成都网站建设设计

将想法与焦点和您一起共享

AGG第四十三课例子image1从椭圆到矩形替换问题

I am basing my code on the p_w_picpaths1 example and I have changed

鲁山ssl适用于网站、小程序/APP、API接口等需要进行数据传输应用场景,ssl证书未来市场广阔!成为创新互联建站的ssl证书销售渠道,可以享受市场价格4-6折优惠!如果有意向欢迎电话联系或者加微信:028-86922220(备注:SSL证书合作)期待与您的合作!

the p_w_picpath 'partner' shape from an ellipse to a rectangle.

The partner rectangle comes out at X,Y and scales and rotates,

but the top left-hand corner of the p_w_picpath is always stuck at

(x,y)=3D(0,0). Only the part of the p_w_picpath that overlaps the=20

rectangle is visible, but that part scales and rotates properly.=20

When there is no overlap, there is no p_w_picpath.

I do not understand much of the the p_w_picpath2 example, so I am

lost as to what might be.the cause. Code is attached.

Would be most grateful for help and/or example code.

void p_w_picpath ( HAGG * h , int x , int y , TCHAR * imgfilename )

   {

   if ( !loadp_w_picpath ( h , imgfilename ) ) // sets p_w_picpath details in h

      {

      return ;

      }

   agg::rendering_buffer rbuf(h->pixels ,

                              h->frame_width ,=20

                              h->frame_height ,=20

                              -(h->frame_width * h->bytesperpixel) ) ;=20

   typedef agg::renderer_base                     renderer_base;

   typedef agg::renderer_base                 =

renderer_base_pre;

   typedef agg::renderer_scanline_aa_solid =

renderer_solid;

   pixfmt            pixf(rbuf);

   pixfmt_pre        pixf_pre(rbuf);

   renderer_base     rb(pixf);

   renderer_base_pre rb_pre(pixf_pre);

   renderer_solid    rs(rb);

   rb.clear(agg::rgba(1.0, 1.0, 1.0));

   agg::rasterizer_scanline_aa<> pf;

   agg::scanline_u8 sl;

   IMGINFO * i =3D &h->imgs [ 0 ] ;

   double imgwd =3D i->width ;       // p_w_picpath width

   double imght =3D i->height  ;     // p_w_picpath height

   agg::trans_affine src_mtx;

   src_mtx *=3D agg::trans_affine_translation(-x,-y);

   src_mtx *=3D agg::trans_affine_rotation(-h->t.angle);  // in radians

   src_mtx *=3D agg::trans_affine_scaling(h->t.scalex , h->t.scaley);

   src_mtx *=3D agg::trans_affine_translation(x,y);

   agg::trans_affine img_mtx;

   img_mtx *=3D agg::trans_affine_translation(-x,-y);

   img_mtx *=3D agg::trans_affine_rotation(-h->t.angle);

   img_mtx *=3D agg::trans_affine_scaling(h->t.scalex , h->t.scaley);

   img_mtx *=3D agg::trans_affine_translation(x,y);

   img_mtx.invert();

   typedef agg::span_allocator span_alloc_type;

   span_alloc_type sa;

   typedef agg::span_interpolator_linear<> interpolator_type;

   interpolator_type interpolator(img_mtx);

   // "hardcoded" bilinear filter

   typedef agg::span_p_w_picpath_filter_rgb_bilinear

component_order,=20

                                               interpolator_type> =

span_gen_type;

   typedef agg::renderer_scanline_aa =

renderer_type;

   agg::rendering_buffer rbuf_img(i->pixels ,

                                  (int)imgwd ,=20

                                  (int)imght ,=20

                                  -i->stride ) ;=20

   span_gen_type sg(sa,=20

                    rbuf_img,  // rendering buf with p_w_picpath pixels

                    agg::rgba_pre(0, 0.4, 0, 0.5),

                    interpolator);

   renderer_type ri(rb_pre, sg);

   agg::path_storage path; // partner rectangle

   path.move_to( x,y);

   path.line_to( x+imgwd, y );

   path.line_to( x+imgwd, y+imght);

   path.line_to( x, y+imght);

   path.close_polygon();

   agg::conv_transform tr(path, src_mtx);

