Consider the following code:
class CustomClass
{
public CustomClass(string value)
{ m_value = value; }
public static bool operator ==(CustomClass a, CustomClass b)
{ return a.m_value == b.m_value; }
public开发者_StackOverflow static bool operator !=(CustomClass a, CustomClass b)
{ return a.m_value != b.m_value; }
public override bool Equals(object o)
{ return m_value == (o as CustomClass).m_value; }
public override int GetHashCode()
{ return 0; /* not needed */ }
string m_value;
}
class G
{
public static bool enericFunction1<T>(T a1, T a2) where T : class
{ return a1.Equals(a2); }
public static bool enericFunction2<T>(T a1, T a2) where T : class
{ return a1==a2; }
}
Now when I call both generic functions, one succeeds and one fails:
var a = new CustomClass("same value");
var b = new CustomClass("same value");
Debug.Assert(G.enericFunction1(a, b)); // Succeeds
Debug.Assert(G.enericFunction2(a, b)); // Fails
Apparently, G.enericFunction2 executes the default operator== implementation instead of my override. Can anybody explain why this happens?
From Constraints on Type Parameters (C# Programming Guide):
When applying the where T : class constraint, avoid the == and != operators on the type parameter because these operators will test for reference identity only, not for value equality. This is the case even if these operators are overloaded in a type that is used as an argument. (...) The reason for this behavior is that, at compile time, the compiler only knows that T is a reference type, and therefore must use the default operators that are valid for all reference types.
If I change the enericFunction2
to:
public static bool enericFunction2<T>(T a1, T a2) where T : class
{
object aa = a1;
CustomClass obj1 = (CustomClass)aa;
object bb = a2;
CustomClass obj2 = (CustomClass)bb;
return obj1 == obj2;
}
Then everything works fine. But I am afraid I can't explain it. I mean a1
and a2
know their type. Why a cast is needed to the CustomClass
, so the operators are called?
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