the MSDN help for more information.

You want to create a statistically random number quickly.

Create an instance of the *System.Random* class, and call the *Next* or *NextDouble* method.

The *Random* class uses a pseudorandom number generator, which means that it uses an algorithm to generate numbers that are statistically random when viewed in sequence.

To use the *Random* class, you simply create an instance and call either *Next* or *NextDouble*. *NextDouble* returns a double-precision floating-point number greater than or equal to 0.0, and less than 1.0. *Next* generates an integer within the maximum and minimum range you specify.

Dim RandomGenerator As New Random() ' Retrieve a random fraction number from 0.0 to 1.0. Dim RandomDouble As Double = RandomGenerator.NextDouble() ' Retrieve a random integer number from 1 to 6. Dim RandomInt As Integer = RandomGenerator.Next(1, 7) ' Retrieve another random integer from 1 to 6. RandomInt = RandomGenerator.Next(1, 7)

Notice that the maximum bound for the *Next* method is always one higher than the maximum integer in your range.

The *Random* class is ideal when you need to generate a quick random value for a game, simulation, or test. However, it's not suitable for use with cryptography, because an attacker can guess the "random" number you will generate by examining previous random values and determining how you are seeding the random generator. If you need cryptographically secure random numbers, you can use the *System.Security.Cryptography.RNGCryptoServiceProvider* class, which is described in Chapter 18 (see recipe 18.15). However, this class is much slower than *Random*, which will become noticeable if you need to generate thousands of random numbers rapidly.

Note |
By default, the ' Because this loop executes so quickly, it's easy to create two ' Random objects with the same seed and end up with ' short sequences of identical numbers (like 8888333322). Dim i As Integer For i = 0 To 10 Dim RandomGenerator As New Random() Console.WriteLine(RandomGenerator.Next(1, 7)) Next |

You want to convert a base 10 number to a hexadecimal, octal, or binary number (or vice versa).

Use the overloaded *Convert.ToString* and *Convert.ToInt32* shared methods that accept a number indicating the base.

Although you can't work directly with non–base 10 numbers in Visual Basic .NET, you can easily convert base 10 values into a string representation that uses a base 2 (binary), base 8 (octal), base 10, or base 16 (hexadecimal). To do so, you use the overloaded *Convert.ToString* method that accepts two parameters: the base 10 number and the base that should be used for the converted number (which must be 2, 8, 10, or 16).

Dim Number As Integer = 3023 Console.WriteLine("Binary: " & Convert.ToString(Number, 2)) Console.WriteLine("Octal: " & Convert.ToString(Number, 8)) Console.WriteLine("Hexadecimal: " & Convert.ToString(Number, 16))

The output for this code is:

Binary: 101111001111 Octal: 5717 Hexadecimal: bcf

You can also use the shared *Convert.ToInt32*, *Convert.ToInt16*, or *Convert.ToInt64* methods to convert a non–base 10 number from a string into an integer type:

Dim Binary As String = "01" Dim Number As Integer = Convert.ToInt32(Binary, 2) ' Double the number. Number *= 2 ' Convert it back to binary. Binary = Convert.ToString(Number, 2) ' Binary is now "10".

If you need to perform calculations with non–base 10 numbers, you have two choices. You could use the *ToInt32* number to convert your numbers to decimal, perform the calculation, and then use *ToString* to convert the number back into its native representation, as shown above. Alternatively, you could create a custom class that represents the number and provides dedicated methods such as *Add*, *Subtract*, *Multiply*, and so on, which perform native calculations. Recipes 2.5 and 2.6 show similar techniques with classes that represent complex numbers and vectors.

You need to perform calculations with complex numbers (numbers that involve i, the square root of –1).

Create your own complex number class.

The .NET Framework does not include any built-in support for complex number calculations. However, it's quite easy to create a class to represent complex numbers. This class will include methods such as *Add*, *Subtract*, *Multiply*, and *DivideBy*, and a few complex-number helper functions such as *GetModulus* and *GetConjugate*. In addition, the class will support cloning and comparing (useful for sorting arrays of complex numbers), and it will override the *Equals* method to perform value equality testing and *ToString* to provide an appropriate string representation. Chapter 4 provides recipes that allow you to implement these refinements in your own custom classes.

The full code for the *ComplexNumber* class is shown here:

