Ytukyg

In python (3.5.2), I was expecting the repr(obj) function to call the magic method __repr__() of obj's class.
However calling both of them do not seem to yield the same result. Can anyone explain why ?

Sample code :

class parent:



    def __init__(self):

        self.a = "haha"



    def __repr__(self):

        return repr(self.a)



class child(parent):

    def __init__(self):

        super().__init__()

        self.b="bebe"



    def __repr__(self):

        return "("+super().__repr__()+", "+repr(super())+", "+self.b+")"



    def print1(self):

        print("super().__repr__() returns:", super().__repr__())

        print("repr(super()) returns:", repr(super()))

        print("plom(super()).__repr__() returns:", plom(super()).__repr__())

        print("repr(plom(super())) returns:", repr(plom(super())))



def plom(var):

    return var



t=child()

print(t.__repr__())

print(repr(t))

print('-----')

t.print1()

print('-----')

print(plom(t).__repr__())

print(repr(plom(t)))

result :

>>> 

 RESTART: test super.py 

('haha', <super: <class 'child'>, <child object>>, bebe)

('haha', <super: <class 'child'>, <child object>>, bebe)

-----

super().__repr__() returns: 'haha'

repr(super()) returns: <super: <class 'child'>, <child object>>

plom(super()).__repr__() returns: 'haha'

repr(plom(super())) returns: <super: <class 'child'>, <child object>>

-----

('haha', <super: <class 'child'>, <child object>>, bebe)

('haha', <super: <class 'child'>, <child object>>, bebe)

>>> 

Calling repr(super()) directly accesses the __repr__ on the super class (technically, the tp_repr of the C PyTypeObject struct defining the super type). Most special dunder methods behave this way when called implicitly (as opposed to explicitly calling them as methods). repr(x) isn't equivalent to x.__repr__(). You can think of repr as being defined as:

def repr(obj):

    return type(obj).__repr__(obj)  # Call unbound function of class with instance as arg

while you were expecting it to be:

def repr(obj):

    return obj.__repr__()  # Call bound method of instance

This behavior is intentional; one, customizing dunder methods per-instance makes little sense, and two, prohibiting it allows for much more efficient code at the C level (it has much faster ways of doing what the illustrative methods above do).

By contrast, super().__repr__() looks up the method on the super instance, and super defines a custom tp_getattro (roughly equivalent to defining a custom __getattribute__ method), which means lookups on the instance are intercepted before they find the tp_repr/__repr__ of the class, and instead are dispatched through the custom attribute getter (which performs the superclass delegation).

If you consult the docs, you'll see that super returns a proxy object which delegates method calls to the appropriate class according to method resolution order.

So repr(super()) gets you the representation of the proxy object. Whereas the method call super().__repr__() gives you the representation defined by the next class in the method resolution order.

If you want the superclass itself, try

my_object.__mro__[1]

In super().__repr__() you're calling the repr class of the super object so you get 'haha'

In the second, you're calling repr of the super(). what does super() output? <super: <class 'child'>, <child object>> so you're effectively calling repr on some class hierarchy