You are looking at a recursive method, as you can see with the line draw(n-1) inside the draw(n) method. 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).
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.
You should keep in mind that everytime a draw(n-1) is called, the current method is “paused” until this call return.
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.
You are not logged in. However you can subscribe from another Fediverse account, for example Lemmy or Mastodon. To do this, paste the following into the search field of your instance: !programming@programming.dev
Welcome to the main community in programming.dev! Feel free to post anything relating to programming here!
Cross posting is strongly encouraged in the instance. If you feel your post or another person’s post makes sense in another community cross post into it.
Hope you enjoy the instance!
Rules
Rules
Follow the programming.dev instance rules
Keep content related to programming in some way
If you’re posting long videos try to add in some form of tldr for those who don’t want to watch videos
You are looking at a recursive method, as you can see with the line draw(n-1) inside the draw(n) method. 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).
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 #.
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.
You should keep in mind that everytime a draw(n-1) is called, the current method is “paused” until this call return.
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().
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 #.