callback - Python, how to pass an argument to a function pointer parameter?

Question

I only just started learning Python and found out that I can pass a function as the parameter of another function. Now if I call foo(bar()) it will not pass as a function pointer but the return value of the used function. Calling foo(bar) will pass the function, but this way I am not able to pass any additional arguments. What if I want to pass a function pointer that calls bar(42)?

I want the ability to repeat a function regardless of what arguments I have passed to it.

def repeat(function, times):
    for calls in range(times):
        function()

def foo(s):
        print s

repeat(foo("test"), 4)

In this case the function foo("test") is supposed to be called 4 times in a row. Is there a way to accomplish this without having to pass "test" to repeat instead of foo?

Answer 1

You can either use a lambda:

repeat(lambda: bar(42))

Or functools.partial:

from functools import partial
repeat(partial(bar, 42))

Or pass the arguments separately:

def repeat(times, f, *args):
    for _ in range(times):
        f(*args)

This final style is quite common in the standard library and major Python tools. *args denotes a variable number of arguments, so you can use this function as

repeat(4, foo, "test")

or

def inquisition(weapon1, weapon2, weapon3):
    print("Our weapons are {}, {} and {}".format(weapon1, weapon2, weapon3))

repeat(10, inquisition, "surprise", "fear", "ruthless efficiency")

Note that I put the number of repetitions up front for convenience. It can't be the last argument if you want to use the *args construct.

(For completeness, you could add keyword arguments as well with **kwargs.)