Question

float pfIdentity[] =
{
    -1.0f,0.0f,0.0f,0.0f,
    0.0f,1.0f,0.0f,0.0f,
    0.0f,0.0f,1.0f,0.0f,
    0.0f,0.0f,0.0f,1.0f
};

=============================================== ===================================

    const char* pszVertShader = "\
    attribute highp vec4    myVertex;\
    uniform mediump mat4    myPMVMatrix;\
    invariant gl_Position;\
    void main(void)\
    {\
        gl_Position = myPMVMatrix * myVertex;\
}";

=============================================== ======================

for(int i = 0; i < 80000; ++i)
{
    glClear(GL_COLOR_BUFFER_BIT);
    int i32Location = glGetUniformLocation(uiProgramObject, "myPMVMatrix");
    glUniformMatrix4fv( i32Location, 1, GL_FALSE, pfIdentity);
    glEnableVertexAttribArray(VERTEX_ARRAY);
    glVertexAttribPointer(VERTEX_ARRAY, 3, GL_FLOAT, GL_FALSE, 0, 0);
    glDrawArrays(GL_TRIANGLES, 0,i);
    eglSwapBuffers(eglDisplay, eglSurface);
}

return 0;
}

p.s：我正在使用kronos头文件在ubuntu 10.10中执行opengl-es，它是linux中opengl-es 2.0的模拟器。

Answer 1

根本没有投影。您将myPMVMatrix制服设置为的投影 - 模型 - 视图矩阵是

float pfIdentity[] =
{
    -1.0f,0.0f,0.0f,0.0f,
    0.0f,1.0f,0.0f,0.0f,
    0.0f,0.0f,1.0f,0.0f,
    0.0f,0.0f,0.0f,1.0f
};

/* ... */

glUniformMatrix4fv( i32Location, 1, GL_FALSE, pfIdentity);

BTW：制服的想法是，你没有在每次原始迭代中设置它们。

无论如何，这是一个单位矩阵，因为它是唯一应用的变换，它只会通过顶点传递到片段阶段。您的问题的解决方案是对其应用正交投影，即将该矩阵与正投影矩阵相乘并使用该操作的结果。 http://www.songho.ca/opengl/gl_projectionmatrix.html

Opengl-es 2.0中的正交投影矩阵

1 个答案: