Question

我一直在尝试画一个小屋（作为一个顶部有一个圆锥的圆柱体），并在墙上添加一个砖块，在屋顶上添加一个屋顶瓦片纹理。但是，我只获得了第一个加载的纹理（砖块）。

这是我的代码（请注意我尝试使用glActiveTexture在纹理之间切换）：

void drawCylinder()
{
int width, height;

unsigned char * data_for_wall = SOIL_load_image("./bricks.jpg", &width, &height, NULL, 0);

glDisable(GL_LIGHTING);

glGenTextures( 2, textures );

glActiveTexture(GL_TEXTURE0);

glEnable( GL_TEXTURE_2D );

glBindTexture(GL_TEXTURE_2D, textures[0]);

glTexParameterf( GL_TEXTURE_2D, GL_TEXTURE_MAG_FILTER, GL_LINEAR_MIPMAP_LINEAR);
glTexParameterf( GL_TEXTURE_2D, GL_TEXTURE_MIN_FILTER, GL_LINEAR);

glTexParameterf( GL_TEXTURE_2D, GL_TEXTURE_WRAP_S, GL_REPEAT);
glTexParameterf( GL_TEXTURE_2D, GL_TEXTURE_WRAP_T, GL_REPEAT);

// Generate mipmaps, by the way.
glGenerateMipmap(GL_TEXTURE_2D);

gluBuild2DMipmaps(GL_TEXTURE_2D, GL_RGB, width, height, GL_RGB, GL_UNSIGNED_BYTE, data_for_wall);
for(float theta = 0.0; theta <= 360.0; theta += 10.0 )
{
    //colors[k] = color4(0.69, 0.35, 0.0, 1.0); //This color is brown.
    tex_coords[global_index]=vec2(theta*DegreesToRadians, 0.0);
    float x = 0.15*sin(theta*DegreesToRadians);
    float y = 0.15*cos(theta*DegreesToRadians);
    points[global_index] = Translate(0.6, 0.0, 0.35)*point4(x, 0.0, y, 1.0);

     // This is the
    // bottom of the cylinder. The points are plotted in a full circle. The first three numbers are the x,y and z values
    // respectively. The last number (ie. 1.0) is not important - it just converts to homogeneous coordinates.
    ++global_index;
    tex_coords[global_index] = vec2(theta*DegreesToRadians, 0.25);
    points[global_index] = Translate(0.6, 0.0, 0.35)*point4(x, 0.25, y, 1.0);
    // This is the
    // top of the cylinder
    ++global_index;
}
}

//The roof of the hut
void drawCone()
{

int width, height;

unsigned char * data_for_roof = SOIL_load_image("./roof_tiles.jpg", &width, &height, NULL, 0);
glDisable(GL_TEXTURE_2D);


glActiveTexture(GL_TEXTURE1);


glBindTexture( GL_TEXTURE_2D, textures[1] );

glTexParameteri( GL_TEXTURE_2D, GL_TEXTURE_MAG_FILTER, GL_LINEAR_MIPMAP_LINEAR);
glTexParameteri( GL_TEXTURE_2D, GL_TEXTURE_MIN_FILTER, GL_LINEAR);

glTexParameterf( GL_TEXTURE_2D, GL_TEXTURE_WRAP_S, GL_REPEAT);
glTexParameterf( GL_TEXTURE_2D, GL_TEXTURE_WRAP_T, GL_REPEAT);

// Generate mipmaps, by the way.
glGenerateMipmap(GL_TEXTURE_2D);

gluBuild2DMipmaps(GL_TEXTURE_2D, GL_RGB, width, height, GL_RGB, GL_UNSIGNED_BYTE, data_for_roof);
int index = 0;
    int l = 0;
    for(float theta = 0.0; theta <= 360.0; theta += 10.0)
    {
    tex_coords[global_index]=vec2(theta*DegreesToRadians, 0.25);

    points[global_index] = Translate(0.6, 0.0, 0.35)*point4(0.0, 0.5, 0.0, 1.0); // This is the top of the cone.
    ++global_index;

    tex_coords[global_index] = vec2(theta*DegreesToRadians, 0.5);

    points[global_index] = Translate(0.6, 0.0, 0.35)*point4(0.25*sin(theta*DegreesToRadians), 0.25, 0.25*cos(theta*DegreesToRadians), 1.0);
     // This is the
    // bottom of the cone.
    ++global_index;

    }
}

这是显示功能：

void
display( void )
{
mat4 mv = Translate(0.0, -0.065, -rad)*RotateX(Theta[0])*RotateY(Theta[1])*RotateZ(Theta[2]);
mat4 p = Perspective(10.0, aspectRatio, zNear, zFar);
glUniformMatrix4fv( matrix_loc, 1, GL_TRUE, mv );
glUniformMatrix4fv( projection_loc, 1, GL_TRUE, p );
glClear( GL_COLOR_BUFFER_BIT | GL_DEPTH_BUFFER_BIT );

glUniform3fv( theta, 1, Theta );

glActiveTexture(GL_TEXTURE0);
glBindTexture(GL_TEXTURE_2D, textures[0]);
glDrawArrays( GL_TRIANGLE_STRIP, 0, 74 );

glActiveTexture(GL_TEXTURE1);
glBindTexture(GL_TEXTURE_2D, textures[1]);
glDrawArrays( GL_TRIANGLE_STRIP, 74, 74);

glutSwapBuffers();
}

我不确定包含片段着色器是否很重要但我仍会这样做： #version 150

in  vec2 texCoord;

out vec4 fColor;

uniform sampler2D texture;

void main()
{
fColor = texture2D( texture, texCoord );
}

我一直在努力争取这个权利。有谁知道我做错了什么？

Answer 1

glActiveTexture（）用于设置在多纹理处理时绑定纹理的纹理槽（在单个通道中渲染多个纹理）。例如，您可能有一个纹理贴图纹理，另一个纹理贴图，等等。

但是，您不是多纹理，因为您在单独的传递中渲染墙和圆锥（即两次单独调用glDrawArrays（）），并且着色器每次传递仅使用一个纹理。因此，在每个传递中，您只渲染单个纹理，该纹理将位于插槽0中。

如果将活动纹理设置为GL_TEXTURE1然后调用glBindTexture（），则纹理将绑定到插槽1，插槽0将保持不变。

因此，将活动纹理槽两次设置为0。从display（）函数中删除此行：

glActiveTexture(GL_TEXTURE1);

opengl - 只显示一个纹理

1 个答案: