Does python coerce types when doing operator overloading?

Question

I have the following code:

a = str('5')
b = int(5)
a == b
# False

But if I make a subclass of int, and reimplement __cmp__:

class A(int):
    def __cmp__(self, other):
        return super(A, self).__cmp__(other)
a = str('5')
b = A(5)
a == b
# TypeError: A.__cmp__(x,y) requires y to be a 'A', not a 'str'

Why are these two different? Is the python runtime catching the TypeError thrown by int.__cmp__(), and interpreting that as a False value? Can someone point me to the bit in the 2.x cpython source that shows how this is working?

Answer 1

The documentation isn't completely explicit on this point, but see here:

If both are numbers, they are converted to a common type. Otherwise, objects of different types always compare unequal, and are ordered consistently but arbitrarily. You can control comparison behavior of objects of non-built-in types by defining a __cmp__ method or rich comparison methods like __gt__, described in section Special method names.

This (particularly the implicit contrast between "objects of different types" and "objects of non-built-in types") suggests that the normal process of actually calling comparison methods is skipped for built-in types: if you try to compare objects of two dfferent (and non-numeric) built-in types, it just short-circuits to an automatic False.