python中是否可以实现'tail-call elimination'的优化

liyeho

今天看 python in nutshell 时看到了这段
Reader who are familiar Lisp, Scheme, or functional-programming languages must in particular be aware that Python does not implement the optimization of "tail-call elimination," which is so important in these languages. In Python, any call, recursive or not, has the same cost in terms of both time and memory space, dependent only on the number of arguments: the cost does not change, whether the call is a "tail-call" (meaning that the call is the last operation that the caller executes) or any other, nontail call.
说python中不能实现optimization of "tail-call elimination,"可是我google中搜索了一下发现了这个
http://aspn.activestate.com/ASPN/Cookbook/Python/Recipe/474088
可是由于不太了解functional-programming 　所以the optimization of "tail-call　elimination"　也不知道确切的意思，所以也不知道下面的代码写的对不对。
下面的代码是上面的链接里面的那个人写的.

#!/usr/bin/env python2.4
# This program shows off a python decorator(
# which implements tail call optimization. It
# does this by throwing an exception if it is
# it's own grandparent, and catching such
# exceptions to recall the stack.
import sys
class TailRecurseException:
def __init__(self, args, kwargs):
self.args = args
self.kwargs = kwargs
def tail_call_optimized(g):
"""
This function decorates a function with tail call
optimization. It does this by throwing an exception
if it is it's own grandparent, and catching such
exceptions to fake the tail call optimization.
This function fails if the decorated
function recurses in a non-tail context.
"""
def func(*args, **kwargs):
f = sys._getframe()
if f.f_back and f.f_back.f_back \
and f.f_back.f_back.f_code == f.f_code:
raise TailRecurseException(args, kwargs)
else:
while 1:
try:
return g(*args, **kwargs)
except TailRecurseException, e:
args = e.args
kwargs = e.kwargs
func.__doc__ = g.__doc__
return func
@tail_call_optimized
def factorial(n, acc=1):
"calculate a factorial"
if n == 0:
return acc
return factorial(n-1, n*acc)
print factorial(10000)
# prints a big, big number,
# but doesn't hit the recursion limit.
@tail_call_optimized
def fib(i, current = 0, next = 1):
if i == 0:
return current
else:
return fib(i - 1, next, current + next)
print fib(10000)
# also prints a big number,
# but doesn't hit the recursion limit.