如何绘制一组水平线?
我是 OpenGL 新手,作为学习练习我决定从包含顶点位置的 mxn 矩阵网格中绘制一组水平线
这就是我所拥有的
如果我使用 LINE_STRIP
使用顶点数组和索引的代码片段会很棒,我似乎无法仅从教科书中获取概念,我需要查看并使用代码示例 任何帮助将不胜感激!
@Thomas 使用以下代码让它工作
totalPoints = GRID_ROWS * 2 * (GRID_COLUMNS - 1);
indices = new int[totalPoints];
points = new GLModel(this, totalPoints, LINES, GLModel.DYNAMIC);
int n = 0;
points.beginUpdateVertices();
for ( int row = 0; row < GRID_ROWS; row++ ) {
for ( int col = 0; col < GRID_COLUMNS - 1; col++ ) {
int rowoffset = row * GRID_COLUMNS;
int n0 = rowoffset + col;
int n1 = rowoffset + col + 1;
points.updateVertex( n, pointsPos[n0].x, pointsPos[n0].y, pointsPos[n0].z );
indices[n] = n0;
n++;
points.updateVertex( n, pointsPos[n1].x, pointsPos[n1].y, pointsPos[n1].z );
indices[n] = n1;
n++;
}
}
points.endUpdateVertices();
然后我通过执行更新和绘制
points.beginUpdateVertices();
for ( int n = 0; n < totalPoints; n++ ) {
points.updateVertex( n, pointsPos[indices[n]].x, pointsPos[indices[n]].y, pointsPos[indices[n]].z );
}
points.endUpdateVertices();
这是结果
通过更改来修复它嵌套的 for 循环
for ( int col = 0; col < GRID_COLUMNS; col++ ) {
for ( int row = 0; row < GRID_ROWS - 1; row++ ) {
int offset = col * GRID_ROWS;
int n0 = offset + row;
int n1 = offset + row + 1;
indices[n++] = n0;
indices[n++] = n1;
}
}
现在我可以拥有任意数量的行和列,
再次感谢!
I am new to OpenGL as learning exercise I decided to draw a set of horizontal lines from a grid of m x n matrix containing the vertices locations
This is what I have
and If I use LINE_STRIP
A code snippet using vertex arrays and indices will be great, I cant seem to be able to get the concept just from a text book I need to see and play with a code example
Any help will be much appreciated!
@Thomas
Got it working with the following code
totalPoints = GRID_ROWS * 2 * (GRID_COLUMNS - 1);
indices = new int[totalPoints];
points = new GLModel(this, totalPoints, LINES, GLModel.DYNAMIC);
int n = 0;
points.beginUpdateVertices();
for ( int row = 0; row < GRID_ROWS; row++ ) {
for ( int col = 0; col < GRID_COLUMNS - 1; col++ ) {
int rowoffset = row * GRID_COLUMNS;
int n0 = rowoffset + col;
int n1 = rowoffset + col + 1;
points.updateVertex( n, pointsPos[n0].x, pointsPos[n0].y, pointsPos[n0].z );
indices[n] = n0;
n++;
points.updateVertex( n, pointsPos[n1].x, pointsPos[n1].y, pointsPos[n1].z );
indices[n] = n1;
n++;
}
}
points.endUpdateVertices();
Then I update and draw by doing
points.beginUpdateVertices();
for ( int n = 0; n < totalPoints; n++ ) {
points.updateVertex( n, pointsPos[indices[n]].x, pointsPos[indices[n]].y, pointsPos[indices[n]].z );
}
points.endUpdateVertices();
This is the result
Fix it by changing the nested for loop
for ( int col = 0; col < GRID_COLUMNS; col++ ) {
for ( int row = 0; row < GRID_ROWS - 1; row++ ) {
int offset = col * GRID_ROWS;
int n0 = offset + row;
int n1 = offset + row + 1;
indices[n++] = n0;
indices[n++] = n1;
}
}
Now I can have any number of rows and columns
Thanks agin!
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您需要为每个段绘制一条线并重新使用索引,即对于第一部分,您将为 (0,1)、(1,2)、(2,3) 等绘制一条线。
编辑:
假设您有一个 4x5 数组(4 行,每行 5 个顶点)。然后,您可以像这样计算索引(伪代码):
然后发出
numrows * (numcols - 1)
线段 (GL_LINES) 的绘制调用,即示例中的 16。请注意,addLineIndices 是一个函数,它将一条线段的索引对添加到索引数组中,然后将其提供给绘制调用。You need to draw a line for each segment and resuse an index, i.e. for the first part you'd draw a line for (0,1), (1,2), (2,3) and so on.
Edit:
Suppose you have a 4x5 array (4 lines, 5 vertices per line). You could then calculate the indices like this (pseudo code):
Then issue the draw call for
numrows * (numcols - 1)
linesegments (GL_LINES), i.e. 16 in the example. Note thataddLineIndices
would be a function that adds the index pair for one line segment to an index array which is then supplied to the draw call.