Public Class ComplexNumber Implements ICloneable, IComparable ' The real and imaginary component. Public Real As Double Public Imaginary As Double ' Create a new complex number. Public Sub New(ByVal real As Double, ByVal imaginary As Double) Me.Real = real Me.Imaginary = imaginary End Sub ' Add a complex number to the current one. Public Function Add(ByVal complexNumber As ComplexNumber) As ComplexNumber Return New ComplexNumber(Me.Real + complexNumber.Real, _ Me.Imaginary + complexNumber.Imaginary) End Function ' Add a real number to the current complex number. Public Function Add(ByVal real As Double) As ComplexNumber Return New ComplexNumber(Me.Real + real, Me.Imaginary) End Function ' Subtract a complex number from the current one. Public Function Subtract(ByVal complexNumber As ComplexNumber) _ As ComplexNumber Return New ComplexNumber(Me.Real - complexNumber.Real, _ Me.Imaginary - complexNumber.Imaginary) End Function ' Subtract a real number from the current complex number. Public Function Subtract(ByVal real As Double) As ComplexNumber Return New ComplexNumber(Me.Real - real, Me.Imaginary) End Function ' Multiply a complex number by the current one. Public Function Multiply(ByVal complexNumber As ComplexNumber) _ As ComplexNumber Dim x, y, u, v As Double x = Me.Real : y = Me.Imaginary u = complexNumber.Real : v = complexNumber.Imaginary Return New ComplexNumber(x * u - y * v, x * v + y * u) End Function ' Multiply the current number by a real number. Public Function Multiply(ByVal real As Double) As ComplexNumber Return New ComplexNumber(Me.Real * real, Me.Imaginary * real) End Function ' Divide the current number by another complex number. Public Function DivideBy(ByVal complexNumber As ComplexNumber) _ As ComplexNumber Dim x, y, u, v As Double x = Me.Real : y = Me.Imaginary u = complexNumber.Real : v = complexNumber.Imaginary Dim Sum As Double = u * u + v * v Return New ComplexNumber((x * u + y * v) / Sum, (y * u - x * v) / Sum) End Function ' Divide the current number by a real number. Public Function DivideBy(ByVal real As Double) As ComplexNumber Return New ComplexNumber(Me.Real / real, Me.Imaginary / real) End Function ' Test for value equality between the number and another complex number. Public Overloads Overrides Function Equals(ByVal obj As Object) As Boolean If Not TypeOf obj Is ComplexNumber Then Return False Dim Compare As ComplexNumber = CType(obj, ComplexNumber) Return (Me.Real = Compare.Real And Me.Imaginary = Compare.Imaginary) End Function ' Test for value equality between two complex numbers. Public Overloads Shared Function Equals(ByVal objA As Object, _ ByVal objB As Object) As Boolean If Not (TypeOf objA Is ComplexNumber) Or _ Not (TypeOf objB Is ComplexNumber) Then Return False Dim ComplexA As ComplexNumber = CType(objA, ComplexNumber) Dim ComplexB As ComplexNumber = CType(objB, ComplexNumber) Return (ComplexA.Real = ComplexB.Real _ And ComplexA.Imaginary = ComplexB.Imaginary) End Function ' Define some helper methods. Public Function GetModulus() As Double Return Math.Sqrt(Me.Real ^ 2 + Me.Imaginary ^ 2) End Function Public Function GetModulusSquared() As Double Return Me.Real ^ 2 + Me.Imaginary ^ 2 End Function Public Function GetArgument() As Double Return Math.Atan2(Me.Imaginary, Me.Real) End Function Public Function GetConjugate() As ComplexNumber Return New ComplexNumber(Me.Real, -Me.Imaginary) End Function ' Return a string representation of a complex number Public Overrides Function ToString() As String Return Me.Real.ToString() & ", " & Me.Imaginary.ToString() & "i" End Function ' Copy a complex number. Public Function Clone() As Object Implements System.ICloneable.Clone Return New ComplexNumber(Me.Real, Me.Imaginary) End Function ' Compare two complex numbers (allows array sorting). Public Function CompareTo(ByVal obj As Object) As Integer _ Implements System.IComparable.CompareTo If Not (TypeOf obj Is ComplexNumber) Then Return 0 Dim Compare As ComplexNumber = CType(obj, ComplexNumber) Return Me.GetModulus().CompareTo(Compare.GetModulus()) End Function End Class

Notice that Visual Basic .NET does not support operator overloading, so you must use methods to perform operations on complex numbers (*Add*, *Subtract*, and so on) rather than predefined operators (such as + and -). This is only a minor inconvenience.

A simple complex number test is shown below.

Dim c1 As New ComplexNumber(3, 3) Dim c2 As New ComplexNumber(1, -4) Dim c3 As New ComplexNumber(3, 3) If c1.Equals(c3) ' This will succeed, as c1 = c3. Console.WriteLine("Passed value equality test.") End If c1 = c1.Multiply(c2) c1 = c1.Add(c3) ' This displays "18, -6i" Console.WriteLine(c1.ToString())

You need to perform calculations with three-dimensional vectors.

Create your own simple vector class.

The .NET Framework does not include any built-in support for vector calculations. However, it's quite easy to create a class to represent vectors. This class will include methods such as *Add*, *Subtract*, *Multiply*, and *DivideBy*, and well as a few vector-specific functions such as *GetCrossProduct* and *GetDotProduct*. In addition, the class will support cloning and comparing (useful for sorting arrays of vectors), and it will override the *Equals* method to perform value equality testing and *ToString* to provide an appropriate string representation. Chapter 4 provides recipes that allow you to implement these refinements in your own custom classes.

The full code for the *Vector* class is shown here:

Public Class Vector Implements ICloneable, IComparable ' The coordinates. Public x, y, z As Double ' Create a new vector. Public Sub New(ByVal x As Double, ByVal y As Double, ByVal z As Double) Me.x = x Me.y = y Me.z = z End Sub ' Add a vector to the current vector. Public Function Add(ByVal vector As Vector) As Vector Return New Vector(Me.x + vector.x, Me.y + vector.y, Me.z + vector.z) End Function ' Subtract a vector from the current vector. Public Function Subtract(ByVal vector As Vector) As Vector Return New Vector(Me.x - vector.x, Me.y - vector.y, Me.z - vector.z) End Function ' Multiply the current vector by a scalar. Public Function Multiply(ByVal n As Double) As Vector Return New Vector(Me.x * n, Me.y * n, Me.z * n) End Function ' Divide the current vector by a scalar. Public Function DivideBy(ByVal n As Double) As Vector Return New Vector(Me.x / n, Me.y / n, Me.z / n) End Function ' Define some helper methods. Public Function GetCrossProduct(ByVal vector As Vector) As Vector Return New Vector(Me.y * vector.z - Me.z * vector.y, _ -Me.x * vector.z + Me.z * vector.x, _ Me.x * vector.y - Me.y * vector.x) End Function Public Function GetDotProduct(ByVal vector As Vector) As Double Return (Me.x * vector.x + Me.y * vector.y + Me.z * vector.z) End Function Public Function Length() As Double Return Math.Sqrt(Me.x * Me.x + Me.y * Me.y + Me.z * Me.z) End Function Public Function Normalize() As Vector Dim nLength As Double nLength = Length() If nLength = 0 Then Throw New DivideByZeroException() Return New Vector(Me.x / nLength, Me.y / nLength, Me.z / nLength) End Function ' Test for value equality between the current vector and another. Public Overloads Function Equals(ByVal obj As Object) As Boolean If Not (TypeOf obj Is Vector) Then Return False Dim Compare As Vector = CType(obj, Vector) Return (Me.x = Compare.x And Me.y = Compare.y And Me.z = Compare.z) End Function ' Test for value equality between two vectors. Public Overloads Shared Function Equals(ByVal objA As Object, _ ByVal objB As Object) As Boolean If Not (TypeOf objA Is Vector) Or _ Not (TypeOf objB Is Vector) Then Return False Dim VectorA As Vector = CType(objA, Vector) Dim VectorB As Vector = CType(objB, Vector) Return (VectorA.x = VectorB.x And VectorA.y = VectorB.y _ And VectorA.z = VectorB.z) End Function Public Overrides Function ToString() As String Return "(" & Me.x.ToString() & ", " & _ Me.y.ToString() & ", " & Me.z.ToString() & ")" End Function ' Copy a vector. Public Function Clone() As Object Implements System.ICloneable.Clone Return New Vector(Me.x, Me.y, Me.z) End Function ' Compare two vectors (allows array sorting). Public Function CompareTo(ByVal obj As Object) As Integer _ Implements System.IComparable.CompareTo If Not (TypeOf obj Is Vector) Then Return 0 Dim Compare As Vector = CType(obj, Vector) Return Me.Length().CompareTo(Compare.Length()) End Function End Class

Remember that Visual Basic .NET does not support operator overloading, so you must use methods to perform operations on vectors (*Add*, *Subtract*, and so on) rather than predefined operators (such as + and -). This is only a minor inconvenience.

The following code shows a simple vector test:

Dim v1 As New Vector(10, 2, -3) Dim v2 As New Vector(1, -2, -4) Dim v3 As New Vector(10, 2, -3) If v1.Equals(v3) Then ' This will succeed, as c1 = c3. Console.WriteLine("Passed value equality test.") End If v1 = v1.GetCrossProduct(v2) ' This displays "(-14, 37, -22)" Console.WriteLine(v1.ToString())

You want to use matrix calculations (matrix multiplications, additions, normalizations, and so on).

Download a free component, such as Lutz Roeder's Mapack.

The .NET Framework does not include any built-in support for matrix manipulation, aside from a *Matrix* class in the *System.Drawing.Drawing2D* namespace, which is intended for graphical operations and only supports a 3-by-3 matrix. Writing your own matrix code from scratch would be quite a chore, but fortunately there are prebuilt matrix components available for free, like Mapack. (See *http://www.aisto.com/ roeder* or the Web site for this book.) Mapack is a fully featured library for matrix manipulation, with complete C# source code. You can use it to inspire your own matrix classes, or you can use it as is in any application (in which case it is provided without any warranty or support).

Once you add a reference to the Mapack.dll library, you can create matrixes of any size, set their values individually, and call methods to perform tasks such as multiplying, transposing, and inverting. Here's an example:

' Create a 1x2 matrix and set values. Dim MatrixA As New Mapack.Matrix(1, 2) MatrixA(0, 0) = 5 : MatrixA(0, 1) = 2 ' Create a 2x1 matrix and set values. Dim MatrixB As New Mapack.Matrix(2, 1) MatrixB(0, 0) = 5 : MatrixB(1, 0) = 2 ' Multiply the matrixes and display the result. Dim Result As Mapack.Matrix = _ CType(MatrixA.Multiply(MatrixB), Mapack.Matrix) Console.WriteLine(Result.ToString()) ' Displays 29.

You want to perform mathematical operations with fractions without converting to decimal notation (and possibly introducing rounding errors).

Create your own simple fraction class.

The .NET Framework does not include any built-in support for fraction calculations. However, it's quite easy to create a class to represent vectors. This class will include methods such as *Add*, *Subtract*, *Multiply*, and *DivideBy*. It will also include a *Normalize* function that will reduce a fraction to lowest terms using Euclid's algorithm and adjust its sign. In addition, the class will support cloning and comparing (useful for sorting arrays of fractions), and it will override the *Equals* method to perform value equality testing and *ToString* to provide an appropriate string representation. Chapter 4 provides recipes that allow you to implement these refinements in your own custom classes.

The full code for the *Fraction* class is shown here:

Public Class Fraction Implements ICloneable, IComparable ' The two components of any fraction. Public Denominator As Integer Public Numerator As Integer ' Create a new fraction. Public Sub New(ByVal numerator As Integer, ByVal denominator As Integer) Me.Numerator = numerator Me.Denominator = denominator End Sub ' Add fraction to current fraction. Public Function Add(ByVal fraction As Fraction) As Fraction Return New Fraction(Me.Numerator * fraction.Denominator + _ fraction.Numerator * Me.Denominator, _ Me.Denominator * fraction.Denominator).Normalize() End Function ' Subtract a fraction from current fraction. Public Function Subtract(ByVal fraction As Fraction) As Fraction Return New Fraction(Me.Numerator * fraction.Denominator - _ fraction.Numerator * Me.Denominator, _ Me.Denominator * fraction.Denominator).Normalize() End Function ' Multiply a fraction by the indicated fraction. Public Function Multiply(ByVal fraction As Fraction) As Fraction Return New Fraction(Me.Numerator * fraction.Numerator, _ Me.Denominator * fraction.Denominator).Normalize() End Function ' Divide a fraction by the indicated fraction. Public Function DivideBy(ByVal fraction As Fraction) As Fraction Return New Fraction(Me.Numerator * fraction.Denominator, _ Me.Denominator * fraction.Numerator).Normalize() End Function ' Reduces a fraction and adjusts its sign Public Function Normalize() As Fraction Dim NormalizedFraction As Fraction = CType(Me.Clone(), Fraction) If (NormalizedFraction.Numerator <> 0) And _ (NormalizedFraction.Denominator <> 0) Then ' Fix signs If NormalizedFraction.Denominator < 0 Then NormalizedFraction.Denominator *= -1 NormalizedFraction.Numerator *= -1 End If Dim divisor As Integer = GCD(NormalizedFraction.Numerator, _ NormalizedFraction.Denominator) NormalizedFraction.Numerator = divisor NormalizedFraction.Denominator = divisor End If Return NormalizedFraction End Function ' Returns the greatest common divisor using Euclid's algorithm Private Function GCD(ByVal x As Integer, ByVal y As Integer) As Integer Dim temp As Integer x = Math.Abs(x) y = Math.Abs(y) Do While (y <> 0) temp = x Mod y x = y y = temp Loop Return x End Function ' Convert the fraction to decimal notation. Public Function GetDouble() As Double Dim Reduced As Fraction = CType(Me.Clone(), Fraction).Normalize() Return CType(Reduced.Numerator, Double) / _ CType(Reduced.Denominator, Double) End Function ' Test for value equality between the current fraction and another. Public Overloads Function Equals(ByVal obj As Object) As Boolean If Not (TypeOf obj Is Fraction) Then Return False Dim Compare As Fraction = CType(obj, Fraction) Return (Me.GetDouble() = Compare.GetDouble()) End Function ' Test for value equality between two fractions. Public Overloads Shared Function Equals(ByVal objA As Object, _ ByVal objB As Object) As Boolean If Not (TypeOf objA Is Fraction) Or _ Not (TypeOf objB Is Fraction) Then Return False Dim FractionA As Fraction = CType(objA, Fraction) Dim FractionB As Fraction = CType(objB, Fraction) Return (FractionA.GetDouble() = FractionB.GetDouble()) End Function ' Get a string representation of the fraction. Public Overrides Function ToString() As String Return Me.Numerator.ToString & "/" & Me.Denominator.ToString End Function ' Copy a fraction. Public Function Clone() As Object Implements System.ICloneable.Clone Return New Fraction(Me.Numerator, Me.Denominator) End Function ' Compare two fractions (allows array sorting). Public Function CompareTo(ByVal obj As Object) As Integer _ Implements System.IComparable.CompareTo If Not (TypeOf obj Is Fraction) Then Return 0 Dim Compare As Fraction = CType(obj, Fraction) Return Me.GetDouble().CompareTo(Compare.GetDouble()) End Function End Class

Currently, the methods such as *Add*, *Multiply*, and so on only support other fractions. If you want to be able to multiply using an integer data type without converting it to a fraction, you could add overloads of these methods. Because Visual Basic .NET does not support operator overloading, you must use the methods (*Add*, *Subtract*, and so on) rather than the predefined operators (such as + and -).

A simple fraction test is shown here:

Dim f1 As New Fraction(2, 3) Dim f2 As New Fraction(1, 2) Dim f3 As New Fraction(2, 3) If f1.Equals(f3) Then ' This will succeed, as f1 = f3. Console.WriteLine("Passed value equality test.") End If f1 = f1.Add(f2) 'Fraction is now 7/6 f1 = f1.DivideBy(f2) 'Fraction is now 7/3 ' This displays "7/3" Console.WriteLine(f1.ToString())

You want to evaluate a mathematical expression that is specified as a string (as in "2 + 3").

Create a component that wraps the expression evaluator provided with Microsoft JScript, and consume this component from any application that needs this functionality. Or, use the Microsoft Script Control COM component.

The JScript .NET engine provides an expression evaluator that can evaluate a mathematical expression or any JScript code (including functions) in a string. Multiple lines can be separated with line-break characters. However, you can't access this functionality directly from another language. Instead, you need to make a simple JScript wrapper to expose this functionality:

package JScriptUtil { class ExpressionEvaluator { public function Evaluate(expr : String) : String { return eval(expr); } } }

Save this in a .js text file, and compile it using the JScript compiler (jsc.exe) using the following command line:

jsc.exe /t:library JScriptUtil.js

Finally, add a reference to the compiled assembly to your Visual Basic .NET application, and a reference to the *Microsoft.JScript* assembly. You can then create instances of your custom *ExpressionEvaluator* class and call the *Evaluate* method with a string. You might need to refer to the MSDN reference to determine what operators are used in the JScript language, although the standard mathematical ones are obvious.

The following example shows a Console application that puts the JScript expression evaluator to work with a mathematical calculation:

Imports JScriptUtil Public Module ExpressionTest Public Sub Main() Dim Expression As String = "2 * (5 + 1) / 3" Dim Eval As New JScriptUtil.ExpressionEvaluator() Dim Result As String = Eval.Evaluate(expression) Console.WriteLine(Expression & " = " & Result) ' Displays "2 * (5 + 1) / 3 = 4" Console.ReadLine() End Sub End Module

Although extremely powerful, the dynamic expression evaluation provided by JScript .NET is unsuitable for some situations. Because it has the ability to execute any JScript code in the context of the caller, it could create a security risk for applications. For that reason, you might want to consider creating a custom expression evaluator written in C# or Visual Basic .NET code. One example, written using regular expressions, is provided in *Programming Microsoft Visual Basic .NET (Core Reference)*, by Francesco Balena, and is included with the samples for this chapter. It can be used as a starting point to developing your own expression evaluator.

