Yes, to better understand this you have to understand the “flow” of the program. Meaning the order at which the instructions are executed and not written.
Here you have the flow of the program starting from n =3 until the recursion reach draw(0), note that none of the for loop have been executed yet. At this point it reach the first “return” instruction and go finish the call to draw(0).
Then the flow go back to where it previously was: inside the draw(1) call just after the line calling draw(0). And it start executing the next lines of the draw(1): the for loop.
Then it reach the second “return” and proceed again until the whole program is over.
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Yes, to better understand this you have to understand the “flow” of the program. Meaning the order at which the instructions are executed and not written.
Here you have the flow of the program starting from n =3 until the recursion reach draw(0), note that none of the for loop have been executed yet. At this point it reach the first “return” instruction and go finish the call to draw(0).
Then the flow go back to where it previously was: inside the draw(1) call just after the line calling draw(0). And it start executing the next lines of the draw(1): the for loop.
Then it reach the second “return” and proceed again until the whole program is over.
Yes, that helps. Thanks. I see now how n goes from 1 to 2 to 3…etc. Now not so sure how i = 1 when the for loop starts.
When called with n=1 ? It’s from i=0 to i<1, so it will do only one iteration with i=0 and print one #.