How does operator overload work?

Question

How does operator overload work?

How does this code work?

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))

2

python classes sobrecarga operadores

Author: Maniero, 2020-05-16

Source

1 answers

score 3 · Answer 1

Doesn't have much secret, the operator used in an expression is actually a syntactic sugar, so it's not quite what you're seeing. Actually the line

Value3 = Value1 + Value2

Will execute something like

Value3 = Value1.__add__(Value2)

Then only the method described in the class is called.

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

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