entityMap|blocks|key|3068087|text|OpenGL可以很好地进行图像(纹理)坐标映射。您可能想要使用某种形式的OpenGL。Qt提供了一些到OpenGL的绑定，可以帮助您解决问题。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3068088^0|0^^$0|$]|1|@$2|3|4|5|6|7|8|D|9|@]|A|@]|B|$]]|$2|C|4|-4|6|7|8|E|9|@]|A|@]|B|$]]]]

OpenGL does image (texture) coordinate mapping really well. You probably want to use some form of OpenGL. Qt has some binding to OpenGL that can help you out.

entityMap|blocks|key|212585|text|一种方法是使用从QGLWidget继承的类，而不是QGraphicsScene/QGraphicsView组合。不幸的是，OpenGL的学习曲线一开始有点陡峭。然而，它将是非常快的，因为它将直接发生在显卡上，显卡针对这种操作进行了优化。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|212586|您将加载图像QGLWidget::bindTexture()。|offset|length|style|CODE|212587|您将把图像中的点与您的三角形网格相关联，并将它们全部发送到显卡。在遗留版本的OpenGL中(在我看来，它比新版本的应用程序接口更容易使用)，它看起来像这样：|212588|glEnable(GL_TEXTURE_2D);

glBegin(GL_TRIANGLES);
for+(int+ii=0;ii<triangle.size();%2B%2Bii)+{
++for+(int+jj=0;jj<3;%2B%2Bjj)+{
++++glTexCoord2d(triangle[ii].tex[jj][0],triangle[ii].tex[jj][1]);
++++glVertex2d(triangle[ii].point[jj[0],triangle[ii].point[jj][1]);
++}
}
glEnd();|code-block|syntax|javascript|212589|其中，triangle是您创建的一些数据结构，用于保存三角形顶点和与图像相关的映射。显卡将为您处理像素插值。|212590^0|0|6|O|0|0|0|3|8|0^^$0|$]|1|@$2|3|4|5|6|7|8|S|9|@]|A|@]|B|$]]|$2|C|4|D|6|7|8|T|9|@$E|U|F|V|G|H]]|A|@]|B|$]]|$2|I|4|J|6|7|8|W|9|@]|A|@]|B|$]]|$2|K|4|L|6|M|8|X|9|@]|A|@]|B|$N|O]]|$2|P|4|Q|6|7|8|Y|9|@$E|Z|F|10|G|H]]|A|@]|B|$]]|$2|R|4|-4|6|7|8|11|9|@]|A|@]|B|$]]]]

One way to do this would be to use a class inheriting from QGLWidget instead of the QGraphicsScene/QGraphicsView combo. Unfortunately, the learning curve for OpenGL starts out a little steep. However, it will be very fast because it will happen directly on the graphics card which is optimized for just this kind of operation. 
You'll load the image <code>QGLWidget::bindTexture()</code>. 
You'll associate points in the image with your triangle mesh and send them all to your graphics card. In the legacy version of OpenGL (which is easier to use than the newer API in my opinion), it would look something like this:

<pre><code>glEnable(GL_TEXTURE_2D);

glBegin(GL_TRIANGLES);
for (int ii=0;ii&lt;triangle.size();++ii) {
 for (int jj=0;jj&lt;3;++jj) {
 glTexCoord2d(triangle[ii].tex[jj][0],triangle[ii].tex[jj][1]);
 glVertex2d(triangle[ii].point[jj[0],triangle[ii].point[jj][1]);
 }
}
glEnd();
</code></pre>

Where <code>triangle</code> is some data structure that you've made holding the triangle vertices and associated mappings into the image. The graphics card will handle the pixel interpolation for you.

entityMap|blocks|key|2854663|text|非OpenGl解决方案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2854664|对目标图像使用RGB缓冲区。像以前一样完成你的前3个步骤。找到位置和像素颜色后，您可以在此缓冲区上设置它。然后你可以使用|2854665|QImage::QImage+(+uchar+*+data,+int+width,+int+height,+Format+format+)|code-block|syntax|javascript|2854666|以基于先前的缓冲区构建图像。它与您提供的解决方案很接近，并且将比您当前拥有的解决方案快得多。|2854667^0|0|0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|M|9|@]|A|@]|B|$]]|$2|C|4|D|6|7|8|N|9|@]|A|@]|B|$]]|$2|E|4|F|6|G|8|O|9|@]|A|@]|B|$H|I]]|$2|J|4|K|6|7|8|P|9|@]|A|@]|B|$]]|$2|L|4|-4|6|7|8|Q|9|@]|A|@]|B|$]]]]

Non-OpenGl solution :

Use a RGB buffer for the destination image. Work through your 3 first steps as you did before.
Once you have found the position and the pixel color, you set it on this buffer. Then you use 

<pre><code>QImage::QImage ( uchar * data, int width, int height, Format format )
</code></pre>

to construct the image based on the previous buffer. It is close to the solution you provided
and will be much faster than what you currently have.

