Python: default comparison

Question

In Python 2.7, I define an empty new-style class:

In [43]: class C(object): pass
   ....:

then create a list of instances of the new class:

In [44]: c = [C() for i in xrange(10)]

then attempt to sort the list:

In [45]: sorted(c)
Out[45]:
[<__main__.C object at 0x1950a490>,
 <__main__.C object at 0x1950a4d0>,
 ...
 <__main__.C object at 0x1950aad0>]

What's surprising is that the sort doesn't complain, even though I haven't defined a way to compare instances of C:

In [46]: dir(C())
Out[46]:
['__class__',
 '__delattr__',
 '__dict__',
 '__doc__',
 '__format__',
 '__getattribute__',
 '__hash__',
 '__init__',
 '__module__',
 '__new__',
 '__reduce__',
 '__reduce_ex__',
 '__repr__',
 '__setattr__',
 '__sizeof__',
 '__str__',
 '__subclasshook__',
 '__weakref__']

What exactly is happening there, and what's the rationale for this -- arguably surprising -- behaviour?

Answer 1

I think the only rationale is that it is convenient that objects can be sorted and e.g. used as dictionary keys with some default behavior. The relevant chapter in the language definition is here: https://docs.python.org/2/reference/expressions.html#not-in

"The choice whether one object is considered smaller or larger than another one is made arbitrarily but consistently within one execution of a program."

So the fact that objects are currently compared using the memory address is just an implementation detail that cannot be counted upon. The only guarantee is that the ordering stays consistent during execution.

Answer 2

I'm not exactly sure, but maybe someone can correct me on this.

When you compare objects, it compares their memory address, think of comparing 2 cstrings in C. If you take a look, the sorting sorted the objects from the lowest memory address to the highest memory address (or pointer location).