c - n-th order Bezier Curves? -

i've managed implement quadratic , cubic bezier curves.they pretty straightforward since have formula. want represent n-th order bezier curve using generalization:

where

and

i'm using bitmap library render output, here code:

// binomialcoef(n, k) = (factorial(n) / (factorial(k) * factorial(n- k))) unsigned int binomialcoef(unsigned int n, const unsigned int k) {     unsigned int r = 1;      if(k > n)         return 0;      for(unsigned int d = 1; d <= k; d++)     {         r *= n--;         r /= d;     }      return r; }  void nbeziercurve(bitmap* obj, const point* p, const unsigned int nbpoint, float steps, const unsigned char red, const unsigned char green, const unsigned char blue) {     int bx1 = p[0].x;     int by1 = p[0].y;     int bx2;     int by2;      steps = 1 / steps;      for(float = 0; < 1; += steps)     {         bx2 = by2 = 0;         for(int j = 0; (unsigned int)j < nbpoint; j++)         {             bx2 += (int)(binomialcoef(nbpoint, j) * pow(1 - i, (float)nbpoint - j) * pow(i, j) * p[j].x);             by2 += (int)(binomialcoef(nbpoint, j) * pow(1 - i, (float)nbpoint - j) * pow(i, j) * p[j].y);         }          bresenhamline(obj, bx1, by1, bx2, by2, red, green, blue);          bx1 = bx2;         by1 = by2;     }      // curve must end on last anchor point     bresenhamline(obj, bx1, by1, p[nbpoint - 1].x, p[nbpoint - 1].y, red, green, blue); } 

here's set of points render:

point ncurv[] = {                     20, 200,                     70, 300,                     200, 400,                     250, 200                 }; 

and here's output:

the red curve cubic bezier. blue 1 supposed 4th order bezier, same cubic bezier, in case, not same ?!

edit : forgot note bottom left point (0, 0)

the sum in formula...

...runs 0 n, ie n-th order bezier need n+1 points.

you have 4 points, you're drawing 3rd-order bezier.

the error in code here:

for(int j = 0; (unsigned int)j < nbpoint; j++) 

it should be:

for(int j = 0; (unsigned int)j <= nbpoint; j++) 

otherwise you're iterating 0 n-1.

edit:

out of interest, shape getting same if missing (5th) point @ (0,0), since that's point contribute nothing sum...