在类型转换时常会遇到隐式转换和显式转换。那我们自定义的类型要如何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;        }    }

测试代码

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

从提示可以看到，是因为Int32 i=r1时缺少了显式转换。现在我们添加显示转换，修改后的代码及输出结果如下：

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)，所以必须要指定转换类型Int32。如果将explicit换成implicit(隐式)，原来的代码将可以正常运行。

修改后的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 implicit operator Int32(Rational r)        {            return r.ToInt32();        }        public override string ToString()        {            //return base.ToString();            String s = String.Format("{0}", this._inner_int);            return s;        }    }

测试代码及输出结果

可见explicit和implicit影响着类型的显式转换和隐式转换。

其实在Rational r1=10已经执行了隐式转换，对应的转换代码如下：

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

如果将implicit换成explicit，Rational r1=10也将会报错(可以自行测试)。

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