Can I do the same without the recursion?

[Lyleo]

Hi There
My name is Lyle and i'm new to programming.

I've noticed something in this code:

// recursive function
#include <iostream>
using namespace std;

void myFunction( int counter)
{
if(counter == 1)

return;

else
{
cout <<counter<<endl;
myFunction(--counter);
return;
}
}

int main()
{
myFunction(10);

cin.get();
return 0;
}

The thing that I've noticed is that the actual task was
to decrement counter if it's greater than 1. So I would
like to know if I could go about it without a recursive function,
like this code example below:

//without recursive function
#include <iostream>
using namespace std;

void myFunction( int counter)

{
if(counter == 1)
return;

else
{
cout <<counter<<endl;
cout << --counter;
return;
}
}

int main()
{
myFunction(10);

cin.get();
return 0;
}




but I don't get the same result,
it just decrements until 9 which is not according to my condition.
I would like to know why and were my mistake is.



Thank you
Lyleo

[Kyon]

Recursion can be seen as a LOOP of function calls. It's sometimes advisable to use recursion above a loop, because of complexity. In your case, a simple for loop will do the trick:

void myFunction(int x)
for(;x != 0; x--)
{
    cout << x << endl;
}

Your code, does the following:

void myFunction(int counter)
{
    if(counter == 1) return; // If the input is 1, end the function
    else // In all other cases:
    {
        cout <<counter<<endl; // Show the input
        cout << --counter; // Show the input-1
        return; // End the function
    }
}

This means it will only ever show two values.

[Fresh Grass]

Here are both ways:

void countBack(int num) {
	if(num) {
		cout << num << endl;
		countBack(num-1);
	}
}

int main() {
	//Without recursion:
	for(int top=10; top; --top) {
		cout << top << endl;
	}

	//With recursion:
	countBack(10);
	return 0;
}

Note that you can use the number itself as an expression. That's because your set of actions execute only if expression is any value different than 0.