Note |
There is yet another option—use the Microsoft Script Control. You can add a reference to this COM component from any .NET application (which creates a new reference named Dim sc As New MSScriptControl.ScriptControl() ' Specify the language. sc.Language = "VBScript" Dim Result As String = CType(sc.Eval("2 * (5 + 1) / 3"), String) However, you will need to ensure that this component is installed on registered on any computer that will run your application. This limitation won't apply with the JScript expression evaluator, because it is a core part of the .NET Framework. |

You need to retrieve the current date and time.

Use the *DateTime.Now* shared property.

*DateTime.Now* retrieves the current date and time as a *DateTime* structure. You can then retrieve specific date information from its properties, such as *Month*, *Day*, *Date*, *Hour*, *Millisecond*, and even *DayOfWeek*.

Dim Now As DateTime = DateTime.Now Console.WriteLine("The current date is: " & Now.ToString()) Console.WriteLine("It's a " & Now.DayOfWeek.ToString())

The output for this code is as follows:

The current date is: 2002-10-31 5:24:14 PM It's a Thursday

If you want to retrieve only the date portion of the current day, you can use the *DateTime.Today* shared method. The time portion will be set to 12:00 AM (00:00:00).

You need to perform calculations with dates and times.

Use the *TimeSpan* and *DateTime* structures, both of which provide *Add* and *Subtract* methods.

The .NET Framework provides two structures for manipulating date and time information. The *DateTime* structure stores a reference to a single date and time (such as January 20, 2004 at 12:00 AM). The *TimeSpan* structure stores an interval of time (such as three hours). *TimeSpan* is ultimately measured in ticks (a unit of time equal to 100 nanoseconds), and *DateTime* is stored as the number of ticks since 12:00:00 midnight, January 1, 0001 C.E. The greater a *DateTime* is, the later the date. The greater a *TimeSpan* is, the larger the interval of time.

There are several ways to use *DateTime* and *TimeSpan* to perform calculations. All of the following are valid options:

- Use the
*DateTime.Add*or*DateTime.Subtract*method with a*TimeSpan*value (which returns a new*DateTime*). - Use
*DateTime.Subtract*with a*DateTime*value (which returns a*TimeSpan*representing the difference). - Combine two
*TimeSpan*values using the*TimeSpan.Add*or*TimeSpan.Subtract*method (both of which return a new*TimeSpan*). - Use a higher-level
*DateTime*method such as*AddDays*,*AddHours*,*AddMinutes*, and so on. These methods return a new*DateTime*and can accept negative numbers.

For example, here is how you might check the current time against a fixed expiration date:

If DateTime.Now > (ExpirationDate.AddDays(30)) ' More than thirty days have elapsed since expiration date. End If

Here's how you can benchmark code:

Dim InitialTime As Date = DateTime.Now ' (Insert the code to benchmark here, or make the appropriate function calls.) Dim ElapsedTime As TimeSpan = DateTime.Now.Subtract(InitialTime) Console.WriteLine("Total time: " & ElapsedTime.TotalSeconds.ToString())

Here's one way you could delay code in a loop for a specified interval of time (although using *Thread.Sleep* is a more efficient approach).

Dim InitialTime As DateTime = DateTime.Now Dim WaitSpan As TimeSpan = TimeSpan.FromSeconds(10) Dim LoopTime As TimeSpan ' Wait for 10 seconds. Do LoopTime = DateTime.Now.Subtract(InitialTime) Loop Until TimeSpan.Compare(LoopTime, WaitSpan) = 1

Note that you can't use the comparison operators (< and >) with dates. However, you can retrieve a number that represents the *TimeSpan* interval, using properties such as *TotalHours*, *TotalMinutes*, *TotalMilliseconds*, or *Ticks*. The code below rewrites the time delay loop with a more readable equivalent using the *TimeSpan.Ticks* property.

' Wait for 10 seconds. Do LoopTime = DateTime.Now.Subtract(InitialTime) Loop Until LoopTime.Ticks > WaitSpan.Ticks

You want to determine date information, such as what day of the week a given date falls on, whether a year is a leap year, and how many days are in a month.

Use a *Calendar*-derived class from the *System.Globalization* namespace, such as *GregorianCalendar*, or use the properties of a *DateTime* object.

The *System.Globalization* namespace includes classes that contain culture-related calendar information. These classes derive from the base class *Calendar* and include *GregorianCalendar* (the Western standard), *HebrewCalendar*, *JulianCalendar*, *JapaneseCalendar*, and so on.

The *Calendar* classes define a number of basic methods, including:

*GetDaysInMonth*returns the number of days in a given month.*GetDaysInYear*returns the number of days in a given year.*IsLeapYear*returns*True*if the specified year is a leap year.- Methods that require you to supply a
*DateTime*instance, such as*GetDayOfWeek*,*GetDayOfYear*, and so on. This functionality is also available through*DateTime*properties.

Dim Calendar As New System.Globalization.GregorianCalendar() Console.WriteLine("Days in December 2000: " & _ Calendar.GetDaysInMonth(2000, 12, _ Calendar.CurrentEra).ToString()) Console.WriteLine("Is 2004 a leap year? " & _ Calendar.IsLeapYear(2004)) Console.WriteLine("Days in 2004: " & _ Calendar.GetDaysInYear(2004)) Console.WriteLine("Today is a " & _ Calendar.GetDayOfWeek(DateTime.Now).ToString())

The output is as follows:

Days in December 2000: 31 Is 2004 a leap year? True Days in 2004: 366 Today is a Friday

You want to retrieve the name of a day or month in another language.

Create a *CultureInfo* object for the appropriate culture, and use the *GetDayName* or *GetMonthName* methods.

The *System.Globalization* namespace defines a *DateTimeFormatInfo* type that contains culture-specific date information. You can retrieve the *DateTimeFormatInfo* object for a culture from the *CultureInfo.DateTimeFormat* property. However, you need to create the *CultureInfo* object yourself, using a valid culture name, such as "en-US". (The full list of culture names is provided in the MSDN reference.)

' Create a CultureInfo object representing French - France. Dim Culture As New System.Globalization.CultureInfo("fr-FR") ' Get the corresponding DateTimeFormatInfo object. Dim FormatInfo As System.Globalization.DateTimeFormatInfo FormatInfo = Culture.DateTimeFormat Console.WriteLine(FormatInfo.GetDayName(DayOfWeek.Monday)) ' Displays "lundi" Console.WriteLine(FormatInfo.GetMonthName(1)) ' Displays "janvier"

You want to convert a date or time into a formatted string.

Create a *DateTime* instance to represent the date, and then use the overloaded *ToString* method that accepts a format specifier.

There are two types of date format strings: standard and custom. *Standard format strings* use a preset format according to the current culture and are identified by a single letter. They might retrieve part of the *DateTime* information (just a time, or just a date), and they might format it in a different order or using a different short form. Here are a few common examples:

Dim Now As DateTime = DateTime.Now ' Get just the long time, using the specifier "T" Dim LongTime As String = Now.ToString("T") ' LongTime is now "3:51:24 PM" ' Get just the short date, using the specifier "d" Dim ShortDate As String = Now.ToString("d") ' ShortDate is now "4/10/2003" ' Get just the long date, using the specifier "D" Dim LongDate As String = Now.ToString("D") ' LongDate is now "Tuesday, April 10, 2003"

Table 2-2 lists the standard date format specifiers and their results (assuming the computer is running under the en-US culture).

String |
Name |
Description |
---|---|---|

d |
Short date pattern |
4/10/2003 |

D |
Long date pattern |
Tuesday, April 10, 2003 |

t |
Short time pattern |
3:51 PM |

T |
Long time pattern |
3:51:24 PM |

f |
Full date/time pattern (short time) |
Tuesday, April 10, 2003 3:51 PM |

F |
Full date/time pattern (long time) |
Tuesday, April 10, 2003 3:51:24 PM |

g |
General date/time pattern |
4/10/2003 3:51 PM |

G |
General date/time pattern |
4/10/2003 3:51:24 PM |

M or m |
Month day pattern |
April 10 |

R or r |
RFC1123 pattern |
Tue, 10 Apr 2003 15:51:24 GMT |

s |
Sortable date/time pattern; conforms to ISO 8601 |
2003-04-10T15:51:24 |

u |
Universal sortable date/time pattern |
2003-04-10 15:51:24Z |

U |
Full date/time pattern (long time). This is the same as F, except that it uses universal (GMT) time. |
Tuesday, April 10, 2003 3:51:24 PM |

Y or y |
Year month pattern |
April, 2003 |

If you need to precisely control the format of a date, you must use a *custom format string*. The custom format string is made up of characters that represent the position for various date and time characters, along with any literal values you need. Table 2-3 shows the full list of custom date format specifiers. Notice that where you have the choice of a single or double character ("d" or "dd"), the only difference is how the number will be padded if it's one digit. For example, "d" represents the date number. A date on the ninth day of a month will be converted to "9" with the "d" specifier, or "09" with the "dd" specifier.

Dim Now As DateTime = DateTime.Now ' Get a custom formatted string. Dim Custom As String = Now.ToString("hh:mm, GMT zzz") ' Custom is now "05:13 GMT -09:00" Custom = Now.ToString("dddd MMMM yy gg") ' Custom is now "Thursday April 03 A.D."

Character |
Description |
---|---|

d or dd |
Displays the current day of the month as a number between 1 and 31. |

ddd |
Displays the abbreviated name of the day. |

dddd |
Displays the full name of the day. |

f, ff, fff, ffff, fffff, ffffff, or fffffff |
Displays seconds fractions represented in 1, 2, 3, 4, 5, 6, or 7 digits, respectively. |

g or gg |
Displays the era (A.D., for example) |

h or hh |
Displays the hour in the range 1–12. |

H or HH |
Displays the hour in the range 0–23. |

m or mm |
Displays the minute in the range 0–59. |

M or MM |
Displays the current month as a number between 1 and 12. |

MMM |
Displays the abbreviated name of the month. |

MMMM |
Displays the full name of the month. |

s or ss |
Displays the seconds in the range 0–59. |

t |
Displays the first character of the AM/PM designator. |

tt |
Displays the AM/PM designator. |

y or yy |
Displays the year as a maximum two-digit number. The first two digits of the year are omitted. |

yyyy |
Displays the four-digit year. If the year is less than four digits in length, preceding zeros are appended as necessary to make the displayed year four digits long. |

z or zz |
Displays the time zone offset for the system's current time zone relative to Greenwich mean time, in whole hours only. |

zzz |
Displays the time zone offset for the system's current time zone relative to Greenwich mean time, in hours and minutes. |

: |
Time separator. |

/ |
Date separator. |

' |
Displays the literal value of any string between two single quotation marks ('). This is only required for special characters, such as the slash (/); other literals can be inserted directly. |

You want to convert a date or time into a formatted string that has the same representation, regardless of the globalization settings of the current computer.

Use the overloaded *DateTime.ToString* method that accepts an *IFormatProvider* instance, and supply the *DateTimeFormatInfo.InvariantInfo* object.

In some cases, you want dates to be formatted identically regardless of the settings on the computer running the code. This might be the case if the string value of the date is being inserted into a legacy database. Using a custom format string will offer some protection (if you are careful to only use numeric date information), but to be completely reassured you should specify the culture settings when you create the string.

