Comment on Why does the for loop repeat in this recursion?
Merwyn@sh.itjust.works 9 months ago
You are looking at a recursive method, as you can see with the line draw(n-1). You can search for “recursive function” on internet for a better understanding.
Basically, the method draw is called a first time n = a user input, but then this method call itself with n-1 until it reach 0. So you can think as if function draw(6) will call draw(5) and wait for it to return before continuing, draw(5) call draw(4), ect until draw(0) that return immediately.
So then the order of execution will be draw(1) that print " #\n" and return, then draw(2) will proceed to print “##\n” and return, then draw(3), ect until draw(n).
milon@lemm.ee 9 months ago
Right. I was aware it was recursion as stated in the title of my post. I had two questions specific to where the for loop returns after printing #.
Merwyn@sh.itjust.works 9 months ago
Yes, as I wrote when the method draw(n=1) finish the for loop that print one “#”, this call of the method draw return. Then the process start again from the after the line draw(n-1) of the method draw(n=2), which execute the for loop to print “##” and return. Then again you come back to after the line draw(n-1) of inside the method draw(n=3), ect.
milon@lemm.ee 9 months ago
I see. I guess my understanding was that the recursion was over after the recursive call, but it’s actually for all the code in draw().
Merwyn@sh.itjust.works 9 months ago
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.
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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).
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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.
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Then it reach the second “return” and proceed again until the whole program is over.