32
you are viewing a single comment's thread
view the rest of the comments
view the rest of the comments
this post was submitted on 16 Dec 2023
32 points (97.1% liked)
Programming
16968 readers
293 users here now
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
Wormhole
Follow the wormhole through a path of communities !webdev@programming.dev
founded 1 year ago
MODERATORS
There's the new line after the for loop to make sure that the next recursion starts on a fresh line. Otherwise the next recursion would print on the same line, right where it left off, and you'd just have a line of "#"'s. The for loop itself is just for printing "#"'s.
I think this is a confusion with the recursion. Look at the line with
draw(n - 1);
this happens before any printing of hashes happens, and only continues after its done. And it calls itself as long as it's not less than or equal to 0. To psuedo-code the order of operations here:so n is never incremented as you think, it just calls decremented versions of the draw function before the current one finishes. The i's are purely involved in the for loop, which only prints hashes. Does that make sense?
Yes - I finally caught that part about n as it's just moving in reverse so it gets decremented. Now I'm not sure about i. In the debugger when the program gets to the for loop both n and i are equal to 1. The n I understand but i?
i starts at 0, checks if i is less than n (the first time it will be, no matter what), prints a "#", then increments i by 1 and loops