I used the debugger to examine this code but not understanding a couple areas.
- 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?
- Why is n incremented and not i as stated with i++?
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");
}
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 #.