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 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?
In Python, overloading is achieved by overriding the method which is specifically for that operator, in the user-defined class. For example, __add__(self, x) is a method reserved for overloading + operator, and __eq__(self, x) is for overloading == .
You can redefine or overload the function of most built-in operators in C++. These operators can be overloaded globally or on a class-by-class basis. Overloaded operators are implemented as functions and can be member functions or global functions.
NO. There is no such requirement that you Must overload != If you need to overload == . However,it is a good practice that you Should overload operators related to each other.
You can overload any of these operators, which can be used to compare the objects of a class. Following example explains how a < operator can be overloaded and similar way you can overload other relational operators.
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.
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