在 OpenGL ES 2 中将原点设置为屏幕的左上角 [英] Set origin to top-left corner of screen in OpenGL ES 2

查看:103
本文介绍了在 OpenGL ES 2 中将原点设置为屏幕的左上角的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

我正在使用 OpenGL ES 2.0 开发 Android 应用.它由覆盖在 3D 场景上的 2D HUD 组成.3D 部分一切正常,但我无法弄清楚在 2D 中绘制时如何将原点定位在屏幕的左上角(y 轴朝下).现在我的原点在屏幕的中心,我只是在绘制时从坐标中减去宽度和高度的一半.(我知道这是一个笨拙的解决方案,我想解决这个问题!)

I'm developing an Android app with OpenGL ES 2.0. It consists of a 2D HUD overlaid on a 3D scene. Everything is fine with the 3D part, but I cannot figure out how to position the origin at the top-left corner of the screen (with the y-axis pointing down) when drawing in 2D. Right now I have the origin at the center of the screen and I am just subtracting half the width and height from the coordinates as I draw. (I know this is a hacky solution and I want to fix this!)

相关代码如下:

GLES20.glViewport(0, 0, surfaceWidth, surfaceHeight);
Matrix.orthoM(projectionMatrix, 0, -width / 2, width / 2, height / 2, -height / 2, -1, 1);
Matrix.setLookAtM(viewMatrix, 0, 0, 0, 2f, 0, 0, 0, 0, -1, 0);
Matrix.multiplyMM(mvpMatrix, 0, projectionMatrix, 0, viewMatrix, 0);

以下是相关的着色器:

private final static String vertexShaderCode =
                "uniform mat4 uMVPMatrix;" +
                "attribute vec2 aTexCoordinate;" +
                "varying vec2 vTexCoordinate;" +
                "attribute vec4 aPosition;" +
                "attribute vec4 aColor;" +
                "varying vec4 vColor;" +
                "void main() {" +
                "  vColor = aColor;" +
                "  vTexCoordinate = aTexCoordinate;" +
                "  gl_Position = aPosition * uMVPMatrix;" +
                "}";

private final static String fragmentShaderCode =
                "precision mediump float;" +
                "uniform sampler2D uTexture;" +
                "varying vec2 vTexCoordinate;" +
                "varying vec4 vColor;" +
                "void main() {" +
                "  gl_FragColor = vColor * texture2D(uTexture, vTexCoordinate);" +
                "}";

我尝试过的:

  • 我试过改用这条线:

  • I have tried using this line instead:

Matrix.orthoM(projectionMatrix, 0, 0, width, height, 0, -1, 1);

  • 我尝试将原点定位在左下角,而不是左上角(我不应该这样做,但我还是尝试了).

  • I have tried positioning the origin at the bottom-left, not the top-left (I shouldn't be doing this but I tried anyway).

    我尝试在顶点着色器的倒数第二行中切换 aPositionuMVPMatrix 的顺序.

    I have tried switching the order of aPosition and uMVPMatrix in the second-to-last line of the vertex shader.

    这三个更改的组合都不能解决我的问题(2D 对象不渲染).

    No combination of these three changes solved the problem for me (the 2D objects do not render).

    我尝试的另一件事是将原点设置为偏离中心一个像素,然后是两个像素,然后是三个像素,等等.图像越偏离中心越偏斜,最终变得太薄而无法看到——但是仍然没有告诉我如何解决这个问题.

    The other thing I tried was setting the origin to be one pixel off center, then two, then three, etc. The images kept getting skewed the more I moved off center, eventually becoming too thin to be visible -- but that still doesn't tell me how to fix this.

    问题又来了:如何在 OpenGL ES 2.0 中将原点定位在左上角?

    So again, the question is: how do I position the origin at the top-left in OpenGL ES 2.0?

    附言我的合同(第一份工作!)将于 7 月 29 日到期,所以我很快就需要答复;我不想让下一个人像这样的黑客修复.

    P.S. My contract (first job!) ends on July 29 so I need an answer pretty soon; I don't want to leave the next people with a hacky fix like this.

    推荐答案

    在 OpenGL 中不是只有一个 mvpMatrix 的规则.这很好,因为您可以自定义很多视图.

    Is not a rule have only one mvpMatrix in OpenGL. It is good because you can customize a lot of views.

    如果您的所有应用程序都基于您在此处展示的 mvpMatrix,我建议您为您所要求的提案创建另一个 mvpMatrix:将视图的中心置于左上角.

    If all of your application is based in mvpMatrix that you showed here, I recommend you create another mvpMatrix for the proposal that you are asking: Make the center of the view in top-left corner.

    所以,答案很简单,让我们重新创建每个矩阵.首先,viewMatrix:

    So, the answer is simple, lets recreate every matrix. At first, viewMatrix:

    Matrix.setLookAtM(viewMatrix, 0, 
         0, 0,  6,   //eye
         0, 0,  0,   //center
         0, 1,  0);  //UP *remove your -1
    

    魔法在下面.你必须颠倒底部和顶部:

    The magic is below. You have to invert bottom and top:

    float screenRatio = (float) width / height;
    Matrix.orthoM(projectionMatrix, 0, 
            0, screenRatio, // left, right
            1, 0,           // bottom, top        <----Solution is invert bottom with top
            -1, 10);        // near, far
    

    最后:

        Matrix.multiplyMM(
                viewProjectionMatrix, 0, 
                projectionMatrix, 0, 
                viewMatrix, 0);
    

    P.S:不需要改变GLES20.glViewport(0, 0,surfaceWidth,surfaceHeight);

    OpenGL 有很多解决这个问题的方法,因为你可以尽可能地处理所有矩阵.

    OpenGL have a lot of solution for this problem, because you can handle all matrix as you can.

    另一种解决方案是自定义您自己的矩阵,而不使用 Matrix.orthoM(..)Matrix.frustum(..):

    Another solution is to customize your own matrix, without using Matrix.orthoM(..) or Matrix.frustum(..):

    public static final float[] topLeftMatrix(int width, int height){
        float matrix[] = new float[16];
    
        // This is inverse of viewPort matrix
        matrix[0] = 2.0f/width;
        matrix[1] = 0;
        matrix[2] = 0;
        matrix[3] = 0;
    
        matrix[4] = 0;
        matrix[5] = -2.0f/height;
        matrix[6] = 0;
        matrix[7] = 0;
    
        matrix[8] = 0;
        matrix[9] = 0;
        matrix[10] = 1;
        matrix[11] = 0;
    
        matrix[12] = -1;
        matrix[13] = 1;
        matrix[14] = 0;
        matrix[15] = 1;
    
        return matrix;
    }
    

    上面的这个矩阵适用于处理像素上的对象(如用户界面).

    This matrix above is good for handling objects over pixels (like user-interface).

    为了充分理解,我建议你阅读这篇文章:3D带有 OpenGL 的图形.它对我理解 opengl 的魔力有很大帮助.

    For full understanding, I recommend you read this article: 3D Graphics with OpenGL. It helped me a lot to understand the opengl magic.

    P.S:你的着色器有错误.将 matrix4x4 乘以向量 4x1 的正确方法是:

    P.S: there is an error in your shader. The correct way to multiply matrix4x4 by a vector 4x1 is:

    gl_Position = uMVPMatrix * aPosition;

    这篇关于在 OpenGL ES 2 中将原点设置为屏幕的左上角的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!

  • 查看全文
    登录 关闭
    扫码关注1秒登录
    发送“验证码”获取 | 15天全站免登陆