In order to do this, you must use an overloaded version of the *DateTime.ToString* method that accepts an *IFormatProvider* instance. Rather than construct your own *IFormatProvider*, you can retrieve one that is guaranteed to be invariant over all computers from the *System.Globalization.DateTimeFormatInfo* object. You should still specify a format provider to indicate the format you want, as described in recipe 2.14.

Dim Now As DateTime = DateTime.Now Dim DateString As String DateString = Now.ToString("G", _ System.Globalization.DateTimeFormatInfo.InvariantInfo) ' This string will always be in the form "10/31/2002 18:17:14", ' on any computer.

You'll notice that the defaults applied by the invariant *IFormatProvider* match the default U.S. culture settings, as described in Table 2-2.

You want to convert a user-supplied string containing date information into a *DateTime* instance without introducing the possibility for error.

Use the *DateTime.ParseExact* method with a custom format string.

The *DateTime* structure includes a *Parse* method that creates a *DateTime* instance from a string. However, this method is aggressive and prone to error. It tries everything possible to avoid throwing a *FormatException*, even ignoring unrecognized characters or filling in assumed dates. If this isn't acceptable in your application, you can use the *ParseExact* method, which allows you to define the format you expect for the string. If the format does not match exactly, or there is any discrepancy in the string, an exception is thrown.

*ParseExact* requires three parameters: the string with the date, a standard or custom format specifier describing the format of the string (see recipe 2.13), and a *CultureInfo* object. You can omit the *CultureInfo* parameter to use the machine-specific default, or use the technique shown in recipe 2.14 to use an invariant *CultureInfo*.

Dim DateString As String = "03/17/1977" ' Note that you must use apostrophes to escape the /, ' which is a special character. Dim d2 As DateTime = DateTime.ParseExact(DateString, "MM'/'dd'/'yyyy", _ Nothing) Console.WriteLine("The date parsed as: " & d2.ToString()) ' Displays "The date parsed as: 1977-03-17 12:00:00 AM" ' Note that all DateTime instances have an associated time, ' which defaults to 12:00:00 AM.

When accepting dates from users, it's far better to prevent invalid input rather than deal with it after the fact. Controls such as *DateTimePicker* (for Microsoft Windows applications) and *Calendar* (for Microsoft ASP.NET Web applications) can remove the possibility for error by removing the necessity for a string conversion step.

You want to create a new Globally Unique Identifier (GUID).

Use the shared *System.Guid.NewGuid* method.

A GUID is a 128-bit integer. GUID values are tremendously useful in programming because they're statistically unique. In other words, you can create GUID values continuously with little chance of every creating a duplicate. For that reason, GUIDs are commonly used to uniquely identify queued tasks, user sessions, and other dynamic information. They also have the advantage over random numbers or sequence numbers in the fact that they can't easily be guessed. For example, if you write an XML Web service that assigns a new GUID to each user session, a malicious user won't be able to determine what GUID a user will receive based on what GUIDs other users received earlier.

The .NET Framework provides a *Guid* structure that represents a single GUID value. To create a new GUID, simply call the static *Guid.NewGuid* method, which creates a new GUID with a random value. You can then convert this value to a string by calling the *ToString* method.

Dim NewGuid As Guid = Guid.NewGuid() Console.WriteLine(NewGuid.ToString) ' This writes a value like ""382c74c3-721d-4f34-80e5-57657b6cbc27"

GUID strings are by convention represented in string form as series of lower-case hexadecimal digits in groups of 8, 4, 4, 4, and 12 digits and separated by hyphens (xxxxxxxx-xxxx-xxxx-xxxx-xxxxxxxxxxxx). This is the representation you will receive when you call the *ToString* method, although you can use the "N" format specifier and call *ToString("N")* to retrieve a formatted string with the dashes omitted.

Note |
The random GUID values that the |

You need to convert basic data types to a binary representation, possibly before encrypting them or writing them to a stream.

Use the *BitConverter.GetBytes* method to convert integers, individual characters, and floating point numbers to binary. Use the *System.IO.BinaryWriter* class to convert decimals and strings, or if you need to combine the binary output of multiple values.

The .NET Framework includes the built-in intelligence to convert most basic types into binary representation. If you need to convert an individual value into a byte array, you can use the overloaded *GetBytes* method of the *BitConverter* class. Here's an example:

Dim MyInt As Integer = 100 Dim Bytes() As Byte Bytes = BitConverter.GetBytes(MyInt) Console.WriteLine("An integer requires: " & _ Bytes.Length.ToString() & " bytes.")

If you want to combine the binary output for multiple values into one array, it's easiest to use the *MemoryStream* class in concert with the *BinaryWriter* class (both of which are found in the *System.IO* namespace). You'll also use the *BinaryWriter* class to convert a single decimal value or string into binary representation, because the *BitConverter* class doesn't support these types.

To use the *BinaryWriter* class, call the overloaded *BinaryWriter*.*Write* method with any simple data type. Then, when all values are written, you can convert the entire stream into a single byte array by calling *MemoryStream*.*ToArray*. The code below demonstrates this technique. It assumes you have imported the *System*.*IO* namespace.

' Create a buffer in memory. Dim ms As New MemoryStream ' Create a BinaryWriter that allows you to place binary data in that buffer. Dim w As New BinaryWriter(ms) ' Write the values. Dim MyInt As Integer = 100 Dim MyString As String = "Sample Text" w.Write(MyInt) w.Write(MyString) ' Convert the stream to a byte array. Dim Bytes() As Byte Bytes = ms.ToArray()

To retrieve the information, you can use the *BinaryReader* class. First, you'll have to re-create the *MemoryStream* by using one of the overloaded constructors that accepts a byte array:

' Recreate the memory stream. Dim ms As New MemoryStream(Bytes) Dim r As New BinaryReader(ms) ' Read the values. Dim MyInt As Integer Dim MyString As String MyInt = r.ReadInt32() MyString = r.ReadString()

Reading information with *BinaryReader* is easy. It also saves a good deal of painstaking array copying and offset counting that would be needed if you were using the *BitConverter* class without streams. However, you must read values in the same order that you wrote them, and you must use the method that corresponds to the appropriate type of data.

