Pascal's triangle

[Kelton2]

I need some help with a program that prints out Pascal's triangle; it just doesn't work. For example, with input 4 it should print:How many rows: 4 1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
but instead it prints:

How many rows? 4




  4
 66
444

how do I get it to print the right thing? My code is:

#include <iostream>
#include <iomanip>
using namespace std;
int combinations (int n, int k) {
    if (k == 0 || k == n) {
          return 1;
          }
          else {
               return combinations(n-1,k-1) + combinations(n-1,k);
}
}
int main ( ) {
    int rows;
    cout << "How many rows? ";
    cin >> rows;
    int track = rows;
    cout << endl;
    for(int r = 0; r < rows; r++) {
            track--;
        int spaces = track;


        
        cout << setw(spaces+1);
       for(int amount = 0; amount < r; amount++) {
        cout << combinations(rows, r);
        }
        cout << endl;
    }
}

[whiteflags]

You need to step through your code and realize what it's actually doing. Your problem starts right when this loop first tries to execute.

		for (int amount = 0; amount < r; amount++) {
			cout << combinations(rows, r);
		}

r comes from the outer loop and is initialized to zero, so you end up comparing amount < r or 0 < 0 which is false, so nothing happens.
Then you iterate and r becomes 1.
You enter combinations and the effect looks like this:

combinations(rows, r);
  combinations(4, 1)
     return combinations(4 - 1, 1 - 1) + combinations(4 - 1, 1)
       return 1 + combinations(3, 1)
          combinations(3, 1)
            return combinations(3 - 1, 1 - 1) + combinations(3 - 1, 1)
              return 1 + combinations(2, 1)
                 combinations(2, 1);
                   return 1 + combinations(2 - 1, 1);
                     combinations(1, 1)
                       return 1

So we have four 1 + somethings... that something eventually resolves into this = 1 + 1 + 1 + 1, which is 4, so you print 4.

It just gets worse from there. Try a different approach to the whole problem.