scratchfoundation/paper.js
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Define Curve constructor for 8 parameters and implement Curve#getPart() that returns a new sub curve.

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This commit is contained in:
Jürg Lehni 2011-06-05 12:36:26 +01:00
parent 3a0f43050e
commit 14816a872e
1 changed files with 11 additions and 4 deletions
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15
src/path/Curve.js
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@ -29,7 +29,7 @@ var Curve = this.Curve = Base.extend({
*
* @class The Curve object represents...
*/
initialize: function(arg0, arg1, arg2, arg3) {
initialize: function(arg0, arg1, arg2, arg3, arg4, arg5, arg6, arg7) {
var count = arguments.length;
if (count == 0) {
this._segment1 = new Segment();
@ -45,6 +45,15 @@ var Curve = this.Curve = Base.extend({
} else if (count == 4) {
this._segment1 = new Segment(arg0, null, arg1);
this._segment2 = new Segment(arg3, arg2, null);
} else if (count == 8) {
// An array as returned by getCurveValues
var p1 = Point.create(arg0, arg1),
p2 = Point.create(arg6, arg7);
p1x, p1y, c1x, c1y, c2x, c2y, p2x, p2y
this._segment1 = new Segment(p1, null,
Point.create(arg2, arg3).subtract(p1));
this._segment2 = new Segment(p2,
Point.create(arg4, arg5).subtract(p2), null);
}
},
@ -221,13 +230,11 @@ var Curve = this.Curve = Base.extend({
return length;
},
/* TODO: Implement this, but convert result to new Curve object again?
getPart: function(from, to) {
var args = this.getCurveValues();
args.push(from, to);
return Curve.getPart.apply(Curve, args);
return new Curve(Curve.getPart.apply(Curve, args));
},
*/
/**
* Checks if this curve is linear, meaning it does not define any curve