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

submitted 9 months ago by [email protected] to c/[email protected]

30 comments fedilink hide all child comments

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 &lt;= 0)
    {
        return;
    }

    draw(n - 1);

    for (int i = 0; i &lt; n; i++)
    {
        printf("#");
    }
    printf("\n");
}

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[–] [email protected] 7 points 9 months ago* (last edited 9 months ago) (6 children)

This code has a recursive call (function calls itself) within the function so that has to be taken into account when tracing it

This would make the function execute multiple times so the for loop would end up executing multiple times.

Lets say main calls draw with a height value of 10 (draw(10)). First it sees that n is greater than 0 so it keeps going. Then it calls the draw function with a value of 10 - 1 aka 9. Now its executing in the draw(9) function. Greater than 0 so continues and calls draw(8). etc. all the way down to draw(0) where it sees that n is equal to 0 so returns out of the function due to the return statement.

Now that draw(0) finished executing draw(1) can keep going and goes to the for loop. Here it prints 1 # and then prints a new line (and then returns since it hit the end of the function). Now that draw(1) is done draw(2) can keep going and prints 2 #'s and then prints a new line (and then returns). This keeps going all the way up to the initial draw call, draw(10) which prints 10 #'s and then a new line, returns, and then the main function keeps going but theres nothing after that so it returns and the execution ends.

The effect from coming back after the recursive calls makes it seem like n is increasing but its just different calls to the same function. i is taken into account for but printing the amount of #'s since thats whats within that loop

[–] [email protected] 2 points 9 months ago (5 children)

Why does the for loop return when it hits the end of the function? Isn't the recursive portion already completed in draw(n - 1)? The rest of it is just normal non-recursive code if I understand it correctly.

[–] [email protected] 13 points 9 months ago* (last edited 9 months ago) (1 children)

No, the moment that draw(9) is called, draw(10) goes on pause while draw(9) runs, which pauses when it calls draw(8) … which repeats (or recurses) until draw(0) gets called. Then it returns which returns to draw(1). The draw(1) un-pauses and does the #\n bit and returns to draw(2), which un-pauses and does ##\n and so forth until draw(10) does ##########\n

[–] [email protected] 1 points 9 months ago

Gotcha. Thanks for the explanation.

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