0

This question already has an answer here:

My function:

int num(int x) 
{ 
   if(x>0) 
     num(x-1); 
   else 
     return 0; 
   printf("%d",x); 
}

This function takes a input x and prints the numbers from 1 upto x.I can't seem to understand how this works, can someone please elaborate. Thanks in advance!

marked as duplicate by user40980, Dan Pichelman, World Engineer Feb 24 '15 at 0:17

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • I assume it wasn't you who wrote the function. If I call int x = num (10), there is undefined behaviour. The return type should be void, and it shouldn't return anything in the else part. Awful code. – gnasher729 Feb 23 '15 at 8:47
  • @gnasher729 I originally wrote this code(with some modification) to find the multiplication table of a number n. It worked fine for my purpose but I didn't know the code blows(since I'm in the early learning phase of recursion). Thanks – user2829775 Feb 23 '15 at 14:10
  • Not related to the question, but recursion isn't the right tool for what you're using it for. A loop will handle an arbitrarily-large number without growing the stack on each iteration. – Blrfl Feb 23 '15 at 14:18
  • Thanks,I know,I just wanted to flex my "recursion muscles"since I understand iteration properly. – user2829775 Feb 23 '15 at 14:27
4

To solve this type of thing, it's often best to take a pen and paper and "pretend to be the computer running the code".

So, if we have a call of num(4), it will lead to:

if (4 > 0)    // Yupp, 4 > 0 
  num(3);     // x - 1 = 3
  if (3 > 0)  // Yupp
    num(2)
    if (2 > 0)
      num(1)
      if (1 > 0)
        num(0)
        if (0 > 0) // Nope
        goto else-part:
        return 0;
      skip else-part
      print(1);
    skip else-part
    print(2);
  skip else-part
  print(3)
  skip else-part
print(4)
  • @Malt It's confusing because of I place the print statement inside if statement inside braces just before the function call, my output is from x to 1. But when I place the print statement inside if statement inside braces just after the function call, my output is from 1 to x. Can you please elaborate the functioning? Thanks. – user2829775 Feb 23 '15 at 14:59
  • Follow the logic. If you put the print at the beginning of the function, the calls to num(3), num(2) etc happens after printing, so obviously, values printed will be "high to low". If you place it after the call to num - either in the if or at the end, it will print in "low to high" order. – Mats Petersson Feb 23 '15 at 21:58
3

in my opinion, the easiest way of wrapping your head around recursion is starting from the end.

In every recursive function, there's a stopping condition. In your case it's else return 0; so the method will return 0 if the argument is zero or lower. That's simple. Now let's go backwards.

We know that num(0) returns 0 without doing anything else. So now we can now figure out what num(1) is. num(1) will call num(0) (which will return 0 and do nothing else), and then print x which is 1. Now we can backtrack and look at num(2). num(2) will call num(1), which we already covered, and then prints 2, so we can look at num(3) etc.

So for every number x, num(x) will simply print all numbers from 1 to x.

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