Python generator yield behaviour

Question

So I have the following generator function:

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

for i in gen(3):
    print(i)

The result:

3
None
None

I understand the first result of yield is 3. Because I assigned 3 to function argument n. But where are the None in the second and third yield coming from? Is it because in the for-loop, yield n returns None and this None is assigned to n in this line: n = yield n?

What have you gathered from the generator and `yield` statement documentation? — S3DEV, Nov 25 '22 at 12:27

score 3 · Accepted Answer · answered Nov 25 '22 at 12:40

This is explained in the documentation of yield expressions, especially this part:

The value of the yield expression after resuming depends on the method which resumed the execution. If next() is used (typically via either a for or the next() builtin) then the result is None. Otherwise, if send() is used, then the result will be the value passed in to that method.

As you use a for loop, n just gets None as a value when resuming after the first yield. So, from now on, n is None, and this is what will be yielded the last two times.

Thanks. I understand now. It's interesting `send()` is designed to change the result value of the yield expression. I thought `send()` was to insert a new value in the sequence being yielded. — CyberPlayerOne, Nov 25 '22 at 13:07

score 2 · Answer 2 · answered Nov 25 '22 at 12:30

2

It seems to me that you answered your own question:

because in the for-loop, yield n returns None and None is assigned to n in this line: n = yield n

You can read more at this answer.

answered Nov 25 '22 at 12:30

msailor

340
1
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Python generator yield behaviour

The result:

2 Answers2