How does operator overload work?

Question

How this code works?

class MyClass:
   def __init__(self, *args):
      self.Input = args

   def __add__(self, Other):
      Output = MyClass()
      Output.Input = self.Input + Other.Input
      return Output

   def __str__(self):
      Output = ""
      for Item in self.Input:
         Output += Item
         Output += " "
      return Output

Value1 = MyClass("Red", "Green", "Blue")
Value2 = MyClass("Yellow", "Purple", "Cyan")
Value3 = Value1 + Value2

print("{0} + {1} = {2}"
      .format(Value1, Value2, Value3))

Answer 1

It’s not very secret, the operator used in an expression is actually a syntactic sugar, so it’s not quite what you’re seeing. In fact the line

Value3 = Value1 + Value2

It will perform something like

Value3 = Value1.__add__(Value2)

So it’s just called the method described in the class.

Note that the first operand will determine which class will be used to call the method, since this method may be present in several classes. In the case Value1 is the type MyClass so it’s the method __add__() inside MyClass which will be called, but this is not even about the operator, applies to any method.

It is a little different because the Operator has special features in the language, mainly it can be noticed precedence between operators when there are several operands and need to decide which operator is executed first, if it were a normal method there will be no precedence, only association.