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)。