First, a bit of background. Read the question and accepted answer posted here for a specific scenario for my question. I\'m not sure if other, similar cases exist but this
It's because a zero-integer is implicitly convertible to an enum:
enum SqlDbType
{
Zero = 0,
One = 1
}
class TestClass
{
public TestClass(string s, object o)
{ System.Console.WriteLine("{0} => TestClass(object)", s); }
public TestClass(string s, SqlDbType e)
{ System.Console.WriteLine("{0} => TestClass(Enum SqlDbType)", s); }
}
// This is perfectly valid:
SqlDbType valid = 0;
// Whilst this is not:
SqlDbType ohNoYouDont = 1;
var a1 = new TestClass("0", 0);
// 0 => TestClass(Enum SqlDbType)
var a2 = new TestClass("1", 1);
// => 1 => TestClass(object)
(Adapted from Visual C# 2008 Breaking Changes - change 12)
When the compiler performs the overload resolution 0 is an Applicable function member for both the SqlDbType
and the object
constructors because:
an implicit conversion (Section 6.1) exists from the type of the argument to the type of the corresponding parameter
(Both SqlDbType x = 0
and object x = 0
are valid)
The SqlDbType
parameter is better than the object
parameter because of the better conversion rules:
T1
and T2
are the same type, neither conversion is better.
object
and SqlDbType
are not the same typeS
is T1
, C1
is the better conversion.
0
is not an object
S
is T2
, C2
is the better conversion.
0
is not a SqlDbType
T1
to T2
exists, and no implicit conversion from T2
to T1
exists, C1
is the better conversion.
object
to SqlDbType
existsT2
to T1
exists, and no implicit conversion from T1
to T2
exists, C2
is the better conversion.
SqlDbType
to object
exists, so the SqlDbType
is the better conversionNote that what exactly constitutes a constant 0 has (quite subtly) changed in Visual C# 2008 (Microsoft's implementation of the C# spec) as @Eric explains in his answer.