Re: Recursive generator in Python 3.5

[tpqnnd01]

Thanks for your detail explanation, my problem may be the 'return' keyword. I 
confuse at the point that return come after yield keyword, I wonder what if in 
case this code do not have the 1st return. So, when i try to delete the 1st 
return, the code run with error (the list out of range).

This is actual clear from your example above, when the 1st return is delected, 
the statement flow shall be from (yield '') to (yield i + tmp) when the Next() 
is called. This behaviour shall reach to a statement is: for i in [][0] that 
shall raise an error: list index out of range.
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Re: Recursive generator in Python 3.5

[tpqnnd01]

Hi Chris,

Please see the exampl I just got it below, this is the Tower of Hanoi with 
recursive generator but it do not have the 'return' here, do you have the 
advice for this term:

def A001511():
yield 1
b=1
for x in A001511():
yield x+1
yield 1


trial=A001511()
for i in range(10):
a=next(trial)
print(a)
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Re: Recursive generator in Python 3.5

[tpqnnd01]

Hi ChrisA,

Thank you so much for sharing this point, I just concern to the point that: 
normally, I found that the generator (it may be recursive generator also) do 
not apply the "return" keyword, just the yield only. But when I delete one or 
both of "return" keyword above, the code do not action. What is the concern 
here, could you share more experience on that,
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Recursive generator in Python 3.5

[tpqnnd01]

I have some confuse about the recursive generator where the code mixing Yield 
and return keywork as below. I understand that "return" keywork just raise 
StopIteration exception but can not understanding in depth, so, could some one 
explain me about the actual function of two "return" keyworks in case it have 
other function and mechanism of for running this code segment below:


def fg(args):
  if not args:
yield ""
return
  for i in args[0]:
for tmp in fg(args[1:]):
  yield i + tmp
  return
print(list(fg(['abc', 'xyz', '123'])))

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