Python Generator: confusing result

Question

I'm playing around with generators to better understand how they work, but I'm confused with the result of the following piece of code:

>>> def gen():
...  for i in range(5):
...   yield i
...
>>> count=gen()
>>> for i in count:
...  print count.next()
...
1
3
Traceback (most recent call last):
  File "<stdin>", line 2, in <module>
StopIteration
>>>

What's going on here? Looks like as it hits the "for i in count:" line the generator yields the first value. I'm not sure.

EDIT: I should add that I'm not trying to "get it right". I'm trying to break things to see how Python reacts. I learn more about the language when I generate errors. This error had me stump but it's all clear now. So far all the answers have been short and sweet. Thank you every one!

Answer 1

Because "for" cycle already iterate your generator values, at every step two values are yielded: one in "for" and one in body of cycle.

Answer 2

Your code should be like this

Code:

def gen():
    for i in range(5):
        yield i

count=gen()
for i in count:
    print i

output:

0
1
2
3
4

Notes:

When you loop for i in count you trying to get the next item again here count.next() you trying to get the second element .So basically you are trying to get element +2 from current position

So what happens in you code is are as below:

steps :

1.for i in count here the next value is got which is0

2.print count.next() here again next value is called and printed so 1 is printed

3.for i in count here again next value is called which is 2

4.print count.next() here again next value is called and printed so 3 is printed

5.for i in count here again next value is called which is 4

6.finally here you are calling print count.next() since there is no value in the generator .You have been give the exception StopIteration

Answer 3

for i in count: is already calling next (and assigns the value to i). Your code is then calling next again.

If you just want to print each value

for i in count:
    print i

or the long way round

while True:
    try:
        print count.next()
    except StopIteration, si:
        break