Pythonic way to add method alias in subclass

[Lee Harr]

I thought of several ways to add another name for
a method in a subclass ...
#alias.py
class Foo(object):
def nod(self):
print "nodding"

class Bar(Foo):
def __init__(self):
self.agree = self.nod

class Bar2(Foo):
agree = Foo.nod

class Bar3(Foo):
def agree(self):
Foo.nod(self)

def alias(method):
def dec(m):
return method
return dec

class Bar4(Foo):
@alias(Foo.nod)
def agree(self):
pass
b = Bar()
b.agree()

b2 = Bar2()
b2.agree()

b3 = Bar3()
b3.agree()

b4 = Bar4()
b4.agree()
#####################
I am leaning towards Bar2 since it has the least code.

Any thoughts?

[Steven D'Aprano]

On Sat, 15 Dec 2007 13:03:33 +0000, Lee Harr wrote:

I thought of several ways to add another name for a method in a subclass

....

class Bar2(Foo):
agree = Foo.nod

....

I am leaning towards Bar2 since it has the least code.

Sure, why not?

Out of curiosity, what's wrong with just calling Bar2.nod()?
--
Steven

[Bruno Desthuilliers]

Lee Harr a écrit :

I thought of several ways to add another name for
a method in a subclass ...
#alias.py
class Foo(object):
def nod(self):
print "nodding"

class Bar(Foo):
def __init__(self):
self.agree = self.nod

Will create an instance attribute of type method for each instance of
Bar. Which might or not be a problem FWIW.

class Bar2(Foo):
agree = Foo.nod

The One Obvious Way.

class Bar3(Foo):
def agree(self):
Foo.nod(self)

Are you willing to take the penalty of an extra method call ?

def alias(method):
def dec(m):
return method
return dec

class Bar4(Foo):
@alias(Foo.nod)
def agree(self):
pass

Bad case of arbitrary overcomplification IMHO. Nothing with metaclasses
while we're at it ?-)

(snip)

I am leaning towards Bar2 since it has the least code.

Indeed.

Any thoughts?

cf above.