问题
I have never done any extensive work with overloading operators, especially the implicit and explicit conversions.
However, I have several numeric parameters that are used frequently, so I am creating a struct as a wrapper around a numeric type to strongly type these parameters. Here's an example implementation:
public struct Parameter
{
private Byte _value;
public Byte Value { get { return _value; } }
public Parameter(Byte value)
{
_value = value;
}
// other methods (GetHashCode, Equals, ToString, etc)
public static implicit operator Byte(Parameter value)
{
return value._value;
}
public static implicit operator Parameter(Byte value)
{
return new Parameter(value);
}
public static explicit operator Int16(Parameter value)
{
return value._value;
}
public static explicit operator Parameter(Int16 value)
{
return new Parameter((Byte)value);
}
}
As i was experimenting with my test implementation to get a hang of the explicit and implicit operators, I tried to explicitly cast a Int64
to my Parameter
type and to my surprised it did not throw an exception, and even more surprising, it just truncated the number and moved on. I tried excluding the custom explicit operator and it still behaved the same.
public void TestCast()
{
try
{
var i = 12000000146;
var p = (Parameter)i;
var d = (Double)p;
Console.WriteLine(i); //Writes 12000000146
Console.WriteLine(p); //Writes 146
Console.WriteLine(d); //Writes 146
}
catch (Exception ex)
{
Console.WriteLine(ex.Message); //Code not reached
}
}
So I repeated my experiment with a plain Byte
in place of my struct and has the same exact behavior, so obviously this is expected behavior, but I thought an explicit cast that results in a lose of data would throw an exception.
回答1:
When the compiler is analyzing an explicit user-defined conversion it is allowed to put an explicit built-in conversion on "either side" (or both) of the conversion. So, for example, if you have a user-defined conversion from int to Fred, and you have:
int? x = whatever;
Fred f = (Fred)x;
then the compiler reasons "there is an explicit conversion from int to Fred, so I can make an explicit conversion from int? to int, and then convert int to Fred.
In your example, there is a built-in explicit conversion from long to short, and there is a user-defined explicit conversion from short to Parameter, so converting long to Parameter is legal.
The same is true of implicit conversions; the compiler may insert built-in implicit conversions on either side of a user-defined implicit conversion.
The compiler never chains two user defined conversions.
Building your own explicit conversions correctly is a difficult task in C#, and I encourage you to stop attempting to do so until you have a thorough and deep understanding of the entire chapter of the specification that covers conversions.
For some interesting aspects of chained conversions, see my articles on the subject:
http://blogs.msdn.com/b/ericlippert/archive/2007/04/16/chained-user-defined-explicit-conversions-in-c.aspx
http://blogs.msdn.com/b/ericlippert/archive/2007/04/18/chained-user-defined-explicit-conversions-in-c-part-two.aspx
回答2:
This goal:
so I am creating a struct as a wrapper around a numeric type to strongly type these parameters
And this code:
public static implicit operator Byte(Parameter value)
{
return value._value;
}
public static implicit operator Parameter(Byte value)
{
return new Parameter(value);
}
Are in total contradiction. By adding 2-way implicit operators you annul any type-safety the wrapper might bring.
So drop the implicit conversions. You can change them to explicit ones.
来源:https://stackoverflow.com/questions/8526547/explicit-implicit-operator-with-numeric-types-unexpected-results