Python list iterator behavior and next(iterator)

Question

What you see is the interpreter echoing back the return value of next() in addition to i being printed each iteration:

>>> a = iter(list(range(10)))
>>> for i in a:
...    print(i)
...    next(a)
... 
0
1
2
3
4
5
6
7
8
9

So 0 is the output of print(i), 1 the return value from next(), echoed by the interactive interpreter, etc. There are just 5 iterations, each iteration resulting in 2 lines being written to the terminal.

If you assign the output of next() things work as expected:

>>> a = iter(list(range(10)))
>>> for i in a:
...    print(i)
...    _ = next(a)
... 
0
2
4
6
8

or print extra information to differentiate the print() output from the interactive interpreter echo:

>>> a = iter(list(range(10)))
>>> for i in a:
...    print('Printing: {}'.format(i))
...    next(a)
... 
Printing: 0
1
Printing: 2
3
Printing: 4
5
Printing: 6
7
Printing: 8
9

In other words, next() is working as expected, but because it returns the next value from the iterator, echoed by the interactive interpreter, you are led to believe that the loop has its own iterator copy somehow.

What is happening is that next(a) returns the next value of a, which is printed to the console because it is not affected.

What you can do is affect a variable with this value:

>>> a = iter(list(range(10)))
>>> for i in a:
...    print(i)
...    b=next(a)
...
0
2
4
6
8

I find the existing answers a little confusing, because they only indirectly indicate the essential mystifying thing in the code example: both* the "print i" and the "next(a)" are causing their results to be printed.

Since they're printing alternating elements of the original sequence, and it's unexpected that the "next(a)" statement is printing, it appears as if the "print i" statement is printing all the values.

In that light, it becomes more clear that assigning the result of "next(a)" to a variable inhibits the printing of its' result, so that just the alternate values that the "i" loop variable are printed. Similarly, making the "print" statement emit something more distinctive disambiguates it, as well.

(One of the existing answers refutes the others because that answer is having the example code evaluated as a block, so that the interpreter is not reporting the intermediate values for "next(a)".)

The beguiling thing in answering questions, in general, is being explicit about what is obvious once you know the answer. It can be elusive. Likewise critiquing answers once you understand them. It's interesting...

For those who still do not understand.

>>> a = iter(list(range(10)))
>>> for i in a:
...    print(i)
...    next(a)
... 
0 # print(i) printed this
1 # next(a) printed this
2 # print(i) printed this
3 # next(a) printed this
4 # print(i) printed this
5 # next(a) printed this
6 # print(i) printed this
7 # next(a) printed this
8 # print(i) printed this
9 # next(a) printed this

As others have already said, next increases the iterator by 1 as expected. Assigning its returned value to a variable doesn't magically changes its behaviour.

Something is wrong with your Python/Computer.

a = iter(list(range(10)))
for i in a:
   print(i)
   next(a)

>>> 
0
2
4
6
8

Works like expected.

Tested in Python 2.7 and in Python 3+ . Works properly in both

It behaves the way you want if called as a function:

>>> def test():
...     a = iter(list(range(10)))
...     for i in a:
...         print(i)
...         next(a)
... 
>>> test()
0
2
4
6
8