Recursion: Count from x until n and backwards

Question

I wrote this recursive function to print a number sequence from x to n and n to x back.

Here is one of the ideas I had to improve it, but it didn't work out (use another ternary operator at the end of the first one):

return x < n ? printNumberSequence(x + 1, n) : n = 0  x > n  
       && x != n ? printNumberSequence(x - 1, n) : n;

However, I do have a working solution below:

#include <iostream>
using namespace std;

int printNumberSequence(int x, int n)
{ //counts from x until n
    cout << x << endl;
    x < n ? printNumberSequence(x + 1, n) : n = 0 ;
    return x > n  && x != n ? printNumberSequence(x - 1, n) : n;
}

int main()
{
    return printNumberSequence(0, 5);
}

I want to know if this way to solve the problem is good and how could I improve that code.

Answer 1

If you print a sequence of numbers from \$x\$ through \$n\$ in increasing order, and then the same sequence in decreasing order, you'll get a row of numbers which starts and ends with \$x\$ and contains an analogous double sequence of numbers \$x+1\ldots n\$ between, right?

For example a sequence from 2 to 5 and back is: 2, sequence from 3 to 5 and back, and 2 again:

2 [3 4 5 5 4 3] 2

So the simplest solution would be a recursion like this:

void printNumberSequence(int x, int n)
{
    cout << x << endl;
    if(x < n) printNumberSequence(x + 1, n);
    cout << x << endl;
}

If you want to avoid doubling the greatest number, e.g. for numbers 2 through 5 you want the output like:

2 3 4 5 4 3 2

just execute the second cout << ... under the if(), that is when you're not at the maximum number:

void printNumberSequence(int x, int n)
{
    cout << x << endl;
    if(x < n)
    {
        printNumberSequence(x + 1, n);
        cout << x << endl;
    }
}

Answer 2

Avoid using namespace std

Importing all names of a namespace is a bad habit to get into, and can cause surprise when names like begin and size are in the global namespace. Get used to using the namespace prefix (std is intentionally very short), or importing just the names you need into the smallest reasonable scope.

The exceptions to this rule are namespaces explicitly intended to be imported wholesale, such as the std::literals namespaces.

No need to return a value

The value we return from the recursive function is only ever used for our main()'s exit status. It ultimately returns n, but what we want from main() is 0 for success and non-zero for failure (small positive values work best). So we always end up reporting failure, except when n is zero.

In our case, we don't have any failures we can report, so printNumberSequence() should return void, and main() can always return 0 - we can do that explicitly, or we can just allow execution to run off the end of main() (note that no other function is allowed to do that, though).

Answer 3

I don't know, if it's intended that the sample code doesn't print n twice. If it is, the solution would be:

void printNumberSequence(int x, int n)
{
  cout << x << endl;
  if (x < n) {
    printNumberSequence(x + 1, n);
    cout << x << endl;
  }
}