[AGG] Bitmap pattern question: FlipX, FlipY, FlipXY

[Marlon <ma...@on...>]

Some week ago I asked how to replicate bitmap pattern for filling a polygon 
in a similar way that Gdi+ do, with flags for bitmap replication, for 
example:
<] is a bitmap,
<],<],<] is a normal replication like in Agg,
<],[>,<] is with flag FlipX, etc...
I solved problem, (actually Maxim helped) by creating temporary bitmap cache 
by 2x2 bitmap size with desired replication and using it with 
span_pattern_filter_rgba32_bilinear.
But I think there must be more elegant way;
Question:
can I use custom interpolator to control replication? Something like this:
//########
template<class Transformer = trans_affine, unsigned SubpixelShift = 8>
    class span_interpolator_linear_mirror :
    public span_interpolator_linear<Transformer, SubpixelShift>
     {
    public:
     typedef span_interpolator_linear<Transformer, SubpixelShift> base_type;
      //--------------------------------------------------------------------
      span_interpolator_linear_mirror(const trans_type& trans, unsigned 
nMirrorX=0, unsigned nMirrorY=0) :
            base_type(trans), mirror_x(nMirrorX<<subpixel_shift), 
mirror_y(nMirrorY<<subpixel_shift) { }
      //----------------------------------------------------------------
      void coordinates(int* x, int* y) const
       {
        int zx, zy;
        base_type::coordinates(&zx, &zy);
        if(mirror_x) { *x= 
mirror_x-1-(abs((((zx%(mirror_x<<1))-mirror_x)<<1)+1)>>1); }
        else         { *x=zx; }
        if(mirror_y) { *y= 
mirror_y-1-(abs((((zy%(mirror_y<<1))-mirror_y)<<1)+1)>>1); }
        else         { *y=zy; }
        //this gargantuan formula is actually
        //*x= ((zx%(2*mirror_x)) < mirror_x) ?
        //      zx%mirror_x : //normal
        //      mirror_x-1-(zx%mirror_x); //mirror
       }

    private:
      unsigned mirror_x, mirror_y;
     };
in nMirrorX, nMirrorY is width and height of bitmap, sorry about messy code 
:)
Now it works, but about 30% slower than original.
Is there a better solution?

Thanks,
Marlon
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