C# 程序员参考--用户定义的转换教程

本教程展示如何定义转换以及如何在类或结构之间使用转换。

示例文件

请参见“用户定义的转换”示例以下载和生成本教程中讨论的示例文件。

教程

C# 允许程序员在类或结构上声明转换，以便可以使类或结构与其他类或结构或者基本类型相互进行转换。转换的定义方法类似于运算符，并根据它们所转换到的类型命名。

在 C# 中，可以将转换声明为 implicit（需要时自动转换）或 explicit（需要调用转换）。所有转换都必须为 static，并且必须采用在其上定义转换的类型，或返回该类型。

本教程介绍两个示例。第一个示例展示如何声明和使用转换，第二个示例演示结构之间的转换。

示例 1

本示例中声明了一个 RomanNumeral 类型，并定义了与该类型之间的若干转换。

// conversion.csusing System;struct RomanNumeral{    public RomanNumeral(int value)     {        this.value = value;     }    // Declare a conversion from an int to a RomanNumeral. Note the    // the use of the operator keyword. This is a conversion     // operator named RomanNumeral:    static public implicit operator RomanNumeral(int value)     {       // Note that because RomanNumeral is declared as a struct,        // calling new on the struct merely calls the constructor        // rather than allocating an object on the heap:       return new RomanNumeral(value);    }    // Declare an explicit conversion from a RomanNumeral to an int:    static public explicit operator int(RomanNumeral roman)    {       return roman.value;    }    // Declare an implicit conversion from a RomanNumeral to     // a string:    static public implicit operator string(RomanNumeral roman)    {       return("Conversion not yet implemented");    }    private int value;}class Test{    static public void Main()    {        RomanNumeral numeral;        numeral = 10;// Call the explicit conversion from numeral to int. Because it is// an explicit conversion, a cast must be used:        Console.WriteLine((int)numeral);// Call the implicit conversion to string. Because there is no// cast, the implicit conversion to string is the only// conversion that is considered:        Console.WriteLine(numeral); // Call the explicit conversion from numeral to int and // then the explicit conversion from int to short:        short s = (short)numeral;        Console.WriteLine(s);    }}

输出

10Conversion not yet implemented10

示例 2

本示例定义 RomanNumeral 和 BinaryNumeral 两个结构，并演示二者之间的转换。

// structconversion.csusing System;struct RomanNumeral{    public RomanNumeral(int value)     {        this.value = value;     }    static public implicit operator RomanNumeral(int value)    {        return new RomanNumeral(value);    }    static public implicit operator RomanNumeral(BinaryNumeral binary)    {        return new RomanNumeral((int)binary);    }    static public explicit operator int(RomanNumeral roman)    {         return roman.value;    }    static public implicit operator string(RomanNumeral roman)     {        return("Conversion not yet implemented");    }    private int value;}struct BinaryNumeral{    public BinaryNumeral(int value)     {        this.value = value;    }    static public implicit operator BinaryNumeral(int value)    {        return new BinaryNumeral(value);    }    static public implicit operator string(BinaryNumeral binary)    {        return("Conversion not yet implemented");    }    static public explicit operator int(BinaryNumeral binary)    {        return(binary.value);    }    private int value;}class Test{    static public void Main()    {        RomanNumeral roman;        roman = 10;        BinaryNumeral binary;        // Perform a conversion from a RomanNumeral to a        // BinaryNumeral:        binary = (BinaryNumeral)(int)roman;        // Performs a conversion from a BinaryNumeral to a RomanNumeral.        // No cast is required:        roman = binary;        Console.WriteLine((int)binary);        Console.WriteLine(binary);    }}

输出

10Conversion not yet implemented

代码讨论

• 在上个示例中，语句

  binary = (BinaryNumeral)(int)roman;

  执行从 RomanNumeral 到 BinaryNumeral 的转换。由于没有从 RomanNumeral 到 BinaryNumeral 的直接转换，所以使用一个转换将 RomanNumeral 转换为 int，并使用另一个转换将 int 转换为 BinaryNumeral。

• 另外，语句

  roman = binary;

  执行从 BinaryNumeral 到 RomanNumeral 的转换。由于 RomanNumeral 定义了从 BinaryNumeral 的隐式转换，所以不需要转换。
  

原文转自：http://www.ltesting.net