或更广泛的问题:如何使一个递归函数的蟒蛇,并更名时,仅需要在声明改变了吗?
我发现了一个简单的,工作液。
from functools import wraps
def recfun(f):
@wraps(f)
def _f(*a, **kwa): return f(_f, *a, **kwa)
return _f
@recfun
# it's a decorator, so a separate class+method don't need to be defined
# for each function and the class does not need to be instantiated,
# as with Alex Hall's answer
def fact(self, n):
if n > 0:
return n * self(n-1) # doesn't need to be self(self, n-1),
# as with lkraider's answer
else:
return 1
print(fact(10)) # works, as opposed to dursk's answer
下面是一个(未经测试)的想法:
class Foo(object):
def __call__(self, *args):
# do stuff
self(*other_args)
可以将功能结合到本身,因此它接收到自身的引用作为第一个参数,就好像在一个绑定方法self
:
def bind(f):
"""Decorate function `f` to pass a reference to the function
as the first argument"""
return f.__get__(f, type(f))
@bind
def foo(self, x):
"This is a bound function!"
print(self, x)
我不知道为什么你会想这样做,但尽管如此,你可以使用一个decorator实现这一目标。
def recursive_function(func):
def decorator(*args, **kwargs):
return func(*args, my_func=func, **kwargs):
return decorator
然后你的函数将如下所示:
@recursive_function
def my_recursive_function(my_func=None):
...