newbie function question

Roads (20)

In the tut example below I cannot understand why "a" is reduced by 1 everytime "factorial" is called. It looks like to me if "a" was originally 5 then it would always be 5 because I don't see anything that sets a--.

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#include <iostream>
using namespace std;

long factorial (long a)
{
  if (a > 1)
   return (a * factorial (a-1));
  else
   return (1);
}

int main ()
{
  long number;
  cout << "Please type a number: ";
  cin >> number;
  cout << number << "! = " << factorial (number);
  return 0;
}

Hope my question makes sense.

Zhuge (102)

factorial is a recursive function so it calls itself with the parameter (a - 1). So if you call factorial(5) you actually end up calling:

1 2 3 4 5

factorial(5) // 5 * factorial(4)
factorial(4) // 4 * factorial(3)
factorial(3) // 3 * factorial(2)
factorial(2) // 2 * factorial(1)
factorial(1) // 1

Roads (20)

Yes Zhuge, I understood that. At any rate I think I understand what is going on here. Apparently when this:


(a-1)

is stated, it reassigns the value of "a" to 4. Can I think of it this way? I know that sounds dumb but I was thinking of this in terms of assignment.

firedraco (651)

It's because it doesn't actually do that...If you are using a debugger and stepping through it, then the 'a' you are talking about is the parameter a, not the actual 'a'.

Roads (20)

Fellas I'm sorry I keep hounding this but I've gotten over to structs and realized I'm pretty much lost. So I'm wading back thru the tuts trying to get a better understanding of what's happening.

I kinda, sorta understand what you are saying. Maybe I need to set CodeBlocks up so I can use the debugger. I know it is returning (a-1) every time, It is just not really clear to me from the code, why that is happening. Anyway, both of you thanks.

firedraco (651)

Well, it isn't returning a-1. It's returning:

a * factorial (a - 1) which if you continue to expand it becomes:

a* (a - 1) * (a - 2) * (a - 3) ... since it is calling itself...until it reaches a = 1, then all the functions start returning until you get the answer:

f# is what the function is returning
f1(a * f2(a-1 * f3(a -2)))

Sorry if this isn't making much sense...

Roads (20)

When I first read Zhuge's explanation I replaced the word 'factorial' with the word 'junk' in the program thinking that just maybe the word 'factorial' meant this, a* (a - 1) * (a - 2) * (a - 3)... in C++.

The new version of the program worked just fine. I'm thinking now that this statement:


(a * factorial (a-1))

will always mean a*(a-1) * (a-2) * (a-3).....regardless of what is used for the function name.

Syx (17)

Actually, I don't think that's what it means. I've never used it but looking at the function, I'm assuming if you have like say the number 6.
Your function will be (6 * factorial (6-1)) = (6 * factorial (5))
From what Zhuge said factorial (5) is 5 * factorial (4)
then factorial (4) = 4 * factorial (3) ...

So you will end up with 6 * factorial (5) which is equivalent to 6 * 5 * factorial (4) which again is equivalent to 6 * 5 * 4 * factorial(3) so on and so forth.
Makes sense?

What you're doing is think of factorial like you learn it in math, you would be correct to say a! = a*(a-1) * (a-2) * (a-3) * ... * 1 in a math class. But this factorial is different from the factorial "!" of math. Think of it like Zhuge said: "a recursive function so it calls itself with the parameter (a - 1)."

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Roads (20)

Yep fellas it all makes sense. I understand what factoring a number is. I just don't understand why when looking at the code. It seems me and Mr. C++ has a failure to communicate. :)

I really appreciate all your posts trying to make me understand but it looks like it is time to move on and hope I can work around any problem I encounter with this.

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firedraco (651)

Yeah...recursive functions are a pain to understand sometimes, so don't worry about it *too* much.

helios (823)

Oddly enough, once you understand recursion, it's easier to write a recursive function than an iterative function.

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