Python equivalent of call/cc?

[The Pythonista]

Yesterday, I was hacking around a bit, trying to figure out how to
implement the semantics of call/cc in Python. Specifically, I wanted to
translate this Scheme code to equivalent Python:

####

(define theContinuation #f)

(define (test)
(let ((i 0))
(call/cc (lambda (k) (set! theContinuation k)))
(set! i (+ i 1))
i))

(test)
(theContinuatio n)
(theContinuatio n)

####

Incidentally, those last three lines evaluate to 1, 2, and 3,
respectively.

The best Python translation I could come up with was something like this
(targeted at Python 2.5):

####
import inspect

theContinuation = None

def call_cc (f):
f (inspect.curren tframe().f_back )

def invoke (c):
exec c.f_code in c.f_globals, c.f_locals

def test():
i = 0
call_cc (lambda k: globals().updat e({'theContinua tion' : k }))
i = i + 1
print i

test()
invoke (theContinuatio n)
invoke (theContinuatio n)

####

Now, this code is wrong on a number of levels [I am indeed aware of
exactly how ugly that lambda is...], but, in particular, my
continuations / stack frames don't seem to be resuming at the right
point. I'd expect invoke (theContinuatio n) to restart on the line
immediately following call_cc, but it does not. Not surprisingly, the
output is also wrong. (I get 1, 1, and 1 rather than 1, 2, and 3.)

Can anyone help me by, perhaps pointing out some silly error I made?
Failing that, can someone show me a working implementation of call/cc
(preferably based on some form of stack inspection)?

Thanks!
--
code.py: A blog about life, the universe, and Python

http://pythonista.wordpress.com

** Posted from http://www.teranews.com **

[Larry Bates]

Re: Python equivalent of call/cc?

The Pythonista wrote:

Yesterday, I was hacking around a bit, trying to figure out how to
implement the semantics of call/cc in Python. Specifically, I wanted to
translate this Scheme code to equivalent Python:
>
####
>
(define theContinuation #f)
>
(define (test)
(let ((i 0))
(call/cc (lambda (k) (set! theContinuation k)))
(set! i (+ i 1))
i))
>
(test)
(theContinuatio n)
(theContinuatio n)
>
####
>
Incidentally, those last three lines evaluate to 1, 2, and 3,
respectively.
>
The best Python translation I could come up with was something like this
(targeted at Python 2.5):
>
####
import inspect
>
theContinuation = None
>
def call_cc (f):
f (inspect.curren tframe().f_back )
>
def invoke (c):
exec c.f_code in c.f_globals, c.f_locals
>
def test():
i = 0
call_cc (lambda k: globals().updat e({'theContinua tion' : k }))
i = i + 1
print i
>
test()
invoke (theContinuatio n)
invoke (theContinuatio n)
>
####
>
Now, this code is wrong on a number of levels [I am indeed aware of
exactly how ugly that lambda is...], but, in particular, my
continuations / stack frames don't seem to be resuming at the right
point. I'd expect invoke (theContinuatio n) to restart on the line
immediately following call_cc, but it does not. Not surprisingly, the
output is also wrong. (I get 1, 1, and 1 rather than 1, 2, and 3.)
>
Can anyone help me by, perhaps pointing out some silly error I made?
Failing that, can someone show me a working implementation of call/cc
(preferably based on some form of stack inspection)?
>
Thanks!

While I know absolutely no scheme what you show here is a generator in Python.

def theContinuation ():
n = 0
while 1:
n += 1
yield n


invoke = theContinuation ()

print
print invoke.next()
print invoke.next()
print invoke.next()

Maybe that is what you are looking for?

-Larry

[Aahz]

Re: Python equivalent of call/cc?

In article <be112$4874d38a $18817@news.ter anews.com>,
The Pythonista <none@this.time wrote:

>
>Yesterday, I was hacking around a bit, trying to figure out how to
>implement the semantics of call/cc in Python. Specifically, I wanted to
>translate this Scheme code to equivalent Python:
>
>####
>
>(define theContinuation #f)
>
(define (test)
(let ((i 0))
(call/cc (lambda (k) (set! theContinuation k)))
(set! i (+ i 1))
i))
>
(test)
(theContinuatio n)
(theContinuatio n)
>

Python relies on mutables to do this:

class holder: pass

def call():
x = holder()
x.counter = 0
def cc():
x.counter += 1
return x.counter
return cc

foo = call()

print foo()
print foo()
print foo()
--
Aahz (aahz@pythoncra ft.com) <* http://www.pythoncraft.com/

"as long as we like the same operating system, things are cool." --piranha