C# 程序员参考--用户定义地转换教学文章

 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 
 
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	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

10



Conversion 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。

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