Higher order function Python walkthrough

Question

def tracer(fn):
    def traced(x):
        print('Calling', fn, '(', x, ')')
        result = fn(x)
        print('Got', result, 'from', fn, '(', x, ')')
        return result
    return traced

def fact(n):
    if n == 0:
        return 1
    return n * fact(n-1)

new_fact = tracer(fact)
new_fact(2)

I used this code on pythontutor.com to better understand higher order functions, but i’m having difficulty understanding why new_fact(2) in step 8 is mapped to traced? In other words, how does the traced function know the argument is 2?

Asked By: user2691652

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Answer 1

In Python, functions are objects too. When you call the tracer() function, it returns the nested traced() function; it is in fact creating a new copy of that function:

return traced

You stored that returned function object in new_fact, then called it:

>>> tracer(fact)
<function traced at 0x10644c320>
>>> new_fact = tracer(fact)
>>> new_fact
<function traced at 0x10644c398>
>>> new_fact(2)
('Calling', <function fact at 0x10644c230>, '(', 2, ')')
('Got', 2, 'from', <function fact at 0x10644c230>, '(', 2, ')')
2

You can do this with any function; store a reference to a function in another name:

>>> def foo(): print 'foo'
... 
>>> bar = foo
>>> bar()
foo

Answered By: Martijn Pieters

Higher order function Python walkthrough

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