How do I implement a Bézier curve in C++?

问题:

I'd like to implement a Bézier curve. I've done this in C# before, but I'm totally unfamiliar with the C++ libraries. How should I go about creating a quadratic curve?

void printQuadCurve(float delta, Vector2f p0, Vector2f p1, Vector2f p2);

Clearly we'd need to use linear interpolation, but does this exist in the standard math library? If not, where can I find it?

Update 1:

Sorry, I forgot to mention I'm using Linux.

回答1:

Did you use a C# library earlier?

In C++, no standard library function for Bezier curves is available (yet). You can of course roll your own (CodeProject sample) or look for a math library.

This blogpost explains the idea nicely but in Actionscript. Translation should not be much of a problem.

回答2:

Recently I ran across the same question and wanted to implemented it on my own. This image from Wikipedia helped me:

The following code is written in C++ and shows how to compute a quadratic bezier.

int getPt( int n1 , int n2 , float perc )
{
    int diff = n2 - n1;

    return n1 + ( diff * perc );
}    

for( float i = 0 ; i < 1 ; i += 0.01 )
{
    // The Green Line
    xa = getPt( x1 , x2 , i );
    ya = getPt( y1 , y2 , i );
    xb = getPt( x2 , x3 , i );
    yb = getPt( y2 , y3 , i );

    // The Black Dot
    x = getPt( xa , xb , i );
    y = getPt( ya , yb , i );

    drawPixel( x , y , COLOR_RED );
}

With (x1|y1), (x2|y2) and (x3|y3) being P0, P1 and P2 in the image. Just for showing the basic idea...

For the ones who ask for the cubic bezier, it just works analogue (also from Wikipedia):

This answer provides Code for it.

回答3:

Here is a general implementation for a curve with any number of points.

vec2 getBezierPoint( vec2* points, int numPoints, float t ) {
    vec2* tmp = new vec2[numPoints];
    memcpy(tmp, points, numPoints * sizeof(vec2));
    int i = numPoints - 1;
    while (i > 0) {
        for (int k = 0; k < i; k++)
            tmp[k] = tmp[k] + t * ( tmp[k+1] - tmp[k] );
        i--;
    }
    vec2 answer = tmp[0];
    delete[] tmp;
    return answer;
}

Note that it uses heap memory for a temporary array which is not all that efficient. If you only need to deal with a fixed number of points you could hard-code the numPoints value and use stack memory instead.

Of course, the above assumes you have a vec2 structure and operators for it like this:

struct vec2 {
    float x, y;
    vec2(float x, float y) : x(x), y(y) {}
};

vec2 operator + (vec2 a, vec2 b) {
    return vec2(a.x + b.x, a.y + b.y);
}

vec2 operator - (vec2 a, vec2 b) {
    return vec2(a.x - b.x, a.y - b.y);
}

vec2 operator * (float s, vec2 a) {
    return vec2(s * a.x, s * a.y);
}

回答4:

You have a choice between de Casteljau's method, which is to recursively split the control path until you arrive at the point using a linear interpolation, as explained above, or Bezier's method which is to blend the control points.

Bezier's method is

 p = (1-t)^3 *P0 + 3*t*(1-t)^2*P1 + 3*t^2*(1-t)*P2 + t^3*P3 

for cubics and

 p = (1-t)^2 *P0 + 2*(1-t)*t*P1 + t*t*P2

for quadratics.

t is usually on 0-1 but that's not an essential - in fact the curves extend to infinity. P0, P1, etc are the control points. The curve goes through the two end points but not usually through the other points.

回答5:

• If you just want to display a Bezier curve, you can use something like PolyBezier for Windows.

• If you want to implement the routine yourself, you can find linear interpolation code all over the Intarnetz.

• I believe the Boost libraries have support for this. Linear interpolation, not Beziers specifically. Don't quote me on this, however.