Of course, if you are writing binary data directly to a file, you won't need to use methods such as *ToArray*. Instead, you would simply substitute a *FileStream* object in place of the *MemoryStream*.

' Create a new file. Dim fs As New FileStream("C: est.bin", FileMode.Create) ' Create a BinaryWriter that allows you to place binary data in that file. Dim w As New BinaryWriter(fs) ' Write the values. Dim MyInt As Integer = 100 Dim MyString As String = "Sample Text" w.Write(MyInt) w.Write(MyString) ' Close the file. w.Close() fs.Close()

Note |
You can also convert complex objects into a stream of bytes or byte array, provided they're serializable. For more information on this technique, refer to recipes 4.8 and 4.9 in Chapter 4. |

You want to compare the bytes in two byte arrays to see if they have the same information.

Iterate over the array and compare each byte, or use the *BitConverter.ToString* method to create a string representation of the entire array.

There is no way to directly test to arrays for equal content. You can use the *Is* operator, but this will only return *True* if both variables point to the *same* array. It will fail if the arrays are duplicate copies of identical data.

If Array1 Is Array2 Then ' This is a reference comparison, which only tests whether ' the variables reference the same array object. End If

A shortcut is to use the *BitConverter.ToString* method to put both byte arrays into a standard string format. You can then compare the two strings:

If BitConverter.ToString(Array1) = BitConverter.ToString(Array2) Then ' Compare the string representations. End If

This approach is not recommended for extremely large byte arrays, because memory will be wasted creating the strings. A better (and faster) approach is simply to create a helper function that iterates through bytes arrays and verifies their equality.

Private Function CompareByteArrays(arrayA() As Byte, arrayB() As Byte) _ As Boolean If Not (arrayA.Length = arrayB.Length) Then Throw New ArgumentException("Arrays must be same length") End If Dim i As Integer For i = 0 To arrayA.Length - 1 If Not (arrayA(i) = arrayB(i)) Then Return False End If Next Return True End Function

Now you can use this helper function to compare binary arrays:

Dim BytesA() As Byte = {32, 22, 10} Dim BytesB() As Byte = {32, 12, 10} Console.WriteLine("Math: " & CompareByteArrays(BytesA, BytesB).ToString())

You want to set an enumeration using the string name of a value, not the integer value, or you want to retrieve all the names used for constants in an enumeration.

Use the *Enum.GetNames* method to get an array of all enumeration names and the *Enum.Parse* method to convert a string into the corresponding value from an enumeration.

An enumeration is a group of integer constants with descriptive names. Usually, you'll use enumeration values by name. Sometimes, however, it's necessary to convert enumeration values into strings, and vice versa. One reason might be to provide a user with a list of enumerated values and give them the chance to choose one.

As an example, consider the *System.Drawing.KnownColor* enumeration, which lists known system colors. You could use the *Enum.GetNames* method to retrieve a string array with all the color names, and add them to a listbox.

' Get the names of all enumerated values. Dim ColorNames() As String ColorNames = System.Enum.GetNames(GetType(KnownColor)) ' Add the contents of the entire array to the list box. lstColors.Items.AddRange(ColorNames)

To demonstrate the reverse task, you can handle the *ListBox.SelectionChanged* event, so that every time an entry is clicked in the list, a new color is set in a label. To perform this task, the listbox text must be converted to a *KnownColor* value and then used with the *Color.FromKnownColor* helper method, which creates a *Color* object based on the value.

Private Sub lstColors_SelectedIndexChanged(ByVal sender As System.Object, _ ByVal e As System.EventArgs) Handles lstColors.SelectedIndexChanged ' Find the enumerated value that corresponds to the selected text. Dim ColorEnum As Object ColorEnum = System.Enum.Parse(GetType(KnownColor), lstColors.Text) Dim SelectedColor As KnownColor SelectedColor = CType(ColorEnum, KnownColor) ' Use the enumerated value to set the background color of the label. lbl.BackColor = System.Drawing.Color.FromKnownColor(SelectedColor) End Sub

Figure 2-1 shows this sample program in action. Accomplishing the same result without using the enumeration would involve quite a bit of code—you would have to add each color value manually.

Figure 2-1: Converting enumerated values to strings and back.

Introduction

- Strings and Regular Expressions
- Numbers, Dates, and Other Data Types
- Arrays and Collections
- Objects, Interfaces, and Patterns
- Files and Directories
- XML
- Multithreading
- Network Programming
- Reflection
- Windows Programming
- Windows Controls
- Printing and Drawing with GDI+
- Windows Services
- ADO.NET
- ASP.NET Web Applications
- Web Services
- Remoting and Enterprise Services
- Security and Cryptography
- Useful COM Interop

Microsoft Visual Basic .NET Programmers Cookbook (Pro-Developer)

ISBN: 073561931X

EAN: 2147483647

EAN: 2147483647

Year: 2003

Pages: 376

Pages: 376

Authors: Matthew MacDonald

Similar book on Amazon

- Chapter II Information Search on the Internet: A Causal Model
- Chapter III Two Models of Online Patronage: Why Do Consumers Shop on the Internet?
- Chapter IX Extrinsic Plus Intrinsic Human Factors Influencing the Web Usage
- Chapter XV Customer Trust in Online Commerce
- Chapter XVI Turning Web Surfers into Loyal Customers: Cognitive Lock-In Through Interface Design and Web Site Usability

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