milonDec 16, 2023

Why does the for loop repeat in this recursion?

https://lemm.ee/post/18216408

Why does the for loop repeat in this recursion? - lemm.ee

I used the debugger to examine this code but not understanding a couple areas. 1. Why does the for loop repeat after it exits to print a new line? If it exits the loop, shouldn’t it be done with it? 2. Why is n incremented and not i as stated with i++? ```void draw(int n); int main(void) { int height = get_int(“Height: “); draw(height); } void draw(int n) { if (n <= 0) { return; } draw(n - 1); for (int i = 0; i < n; i++) { printf(”#”); } printf(“\n”); }

Show threadxmunk

So you’re drawing a triangle of # that starts skinny in the top left and grows to the right with a height and width of n?

As an example, Height 3 would be:

# ## ###

I’m not following your questions, but the code seems pretty reasonable.

Dec 16, 2023 at 11:37pmWeb
Show threadmilonDec 16, 2023
It’s supposed to be a pyramid but not my code. It’s an example of a recursive function from a CS50 lecture and I’m just trying to understand how the code works line by line.
Show threadxmunkDec 16, 2023
So is my example of Height 3 in line with your expectations?
Show threadmilonDec 16, 2023
Yep
Show threadxmunkDec 16, 2023
I wrote an equivalent version just using nested loops - reading it might help you understand why the recursion works the way it does.
Show threadmilonDec 16, 2023
Thanks. I did see that. I have a general understanding of how recursion works I think where the function calls itself again and again but I don’t get why the code (for loop) below the draw(n - 1) is recursive.
Show threadxmunkDec 17, 2023

The code below the draw(n - 1) isn’t recursive… the call to draw(n - 1) is the recursion.

Sometimes, it can be helpful to invert recursion. Think about what draw(0) would be and write it down… then compute draw(1) using the value you previously computed for draw(0).

Show threadxmunkDec 16, 2023

An example using a nested loop

$n = 3; for ($i = 0; $i &lt;= $n; $i++) { for ($j = 0; $j &lt; $i; $j++) { echo '#'; } echo "\n"; }

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