entityMap|0|type|LINK|mutability|MUTABLE|data|url|http://doc.qt.digia.com/opencl-snapshot/index.html|blocks|key|213109|text|除了OpenGL之外，另一个选择是使用OpenCL，这对您来说可能更容易。您只需将输入/输出位图内存映射到图形卡，用C语言编写一个处理一个三角形的小内核，然后为每个三角形排队一个内核执行。这将比CPU上的单个内核快100倍。|unstyled|depth|inlineStyleRanges|entityRanges|213110|这里有一个用于OpenCL主机api的Qt包装器：|213111|offset|length|213112^0|0|0|0|1E|0|0^^$0|$1|$2|3|4|5|6|$7|8]]]|9|@$A|B|C|D|2|E|F|O|G|@]|H|@]|6|$]]|$A|I|C|J|2|E|F|P|G|@]|H|@]|6|$]]|$A|K|C|8|2|E|F|Q|G|@]|H|@$L|R|M|S|A|T]]|6|$]]|$A|N|C|-4|2|E|F|U|G|@]|H|@]|6|$]]]]

Another option apart from OpenGL is to use OpenCL, which may be easier for you. You just have to memory map the input/output bitmap to the graphics card, write a little kernel in C that handles one triangle, and then queue a kernel execution for each triangle. This will work as much as 100x as fast as a single core on the CPU.

There is a Qt wrapper for the OpenCL host api here:

<a href="http://doc.qt.digia.com/opencl-snapshot/index.html" rel="nofollow">http://doc.qt.digia.com/opencl-snapshot/index.html</a>

entityMap|blocks|key|3068346|text|一种不同的方法是利用已经在光栅绘制引擎中有效实现的裁剪和转换。只要两个三角形之间的变换可以使用3x3增强变换矩阵来表示，您只需将其设置在目标绘制器上，然后在目标上绘制整个源图像。它将被裁剪和变换，以填充目标三角形。您也可以只绘制源三角形的边界矩形，而不是整个图像，如果分析显示它的优势的话。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3068347|这可以并行化，以便您可以并行处理与CPU核心数量一样多的三角形。|3068348^0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|F|9|@]|A|@]|B|$]]|$2|C|4|D|6|7|8|G|9|@]|A|@]|B|$]]|$2|E|4|-4|6|7|8|H|9|@]|A|@]|B|$]]]]

A different approach is to leverage the clipping and transformations already implemented efficiently within the raster paint engine. As long as the transformation between the two triangles can be expressed using a 3x3 augmented transformation matrix, you merely need to set it on the target painter, and then draw the entire source image on the target. It will get clipped and transformed to fill the target triangle. You can also merely draw the bounding rect of the source triangle, instead of the whole image, if profiling shows an advantage to it.

This can be parallelized so that you process as many triangles in parallel as there are CPU cores.

I have designed a program which, basically, cuts a geometrical shape into many small triangles (in a "left canvas"), applies some simple mathematical transformation to the bunch of triangles, and redraws them in their new configuration. See screen capture below.

<img src="https://i.stack.imgur.com/N1HED.png" alt="screen cap 1">

In order to draw these triangles, I use <code>QPainter::drawPolygon</code>. Each triangle on the right corresponds to a triangle on the left, so I know what color I want to use to draw it.

So far, fine. Even if I draw many more triangles than this (when I use much smaller triangles to cut the shape), this is fast enough.

I've added a feature to my program: I can draw triangles extracted from a picture instead of plain triangles: see following screen capture.

<img src="https://i.stack.imgur.com/eyuUO.png" alt="enter image description here">

The problem is that the way I do this is too slow. Here's how I do it:

<ol>
<li>I run through all the triangles</li>
<li>For each triangle, I compute the coordinates of each pixel that will be displayed.</li>
<li>For each one of these pixels, I compute the coordinates of the corresponding pixel on the picture (this is an easy mathematical operation), and I retrieve that pixel's color.</li>
<li>I use <code>QPainter::setPen(QColor)</code> and <code>QPainter::drawPoint(QPoint)</code> to draw the pixel.</li>
</ol>

I am new to programming in Qt and I know nothing about graphics, so this is what I could come up with. The problem is that it's "unacceptably" too slow (the <code>paintEvent</code> of each canvas takes about 0.15s, compared to 0.01s in the case of plain triangles).

I ran a profiler to try to understand what's going on, I noticed that in the canvas widget's <code>paintEvent</code>,

<ol>
<li>58% of the time is spent in <code>QPainter::drawPoint</code></li>
<li>27% of the time is spent in <code>QPainter::setPen</code></li>
</ol>

It seems that <code>QPainter::drawPoint</code> is far too complicated and slow: I just want it to print a pixel of a given color, that's it.

I may have found a solution to my problem: store a <code>QImage</code> (as a member variable of my canvas widget) that represents the whole thing I want my canvas to display, and define it entirely in my <code>paintEvent</code> pixel by pixel, and then draw it at once at the end of my <code>paintEvent</code> with <code>QPainter::drawImage</code>. I have a hint that this will be much faster. But before I rewrite my code all over again, I'd like to know whether that's really what I want to do.

I hope I didn't bore you to do death! Many thanks in advance for your insights.

Qt/C++ : drawing efficiently

数据结构

内存映射

我设计了一个程序，基本上，它将一个几何形状切割成许多小三角形(在“左画布”中)，对这一堆三角形应用一些简单的数学变换，然后以新的配置重新绘制它们。请参阅下面的屏幕截图。​​为了绘制这些三角形，我使用了QPainter::drawPolygon。右边的每个三角形对应左边的一个三角形，所以我知道我想用什么颜色来画它。到目前...

问Qt/C++：高效绘图
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