Python method-wrapper type?

Question

What is the method-wrapper type in Python 3? If I define a class like so:

class Foo(object):
    def __init__(self, val):
        self.val = val
    def __e         


        

Answer 1

Below some testcode from my variable inspection routines. Defined class Tst, which has a __str__ method. To find out if an instance of a class has either __repr__ or __str__ defined, you can test for hasattr(obj.__str__, '__code__'):

Example:

class Tst(object):
    """ Test base class.
    """
    cl_var = 15

    def __init__(self, in_var):
        self.ins_var = in_var

    def __str__(self):
        # Construct to build the classname ({self.__class__.__name__}).
        # so it works for subclass too.
        result = f"{self.__class__.__name__}"
        result += f"(cl_var: {self.__class__.cl_var}, ins_var: {self.ins_var})"
        return result

    def show(self):
        """ Test method
        """
        print(self.ins_var)


t5 = Tst(299)


print('\n\n----- repr() ------')
print(f"obj.repr(): {repr(t5)}")
print(f"obj.repr.class type: {type(t5.__repr__.__class__)}")
print(f"obj.repr.class.name: {t5.__repr__.__class__.__name__}")
print(f"obj.__repr__ has code: {hasattr(t5.__repr__, '__code__')}")

print('\n\n----- str() ------')
print(f"obj.str(): {str(t5)}")
print(f"obj.str.class type: {type(t5.__str__.__class__)}")
print(f"obj.str.class.name: {t5.__str__.__class__.__name__}")
print(f"obj.__str__ has code: {hasattr(t5.__str__, '__code__')}")

Returns:

----- repr() ------
obj.repr(): <__main__.Tst object at 0x107716198>
obj.repr.class type: 
obj.repr.class.name: method-wrapper
obj.__repr__ has code: False


----- str() ------
obj.str(): Tst(cl_var: 15, ins_var: 299)
obj.str.class type: 
obj.str.class.name: method
obj.__str__ has code: True

Since __repr__ isn't defined on the class, it defaults to the __repr__ on base-class object via the method-wrapper and gives a default output. __str__ is defined (and thus a method), so the test hasattr(t5.__str__, '__code__') results True.