How do I override a method object's call method in Python?

Question

Here is what I am working with so far

def f(n):
    return n

f.__call__ = lambda n: n + 1

print f(2) #I expect an output of 3 but get an output of 2

I am not interested in another way to achieve the desired output. Rather, for educational purposes, I would like to know why overriding the __call__ as I have done, doesn't work as I expect.

@SethMMorton: Try to execute `dir(f)` to see that they have it. — Tadeck, Aug 13 '13 at 01:46
Related: http://stackoverflow.com/questions/9707600/python-functions-and-their-call-attribute — Jared, Aug 13 '13 at 01:51

score 5 · Accepted Answer · edited Aug 13 '13 at 02:45

5

This appears to be due to special-casing of function types in ceval.c, in call_function:

if (PyFunction_Check(func))
    x = fast_function(func, pp_stack, n, na, nk);
else
    x = do_call(func, pp_stack, na, nk);

I'd guess that this is probably for efficiency, since calling regular functions, and ignoring the __call__ attribute, is by far the most common kind of calling that gets done.

edited Aug 13 '13 at 02:45

unutbu

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answered Aug 13 '13 at 02:13

Owen

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I've changed the link to point to hg.python.org since this is the official repository for CPython now. – unutbu Aug 13 '13 at 02:46

How do I override a method object's __call__ method in Python?

1 Answers1

How do I override a method object's call method in Python?