how to make my function be a generator

[Skaperen]

it was my understanding that a function with a yield statement would be a generator. but, what if the function i want to be the generator is really calling another function that does the actual yield? how would it be known that the 1st one is to be a generator (when there is no yield statement coded in it)? how would it be known that the 2nd one is to not be a generator (even though it does have a yield statement to do yields on behalf of its caller)?

[jefsummers]

If I understand correctly -
fnA calls fnB which is a generator. It returns a value to fnA. fnA should then yield that value back to whoever called fnA. It is then a generator.
Missing something?

[Skaperen]

pgmX calls fnA expecting a generator to be returned. i'll identify that generator as genA/ pgmX iterates genA. fnA runs and calls fnB which produces many items on order that genA should yield. fnB cannot invoke yield for fnA as part of genA because Python considers fnB to be a generator since fnB has the yield statement. calling fnB would create genB. so fnA cannot have fnB do yield for it. instead, fnB needs to be implemented to return items to let fnA yield them.

a common reason to create and call a function is having the same big code in many places. you implement a function with that big code and call the function in those many places, instead of many copies that same code. but if that big code includes a yield statement your expectations will not be met unless you realize this turns the function into a generator. also, if the code calling that function is itself a function whose callers expect a generator, moving all its yield statements into another function causes it to no longer be a function.