C# 程序员参考--用户定义的换教程

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.cs

using 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

10

Conversion not yet implemented

10  

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.cs

using 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);

    }

} 软件开发网 www.mscto.com 

输出

10

Conversion not yet implemented  

代码讨论

• 在上个示例中，语句

  binary = (BinaryNumeral)(int)roman; 
  软件开发网 www.mscto.com
   

  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。