capp_lint CPAnimation.j.

Reviewed by me.

AppKit/CPAnimation.j

@@ -312,7 +312,7 @@ ACTUAL_FRAME_RATE = 0;
     [_timingFunction getControlPointAtIndex:1 values:c1];
     [_timingFunction getControlPointAtIndex:2 values:c2];
 
-    return CubicBezierAtTime(t,c1[0],c1[1],c2[0],c2[1],_duration);
+    return CubicBezierAtTime(t, c1[0], c1[1], c2[0], c2[1], _duration);
 }
 
 @end
@@ -320,28 +320,104 @@ ACTUAL_FRAME_RATE = 0;
 // currently used function to determine time
 // 1:1 conversion to js from webkit source files
 // UnitBezier.h, WebCore_animation_AnimationBase.cpp
-var CubicBezierAtTime = function CubicBezierAtTime(t,p1x,p1y,p2x,p2y,duration)
+var CubicBezierAtTime = function CubicBezierAtTime(t, p1x, p1y, p2x, p2y, duration)
 {
-    var ax=0,bx=0,cx=0,ay=0,by=0,cy=0;
+    var ax = 0,
+        bx = 0,
+        cx = 0,
+        ay = 0,
+        by = 0,
+        cy = 0;
     // `ax t^3 + bx t^2 + cx t' expanded using Horner's rule.
-    function sampleCurveX(t) {return ((ax*t+bx)*t+cx)*t;};
-    function sampleCurveY(t) {return ((ay*t+by)*t+cy)*t;};
-    function sampleCurveDerivativeX(t) {return (3.0*ax*t+2.0*bx)*t+cx;};
+    function sampleCurveX(t)
+    {
+        return ((ax * t + bx) * t + cx) * t;
+    }
+
+    function sampleCurveY(t)
+    {
+        return ((ay * t + by) * t + cy) * t;
+    }
+
+    function sampleCurveDerivativeX(t)
+    {
+        return (3.0 * ax * t + 2.0 * bx) * t + cx;
+    }
+
     // The epsilon value to pass given that the animation is going to run over |duration| seconds. The longer the animation, the more precision is needed in the timing function result to avoid ugly discontinuities.
-    function solveEpsilon(duration) {return 1.0/(200.0*duration);};
-    function solve(x,epsilon) {return sampleCurveY(solveCurveX(x,epsilon));};
+    function solveEpsilon(duration)
+    {
+        return 1.0 / (200.0 * duration);
+    }
+
+    function solve(x, epsilon)
+    {
+        return sampleCurveY(solveCurveX(x, epsilon));
+    }
+
     // Given an x value, find a parametric value it came from.
-    function solveCurveX(x,epsilon) {var t0,t1,t2,x2,d2,i;
-        function fabs(n) {if(n>=0) {return n;}else {return 0-n;}};
+    function solveCurveX(x, epsilon)
+    {
+        var t0,
+            t1,
+            t2 = x,
+            x2,
+            d2,
+            i = 0;
+
         // First try a few iterations of Newton's method -- normally very fast.
-        for(t2=x, i=0; i<8; i++) {x2=sampleCurveX(t2)-x; if(fabs(x2)<epsilon) {return t2;} d2=sampleCurveDerivativeX(t2); if(fabs(d2)<1e-6) {break;} t2=t2-x2/d2;}
+        for (; i < 8; i++)
+        {
+            x2 = sampleCurveX(t2) - x;
+
+            if (ABS(x2) < epsilon)
+                return t2;
+
+            d2 = sampleCurveDerivativeX(t2);
+
+            if (ABS(d2) < 1e-6)
+                break;
+
+            t2 = t2 - x2 / d2;
+        }
+
         // Fall back to the bisection method for reliability.
-        t0=0.0; t1=1.0; t2=x; if(t2<t0) {return t0;} if(t2>t1) {return t1;}
-        while(t0<t1) {x2=sampleCurveX(t2); if(fabs(x2-x)<epsilon) {return t2;} if(x>x2) {t0=t2;}else {t1=t2;} t2=(t1-t0)*.5+t0;}
+        t0 = 0.0;
+        t1 = 1.0;
+        t2 = x;
+
+        if (t2 < t0)
+            return t0;
+
+        if (t2 > t1)
+            return t1;
+
+        while (t0 < t1)
+        {
+            x2 = sampleCurveX(t2);
+
+            if (ABS(x2 - x) < epsilon)
+                return t2;
+
+            if (x > x2)
+                t0 = t2;
+
+            else
+                t1 = t2;
+
+            t2 = (t1 - t0) * 0.5 + t0;
+        }
+
         return t2; // Failure.
     };
     // Calculate the polynomial coefficients, implicit first and last control points are (0,0) and (1,1).
-    cx=3.0*p1x; bx=3.0*(p2x-p1x)-cx; ax=1.0-cx-bx; cy=3.0*p1y; by=3.0*(p2y-p1y)-cy; ay=1.0-cy-by;
+    cx = 3.0 * p1x;
+    bx = 3.0 * (p2x - p1x) - cx;
+    ax = 1.0 - cx - bx;
+    cy = 3.0 * p1y;
+    by = 3.0 * (p2y - p1y) - cy;
+    ay = 1.0 - cy - by;
+
     // Convert from input time to parametric value in curve, then from that to output time.
     return solve(t, solveEpsilon(duration));
 };