Functions and Factorials why to type return?

[sime]

If I type return faktorijel(rezultat,broj-1);
for 5 result is just 120
and if I type:
faktorijel(rezultat,broj-1);
without return
for number 5 output is:
120
120
60
20
5

//This is working ok,for example for 5 it is 120
 #include <iostream>
using namespace std;
void faktorijel(int rezultat,int broj)
{
    if(broj>1)
    {
		rezultat*=broj;//5  20   60   120
		return faktorijel(rezultat,broj-1);  //This is my question
    }
   cout<<rezultat<<endl;
}
int main()
{
    int broj;
    cout<<"Unesite broj"<<endl;
    cin>>broj;
    faktorijel(1,broj);
    return 0;
}





//This is not working ok
#include <iostream>
using namespace std;
void faktorijel(int rezultat,int broj)
{
    if(broj>1)
    {
		rezultat*=broj;//5  20   60   120
		faktorijel(rezultat,broj-1);       //without return
    }
   cout<<rezultat<<endl;
}
int main()
{
    int broj;
    cout<<"Unesite broj"<<endl;
    cin>>broj;
    faktorijel(1,broj);
    return 0;
}

[iQChange]

Well. I think if I write another one you'll understand.

#include <iostream>
//In your case, the return make the code stop and call itself again. (because it's void)
//A better application is that:
unsigned factorial(unsigned number)
{
    unsigned result;
    if(number <= 1 ) return 1;
    result = number * factorial(number - 1); // number * number - 1 * number - 2 .. * [number - (number - 1) = 1]
    return result;
}

//So think on this recursion function:

// y = 10
//[sigma] x + y = (0 + 10) + (1 + 10) + (2 + 10) + (3 + 10) ... (10 + 10)
// x = 0

unsigned int sum(unsigned int start_, unsigned int end_)
{
    if(start_ != end_ + 1)
   return start_ + end_ + sum(start_ + 1, end_); //sum will be evaluated on the recursion call
   else return 0;
}

//And try to understand it.

int main(int argc, char* argv[])
{
    std::cout << "Factorial of 3. Expected result = 6\n(3x2x1 = 6)\n\n";
    std::cout << "Calling factorial(3)... " << (factorial(3) == 6? "Passed" : "Failed");
    std::cout << "\n\n" << "Sum of 3. Expected result = 18\n[(0+3) + (1 + 3) + (2 + 3) + (3 + 3)] = 18\n\n";
    std::cout << "Calling sum(0,3)... " << (sum(0,3) == 18? "Passed" : "Failed");
    std::cout << "\n\nFinal result of recursion functions: " << (sum(0,5) == 45 && factorial(5) == 120? "OK." : "Fail.");
}

[coder777]

the return on line 10 prevents the cout on line 12 to be executed. It then jumps out of the function

You can achieve that without return:

void faktorijel(int rezultat,int broj)
{
    if(broj>1)
    {
		rezultat*=broj;//5  20   60   120
		faktorijel(rezultat,broj-1);  // Note: No return
    }
    else // Note: else instead of return
   cout<<rezultat<<endl;
}

[iQChange]

@coder777 that's what I said.

iQChange wrote:

In your case, the return make the code stop and call itself again. (because it's void)

[sime]

Thanks to all.

[coder777]

iQChange wrote:

In your case, the return make the code stop and call itself again. (because it's void)

No, return ends the execution of a function. It has nothing to do with calling itself

[iQChange]

RECURSION, GENIUS!