Python 协程的无限递归的问题

Question

hi, 最近看了关于Python协程的相关文章协程的简单理解，说协程可以无限递归，于是想写一个协程示例练练，于是：

import time


def ping():
    print 'ping 0'
    count = 1
    try:
        pi,po = yield s
        print 'ping ' + str(count)
        count += 1     
        time.sleep(1)
        po.send([po, pi])
    except StopIteration:
        pass


def pong():
    print 'pong 0'
    count = 1
    try:
        po, pi = yield s
        print 'pong ' + str(count)
        count += 1
        time.sleep(1)
        pi.send([pi,po])
    except StopIteration:
        pass


s = ping()
r = pong()
s.next()
r.next()
s.send([s,r])

运行结果是：

ping 0
pong 0
ping 1
pong 1
Traceback (most recent call last):
  File "D:\test\coroutine.py", line 34, in <module>
    s.send([s,r])
  File "D:\test\coroutine.py", line 12, in ping
    po.send([po, pi])
  File "D:\test\coroutine.py", line 25, in pong
    pi.send([pi,po])
ValueError: generator already executing

那篇文章使用了stackless，我想实现一个原始的方法。但是出错，不知道要实现无限递归的话，应该怎么写嘞？

Answer 1

你需要的是用一個 run 函數來管理協程。
在 ping 和 pong 函數內，yield 傳回 run。
在 run 函數內，以 next() 對協程進行調度。

import time

def ping():
    print 'ping 0'
    count = 1
    while 1:
        yield po
        print 'ping ' + str(count)
        count += 1
        time.sleep(1)

def pong():
    print 'pong 0'
    count = 1
    while 1:
        yield pi
        print 'pong ' + str(count)
        count += 1
        time.sleep(1)

def run(co):
    while 1:
        co = co.next() # 这行实现 ping 和 pong 交替运行

pi = ping()
po = pong()
pi.next()
po.next()
run(pi)

運行結果：

ping 0
pong 0
ping 1
pong 1
ping 2
pong 2
ping 3
pong 3
ping 4
pong 4

pong 3

ping 4

pong 4
.........

================================================= =======

帶訊息傳遞版：

import time

def ping():
    print 'ping 0'
    count = 1
    while 1:
        msg = yield (po, 'msg from ping')
        print 'ping get', msg
        print 'ping ' + str(count)
        count += 1
        time.sleep(1)

def pong():
    print 'pong 0'
    count = 1
    while 1:
        msg = yield (pi, 'msg from pong')
        print 'pong get', msg
        print 'pong ' + str(count)
        count += 1
        time.sleep(1)

def run(cur, msg='msg from run'):
    while 1:
        cur, msg = cur.send(msg)

pi = ping()
po = pong()
pi.next()
po.next()
run(pi)

運行結果：

ping 0
pong 0
ping get msg from run
ping 1
pong get msg from ping
pong 1
ping get msg from pong
p 2
pong 1
ping 3
pong get msg from ping
pong 3
ping get msg from pong
ping 4
..............

Answer 2

你寫的確實是有問題，不過最主要的是，Stackless是一個修改版的Python，它的coroutine和標準CPython裡的貌似不太一樣，如果你想要在CPython裡面實現類似的功能，可能需要基於greenlet來做。

Answer 3

http://anandology.com/blog/using-iterators-and-generators/
應該是把generator自身傳遞造成了問題，send自己導致的
要是pi po是全域變數應該不會有問題

Answer 4

生成器的用法明顯錯誤，s 沒有被 yield 掛起 ， 你再去 send/next 當然會報這個錯誤
不知道你具體是要實現什麼， 如果是要 “ping pong” 交替打印， 用一個generator 就行了