Python-3.2 协程 : AttributeError: 'generator' object has no attribute 'next'

Question

这个问题在这里已经有了答案:

there's no next() function in a yield generator in python 3 (2 个回答)

关闭 4个月前。

引自 Python Essential Reference，David Beazley，第 20 页:

Normally, functions operate on a single set of input arguments. However, a function can also be written to operate as a task that processes a sequence of inputs sent to it. This type of function is known as a coroutine and is created by using the yield statement as an expression (yield) as shown in this example:

def print_matches(matchtext):
    print "Looking for", matchtext
    while True:
        line = (yield)       # Get a line of text
        if matchtext in line:
            print line

To use this function, you first call it, advance it to the first (yield), and then start sending data to it using send(). For example:

>>> matcher = print_matches("python")
>>> matcher.next() # Advance to the first (yield)
Looking for python
>>> matcher.send("Hello World")
>>> matcher.send("python is cool")
python is cool
>>> matcher.send("yow!")
>>> matcher.close() # Done with the matcher function call

根据这些信息，我编写了以下代码:

#!/usr/bin/python3.2
import sys

def match_text(pattern):
    line = (yield)
    if pattern in line:
        print(line)

x = match_text('apple')
x.next()

for line in input('>>>> '):
    if x.send(line):
        print(line)

x.close()

但我收到如下错误消息:

Traceback (most recent call last):
  File "xxx", line 9, in <module>
    matcher.next() # Advance to the first (yield)
AttributeError: 'generator' object has no attribute 'next'

为什么这段代码(或者书中的代码，就此而言)在 Python 3.2 中不起作用？似乎应该是协程的东西被视为生成器 - 为什么？这是怎么回事？

Answer 1

你被错误信息吓跑了；在类型方面，Python 没有区别——你可以 .send 到任何使用 yield 的东西，即使它在内部没有对发送的值做任何事情。

在 3.x 中，不再附加 .next 方法；相反，使用内置的自由函数 next:

next(matcher)