         =20

   pf.add_path(tr);

   agg::render_scanlines(pf, sl, ri);

   }

static void drawp_w_picpath ( )

   {

   RECT rt ;

   GetClientRect(hwndmain, &rt);

   int width =3D rt.right - rt.left;

   int height =3D rt.bottom - rt.top;

   HAGG * h =3D gethandle ( mybuf , width , height , 4 ) ;

   settrans_scale ( h , scale ) ;

   settrans_rotate ( h , degrees ) ;

//   p_w_picpath ( h , 20,50 , "bmpeivor.bmp"  ) ; // does not work

   p_w_picpath ( h , 0,0 , "bmpeivor.bmp"  ) ;  // works

   display ( h , hwndmain ) ;  // on screen

   }

作者的回答:

Transforming p_w_picpaths is tricky, especially proper calculation of the affine 

matrix.

But first, if you don't need to transform it you can directly copy or blend 

the p_w_picpath, it will work much faster. See renderer_base<>::copy_from(), 

blend_from().

For the transformer there's a simple way of calculating the matrix as a 

parallelogram, see p_w_picpath_perspective.cpp

// Note that we consruct an affine matrix that transforms

// a parallelogram to a rectangle, i.e., it's inverted.

// It's actually the same as:

// tr(0, 0, img_width, img_height, para); tr.invert();

agg::trans_affine tr(para, 0, 0, img_width, img_height);

Where "para" is double[6] that defines 3 point of the parallelogram.

困惑:

I have replaced

   agg::path_storage path; // partner rectangle

    path.move_to( x,y);

    path.line_to( x+imgwd, y );

    pathmline_to( x+imgwd, y+imght);

    path.line_to( x, y+imght);

    path.close_polygon();

    agg::conv_transform tr(path, src_mtx);

    pf.add_path(tr);

    agg::render_scanlines(pf, sl, ri);

at the and of my p_w_picpath proc (code of the whole proc is at

the end of my original post (and at the end of this email))

by

   double para [ 6 ]

    = { 0,100 ,  0,0 , 100.0 } ; // 3 points (0,100) (0,0) and (100,0)

   agg::trans_affine tr(para, 0, 0, imgwd, imght);

   pf.add_path(tr);

   agg::render_scanlines(pf, sl, ri);

Q1.   is this the right way?

Q2.  what should the para points be expressed as functions of

        p_w_picpath top-left hand corner, p_w_picpath width and p_w_picpath height, i.e.

        x,y, imgwd, imght?

My test cases includes p_w_picpath (x,y)=(0,0), so I defined para points

(0,100), (0,0) and (100,0) just to see what would happen.

but got compilation errors:

..\agg23\include\agg_rasterizer_scanline_aa.h(465) : error C2039: 'rewind' :

is not a member of 'trans_affine'

  ..\agg23\include\agg_trans_affine.h(88) : see declaration of

'trans_affine'

and one more very similar:  'vertex' : is not a member of 'trans_affine'

作者的回答:

> double para [ 6 ] = { 0,100 ,  0,0 , 100,0 } ; // 3 points (0,100) (0,0) 

> and (100,0)

> agg::trans_affine mtx(para, 0, 0, imgwd, imght);

> agg::path_storage path; // partner rectangle

> path.move_to( x,y);

> path.line_to( x+imgwd, y );

> path.line_to( x+imgwd, y+imght);

> path.line_to( x, y+imght);

> path.close_polygon();

> agg::conv_transform trans(path, 

> mtx);

>

> pf.add_path(trans); // Note you add "trans"

>

> Then, if you want your p_w_picpath to fit exactly your parallelogram path (you 

> also may want to do differently!), you need to create a copy of the matrix 

> and invert it:

>

> agg::trans_affine img_mtx(mtx);

> img_mtx.invert();

I'm sorry, Ken, this is not correct; I have confused myself, so, please 

discard the code above. :)

So, suppose you have an p_w_picpath of imgwd, imght and a destination 

parallelogram. To define the parallelogram you need 3 points, x1,y1 - bottom 

left, x2,y2 - bottom right, x3,y3 - top right. The parallelogram can also 

define a 2D affine matrix: rotation, scaling, translation and skewing. You 

can rasterize your destination parallelogram directly:

agg::rasterizer_scanline_aa<> ras;

ras.move_to_d(x1,y1);

ras.line_to_d(x2,y2);

ras.line_to_d(x3,y3);

ras.line_to_d(x1 + x3 - x2, y1 + y3 - y2);

So that, you can draw a solid parallelogram (well, you can also use the 

path_storage if you want).

