Python相当于Ruby的延续

Question

What is the Python equivalent of the following code in Ruby?

def loop
  cont=nil
  for i in 1..4
    puts i
    callcc {|continuation| cont=continuation} if i==2
  end
  return cont
end

> c=loop
1
2
3
4
> c.call
3
4

Reference: Secrets of lightweight development success, Part 9: Continuations-based frameworks

Answer 1

您引用的文章在参考资料部分中包含指向Continuations Made Simple和I​​llustrated的链接，该部分讨论了Python语言中的延续。

Answer 2

take a look at the yield statement to make generators.

I don't speak any ruby, but it seems like you're looking for this:

def loop():
    for i in xrange(1,5):
        print i
        if i == 2:
            yield


for i in loop():
    print "pass"

Edit: I realize this is basically a specialization of real continuations, but it should be sufficient for most purposes. Use yield to return the continuation and the .next() message on the generator (returned by just calling loop() ) to reenter.

Answer 3

Using generator_tools (to install: ' $ easy_install generator_tools '):

from generator_tools import copy_generator

def _callg(generator, generator_copy=None):
    for _ in generator: # run to the end
        pass
    if generator_copy is not None:
        return lambda: _callg(copy_generator(generator_copy))

def loop(c):
    c.next() # advance to yield's expression
    return _callg(c, copy_generator(c))

if __name__ == '__main__':
    def loop_gen():
        i = 1
        while i <= 4:
            print i
            if i == 2:
                yield
            i += 1

    c = loop(loop_gen())
    print("c:", c)
    for _ in range(2):
        print("c():", c())

Output:

1
2
3
4
('c:', <function <lambda> at 0x00A9AC70>)
3
4
('c():', None)
3
4
('c():', None)

Answer 4

There are many weak workarounds which work in special cases (see other answers to this question), but there is no Python language construct which is equivalent to callcc or which can be used to build something equivalent to callcc .

You may want to try Stackless Python or the greenlet Python extension, both of which provide coroutines, based on which it is possible to build one-shot continutations, but that's still weaker than Ruby's callcc (which provides full continuations).

Answer 5

def loop():    
    def f(i, cont=[None]):        
        for i in range(i, 5):
            print i
            if i == 2:
                cont[0] = lambda i=i+1: f(i)
        return cont[0]
    return f(1)

if __name__ == '__main__':
    c = loop()
    c()