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i'm having trouble working out the returned values of the below code pMeasure = PathMeasure, m = Matrix, distCount is the distance along the path

pMeasure.getMatrix(distCount, m, 0x01 | 0x02); 
m.getValues(float[] values)
float[2] & float[5] are position x & y respectively but i can't figure out the rest
any help once again appreciated.
                Honesly, I don't know.. I bet these values are reserved, and the reason why they are present perhaps is to complete the matrix. Here you will find more info on this topic: mobiledevelop.blogspot.com/2009/02/…
– Andrejs Cainikovs
                Jul 14, 2010 at 13:05
                Android uses skia and I checked the source code for androids matrix class, in its C++ core, it simply glues Matrix, with SkMatrix. The matrix is 3x3 row major. Because of this MSCALE_X/Y is not actually scale. As long as no rotations are involved it is scale. If matrix encodes a rotation you need to do extra math to calculate the scale.  MPERSP is used to calculate the "z" coordinate, by default z=1. If z !=1 then x=x/z; y=y/z; z=z/z=1. In general you don't need to worry about it z. It's just like opengl 4x4 for 3D, except on android its 3x3 for 2D
– over_optimistic
                Mar 22, 2013 at 14:49
a,b,c,d encode scale & rotation at the same time. tx/ty encode translation. If you do m.getValues(vals); then vals[2] == tx, vals[5] == ty, and the rest is straight forward. The best way to extract translation is to make a vector 
float[] point = {0, 0};
Then map it and see where it ends up and that's your translation (which is exactly (tx, ty). Under rare circumstances its not
to get scale map a second point
float[] point2 = {1, 0};
Now take the difference rel = (point - point2) and get the length of that vector and that's your scale. To extract rotation, normalize rel and it's simple to get it's standard angle.
        
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