How to access coordinates of Bezier curve control point handles?

Question

There are many examples of how to get the Control Points of a Bezier curve from Python. (For example, with a 2D bezier curve 'b'):

for ctrlPtIndex in range(0, numCtrlPts):  
    print("ptX:", b.data.splines[0].bezier_points[ctrlPtIndex].co[0])
    print("ptY:", b.data.splines[0].bezier_points[ctrlPtIndex].co[1])

Is it possible to access the handle coordinates for each of these control points from bpy (python)?

I know that the handle-types can be changed for each control point in bpy, but can't find how to access the handle coordinates. These are plain curves created in Blender, they aren't connected to any armatures.

Thanks, Jon

Answer 1

Here is a Python function to print X, Y, Z values of the points given a curve object reference parameter.
(based on the answer linked in the comment)

import mathutils.geometry

# The parameters can be found by taking the coordinate and 
# right handle of a point, and the coordinate and
# left handle of the next point in the Bezier points array.
# For cyclic curves the last point and the first point in the array 
# form an extra segment.

def printInterpolatedCurveSegmentPoints(existingCurveObj):

    spline = existingCurveObj.data.splines[0]
    numSegments = len(spline.bezier_points)
    assert(numSegments >= 2)

    r = spline.resolution_u
    if spline.use_cyclic_u:
        numSegments += 1

    points = []
    for i in range(numSegments):
        nextIdx = (i + 1) % numSegments

        knot1 = spline.bezier_points[i].co
        handle1 = spline.bezier_points[i].handle_right
        handle2 = spline.bezier_points[nextIdx].handle_left
        knot2 = spline.bezier_points[nextIdx].co

        _points = mathutils.geometry.interpolate_bezier(knot1, handle1, handle2, knot2, r)
        points.extend(_points)

    assert('3D' == existingCurveObj.data.dimensions)
    for pointVec in points:
        print("Point X:", pointVec[0])
        print("Point Y:", pointVec[1])
        print("Point Z:", pointVec[2])
        print()