If reference type doesn't overload an equality operator ==, then build-in equality operator on Object will be used instead. Why isn't the same true for user-defined structs:
struct A{ } static void Main(string[] args) { A a = new A(); A a1 = new A(); bool equal= (a == a1); //error } Namely, doesn't ValueType ( from which all structs derive ) also overload == operator?
How would such a default == operator work? For reference types, comparing adresses is reasonable, but since that check will never be true for two ValueTypes (since if two ValueTypes are in scope then they are guaranteed to have different locations on the stack,) address comparison is pointless.
As the compiler has helpfully pointed out, ValueType very intentionally does not have a default == operator.
bool equal = (5 == 6);. After all Int32 is a value type as well.
==. It still isn't default.
bool equal = (5 == 6); is that the C# compiler translates the == operator into a ceq IL instruction. System.Int32 doesn't define any operators.Structs probably don't provide a default == operator implementation because, unlike a class instance, a struct has no concept of reference-style identity.
Implementing the Equality Operator (==) on Value Types
In most programming languages there is no default implementation of the equality operator (==) for value types. Therefore, you should overload == any time equality is meaningful.
You should consider implementing the Equals method on value types because the default implementation on System.ValueType will not perform as well as your custom implementation.
Implement == any time you override the Equals method.
However, structs do provide a default Equals Method implementation which will do a memberwise compare using reflection.
