没有当前上下文的opengl es api

Question

我已经查看了解决方案并且没有真正找到解决方案.我收到此错误,因为它似乎执行发生在gl线程之外.但是我不知道如何解决这个问题.代码如下:

public shape()
{
    super();        



    vertexShader = Shader.loadShader(GLES20.GL_VERTEX_SHADER, vertexShaderCode); //<============
    fragmentShader = Shader.loadShader(GLES20.GL_FRAGMENT_SHADER, fragmentShaderCode);      

    ByteBuffer buffer = ByteBuffer.allocateDirect(getCoordinates().length * 4);
    buffer.order(ByteOrder.nativeOrder());
    vertexBuffer = buffer.asFloatBuffer();

    vertexBuffer.put(getCoordinates());
    vertexBuffer.position(0);

    ByteBuffer drawListBuffer = ByteBuffer.allocateDirect(getOrderOfDraw().length * 2);

    drawListBuffer.order(ByteOrder.nativeOrder());

    listBuffer = drawListBuffer.asShortBuffer();
    listBuffer.put(getOrderOfDraw());

    listBuffer.position(0);


     mProgram = GLES20.glCreateProgram();             // create empty OpenGL Program
     GLES20.glAttachShader(mProgram, vertexShader);   // add the vertex shader to program
     GLES20.glAttachShader(mProgram, fragmentShader); // add the fragment shader to program
     GLES20.glLinkProgram(mProgram); 

}

并且调用渲染器是

    Square square = new Square(5, 5);

public void onDrawFrame(GL10 unused) 
{

    unused.glLoadIdentity();
    unused.glClear(GLES20.GL_COLOR_BUFFER_BIT);
    square.Draw();
}

方形从形状延伸

Answer 1

如果那new Square(5,5);不是任何opengl回调的一部分(我假设你使用的是glSurfaceView),那么我认为它不会在OpenGL线程上运行.它将在你创建glSurfaceView时执行,我相信它是在主要的android线程上.

试着在new Square(5,5);里面移动onSurfaceCreated.