To map an p_w_picpath to it you need to create the matrix:

double para[6] = {x1,y1,x2,y2,x3,y3};

agg::trans_affine img_mtx(0, 0, imgwd, imght, para);

img_mtx.invert();

Or, which is the same:

double para[6] = {x1,y1,x2,y2,x3,y3};

agg::trans_affine img_mtx(para, 0, 0, imgwd, imght);

The first one construicts a matrix to transform a rectangle to a a 

parellelogram, the second one - parallelogram to rectangle. The p_w_picpath 

transformer requires namely inverse matrix, so that, you transform your 

parallelogram (destination) to rectangle (p_w_picpath).

Technically that's it. But you may want to apply additional transformations. 

To do that you will need two matrices:

agg::trans_affine master_mtx;

master_mtx *= agg::trans_affine_translation(. . .);

master_mtx *= agg::trans_affine_rotation(. . .);

. . .

agg::rasterizer_scanline_aa<> ras;

agg::path_storage path; // partner rectangle

path.move_to(x1,y1);

path.line_to(x2,y2);

path.line_to(x3,y3);

path.line_to(x1 + x3 - x2, y1 + y3 - y2);

path.close_polygon();

agg::conv_transform trans(path, 

master_mtx);

Then you prepare the p_w_picpath matrix:

double para[6] = {x1,y1,x2,y2,x3,y3};

agg::trans_affine img_mtx(0, 0, imgwd, imght, para);

img_mtx *= master_mtx; //!!!!!!!!!!!!! Integrate the master transforms

img_mtx.invert();

ras.add_path(trans);

. . .Render

Now, whatever transformations you use in the master_mtxà they will be 

synchronized with the p_w_picpath.

Sorry for the delay, I was kinda busy last time. and besides, I'm suffering 

from constant problems with the Internet (Verizon in NYC sucks, I'm 

switching to cable).

Well, I understand everyone is busy, but could someone else answer the 

questions too?

First, you need to understand that a path is the primary thing in AGG. 

Without path you can't draw anything. So that, to rotate an p_w_picpath you need 

to create a respective path as if you wanted to fill this area with a solid 

color. And then, you just substitute an p_w_picpath renderer for your solid fill. 

Obviously, to transform the whole p_w_picpath you need to create a parallelogram 

path (a rectangle in particular). You can do that calculating the points 

manually:

ras.move_to_d(x1, y1);

ras.line_to_d(x2, y2);

. . .

You you can use transformations.

Next, trans_affine doesn't have any "VertexSource" interface, it can't 

generate vertices. It can only transform them: affine.transform(&x, &y); To 

add affine transformer into your pipeline you also need conv_transform:

double para [ 6 ] = { 0,100 ,  0,0 , 100,0 } ; // 3 points (0,100) (0,0) and 

(100,0)

agg::trans_affine mtx(para, 0, 0, imgwd, imght);

agg::path_storage path; // partner rectangle

path.move_to( x,y);

path.line_to( x+imgwd, y );

path.line_to( x+imgwd, y+imght);

path.line_to( x, y+imght);

path.close_polygon();

agg::conv_transform trans(path, mtx);

pf.add_path(trans); // Note you add "trans"

Then, if you want your p_w_picpath to fit exactly your parallelogram path (you 

also may want to do differently!), you need to create a copy of the matrix 

and invert it:

agg::trans_affine img_mtx(mtx);

img_mtx.invert();

Well, I realize that it all is pretty confusing. But this kind of a design 

is most flexible.

摘自:http://sourceforge.net/p/vector-agg/mailman/vector-agg-general/?viewmonth=200511&page=0


当前标题:AGG第四十三课例子image1从椭圆到矩形替换问题
链接分享:http://chengdu.cdxwcx.cn/article/ipodhj.html