我在一些演示代码中有一个 Pixbuf，我目前根据屏幕触摸顺时针或逆时针旋转它。
我这样做使用RotateSimple但仅限于 90 度的倍数。
GDK 中有没有办法将 Pixbuf 缓冲区中的图像旋转 45 度（或更小）？
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我写了一个通用的 pixbuf 旋转器函数。论据：
- src -- pixbuf 旋转
- 弧度 -- 旋转的角度（以弧度表示）
full_size -- 此函数在两种模式下工作。它将生成一个新的 pixbuf，其大小刚好足以包含完全旋转的图像（加上一些 alpha=0 的额外三角形），或者它将生成一个与最大内接（水平/垂直）矩形一样大的 pixbuf旋转后的图像。如果 full_size 为 true，则生成带有（空白角）的较大矩形，如果 full_size 为 false，则生成较小的矩形。当旋转接近 45 度时，小矩形的大小可能为 0，并且将返回 NULL pixbuf。
#include <gtk/gtk.h> /* There are two reasonable sizes for a rotated image-- Either the minimum */ /* bounding box which contains all rotated pixels (and a bunch of white space)*/ /* or the maximum rectangle where all pixels come from the source image (but */ /* where we lose some of the corners) */ /* The first is easy to calculate: The minimum bounding box will have the corners */ /* of the rotated image on its edges, this leaves us with four triangles in */ /* the corners of the bb. Two triangles have edges width*sin(theta), width*cos(theta) */ /* and two have edges height*sin(theta), height*cos(theta) */ /* so the new width height will be the sum of two adjacent triangle edges: */ /* width" = width*cos + height*sin */ /* height"= width*sin + height*cos */ /* Now for the maximum inscribed rectangle we draw a similar picture (except */ /* the unknown rectangle is internal now) and get similar triangles. Here the*/ /* equations are: */ /* width = width'*cos + height'*sin */ /* height= width'*sin + height'*cos */ /* solving for height'... */ /* height' = (width-width'*cos)/sin */ /* height' = (height-width'*sin)/cos */ /* (width-width'*cos)/sin = (height-width'*sin)/cos */ /* width*cos - width'*cos^2 = height*sin - width'*sin^2 */ /* width' * (sin^2-cos^2) = height*sin-width*cos */ /* width' = (height*sin - width*cos)/(sin^2-cos^2) */ /* height'= (width*sin - height*cos)/(sin^2-cos^2) */ /* Note this produces garbage (0/0) when rotated by 45 degrees (135,225,...) */ /* A little experimentation shows that at 45 degrees the only thing with */ /* an internal rectangle is a square, all other aspect ratios have a height */ /* of 0. A square, however, has an internal square with sides 1/sqrt(2) of the original */ /* When creating a full_size image (minimum bounding box) we should return */ /* an image with an alpha channel (whether the original had one or no). */ /* otherwise we should create an alpha channel only if the original had one */ /* A pixel at (x,y) will be rotated to: */ /* ((x-width/2)*cos + (y-height/2)*sin + width'/2 , */ /* =(x-width/2)*sin + (y-height/2)*cos + height'/2 ) */ /* A pixel at (x',y') will have come from: */ /* ((x'-width'/2)*cos - (y'-height'/2)*sin + width/2 , */ /* (x'-width'/2)*sin + (y'-height'/2)*cos + height/2 ) */ static GdkPixbuf *gdk_pixbuf_rotate(GdkPixbuf *src,double radian,gboolean full_size) { double s = sin(radian), c = cos(radian); double as= s<0 ? -s : s, ac= c<0 ? -c : c; int width, height, nwidth, nheight; int hasalpha, nhasalpha; GdkPixbuf *ret; int nr,nc,r,col; double nmodr, nmodc; int alpha=0; guchar *pixels, *npixels, *pt, *npt; int rowstride, nrowstride, pixellen; if ( src==NULL ) return( NULL ); width = gdk_pixbuf_get_width(src); height = gdk_pixbuf_get_height(src); hasalpha = gdk_pixbuf_get_has_alpha(src); rowstride = gdk_pixbuf_get_rowstride(src); pixels = gdk_pixbuf_get_pixels(src); pixellen = hasalpha ? 4 : 3; if ( full_size ) { nwidth = round( ac*width + as*height ); nheight= round( as*width + ac*height ); nhasalpha = TRUE; } else { double denom = as*as - ac*ac; if ( denom<.1e-7 && denom>-1.e-7 ) { if ( width!=height ) return( NULL ); nwidth = nheight = round( width/sqrt(2.0) ); } else { nwidth = round( (height*as - width*ac)/denom ); nheight = round( (width*as - height*ac)/denom ); } if ( nwidth<=0 || nheight<=0 ) return( NULL ); nhasalpha = hasalpha; } ret = gdk_pixbuf_new(GDK_COLORSPACE_RGB,nhasalpha,8,nwidth,nheight); if ( ret==NULL ) return( NULL ); nrowstride = gdk_pixbuf_get_rowstride(ret); npixels = gdk_pixbuf_get_pixels(ret); for ( nr=0; nr<nheight; ++nr ) { nmodr = nr-nheight/2.0; npt = npixels + nr*nrowstride; for ( nc=0; nc<nwidth; ++nc ) { nmodc = nc-nwidth/2.0; /* Where did this pixel come from? */ r = round( height/2 - nmodc*s + nmodr*c ); col = round( width/2 + nmodc*c + nmodr*s ); if ( r<0 || col<0 || r>=height || col>=width ) { alpha = 0; if ( r<0 ) r=0; else if ( r>=height ) r = height-1; if ( col<0 ) col = 0; else if ( col>=width ) col = width-1; } else alpha = 0xff; pt = pixels + r*rowstride + col*pixellen; *npt++ = *pt++; *npt++ = *pt++; *npt++ = *pt++; if ( hasalpha && alpha!=0 ) alpha = *pt; if ( nhasalpha ) *npt++ = alpha; } } return( ret ); }
于 2016-08-24T18:14:45.920 回答
1
不，不是RotateSimple（或gdk_pixbuf_rotate_simple()在底层库中）。根据文档，这仅限于“旋转 90 度的倍数”。
但是，您可以做的一件事是提供多个图像，使其看起来好像您正在旋转一个较小的值。
对于 45 度的具体示例，只需要两个图像。第一个是“直立”图像，您可以使用 90 度旋转（即使用SimpleRotate）来获得八个所需旋转中的四个0，90、180和270。
要获得其他四种可能性，请将图像放入一些图像编辑软件并使用它将其旋转 45 度，将其保存为“倾斜”图像。
这样，您可以通过使用两个图像的各种旋转来获得所有可能性：
Desired rotation Uses image Actual rotation
---------------- ---------- ---------------
0 upright 0
45 tilted 0
90 upright 90
135 tilted 90
180 upright 180
225 tilted 180
270 upright 270
315 tilted 270
对于更细粒度的旋转，您可以做类似的事情，特别是如果旋转的分辨率是 360 倍。而且，由于巴比伦人（或苏美尔人或其他人）的前瞻性天性，我的历史有点生锈），360有相当多的因素。
于 2016-06-22T04:44:56.550 回答