在其他函数python中传递函数作为参数

Question

I have these functions, and I'm getting errors, with the do_twice functions, but I'm having problems debugging it

#!/usr/bin/python
#functins exercise 3.4

def do_twice(f):
    f()
    f()

def do_four(f):
    do_twice(f)
    do_twice(f)

def print_twice(str):
    print str + 'one' 
    print str + 'two'


str = 'spam'
do_four(print_twice(str))

debugger errors

:!python 'workspace/python/functions3.4.py'
spamone
spamtwo
Traceback (most recent call last):
  File "workspace/python/functions3.4.py", line 18, in <module>
    do_four(print_twice(str))
  File "workspace/python/functions3.4.py", line 9, in do_four
    do_twice(f)
  File "workspace/python/functions3.4.py", line 5, in do_twice
    f()
TypeError: 'NoneType' object is not callable

shell returned 1

Answer 1

The problem is that the expression print_twice(str) is evaluated by calling print_twice with str and getting the result that you returned,* and that result is what you're passing as the argument to do_four .

What you need to pass to do_four is a function that, when called, calls print_twice(str) .

You can build such a function manually:

def print_twice_str():
    print_twice(str)
do_four(print_twice_str)

Or you can do the same thing inline:

do_four(lambda: print_twice(str))

Or you can use the higher-order function partial to do it for you:

from functools import partial
do_four(partial(print_twice, str))

The documentation for partial has a pretty nice explanation:

The partial() is used for partial function application which “freezes” some portion of a function's arguments and/or keywords resulting in a new object with a simplified signature. For example, partial() can be used to create a callable that behaves like the int() function where the base argument defaults to two: [snip] basetwo = partial(int, base=2)

---

_{* If you're thinking "But I didn't return anything, so where does that None come from?": Every function always returns a value in Python. If you don't tell it what to return, it returns None .}

Answer 2

The line do_four(print_twice(str)) evaluates the expression in the brackets first before passing it. Since print_twice doesn't return anything, None is assumed, and that gets passed.

Answer 3

Right now print_twice is returning None which is what ends up being passed to do_four as a parameter. In other words, you are passing the result of the function call instead of the function call itself.

Instead you want to wrap that function call in a lamda function like this:

do_four(lambda: print_twice(str))

This will pass the actual function call as a parameter instead of calling the function and passing its result.