C#之自己定义的implicit和explicit转换

在类型转换时常会遇到隐式转换和显式转换。那我们自己定义的类型要怎样去定义隐式转换和显式转换？我们来看一段代码

public class Rational
{
private Int32 _inner_int = 0;

public Rational()
{

}

public Rational(Int32 num)
{
this._inner_int = num;
}

public Int32 ToInt32() { return this._inner_int; }

// Implicitly constructs and returns a Rational from an Int32
public static implicit operator Rational(Int32 num)
{
return new Rational(num);
}

// Explicitly returns an Int32 from a Rational
public static explicit operator Int32(Rational r)
{
return r.ToInt32();
}

public override string ToString()
{
//return base.ToString();
String s = String.Format("{0}", this._inner_int);
return s;
}
}

測试代码

class Program
{
static void Main(string[] args)
{
Rational r1 = 10;
Console.WriteLine(r1);

Int32 i = r1;
Console.WriteLine(i);
Console.ReadLine();
}
}

这时编辑会报错，见下图



从提示能够看到，是由于Int32 i=r1时缺少了显式转换。如今我们加入显示转换，改动后的代码及输出结果例如以下：



结果正常输出为10.

那为什么会这样呢？究其原因是在Rational转换成 Int32时，指定了explicit(显式的)，所以必需要指定转换类型Int32。假设将explicit换成implicit(隐式)，原来的代码将能够正常执行。

改动后的Rational

public class Rational
{
private Int32 _inner_int = 0;

public Rational()
{

}

public Rational(Int32 num)
{
this._inner_int = num;
}

public Int32 ToInt32() { return this._inner_int; }

// Implicitly constructs and returns a Rational from an Int32
public static implicit operator Rational(Int32 num)
{
return new Rational(num);
}

// Explicitly returns an Int32 from a Rational
public static <span style="color:#ff0000;">implicit</span> operator Int32(Rational r)
{
return r.ToInt32();
}

public override string ToString()
{
//return base.ToString();
String s = String.Format("{0}", this._inner_int);
return s;
}
}

測试代码及输出结果



可见explicit和implicit影响着类型的显式转换和隐式转换。

事实上在Rational r1=10已经运行了隐式转换，相应的转换代码例如以下：

// Implicitly constructs and returns a Rational from an Int32
public static implicit operator Rational(Int32 num)
{
return new Rational(num);
}

假设将implicit换成explicit，Rational r1=10也将会报错(能够自行測试)。

转载请注明出处：http://blog.csdn.net/xxdddail/article/details/38057563