C++ 基于土壤的opengl纹理图谱区域提取

我想用opengl使用土壤将纹理图集的不同区域映射到立方体的不同侧面。到目前为止，我成功地将单个图像映射到立方体的一侧：

int LoadTexture(const char*);

int texID;

void DrawCube(float width)
{
    float wd2 = width / 2;
    glColor3f(0, 0, 1);
    glBegin(GL_QUADS);

    //front
    glTexCoord2f(0, 1);
    glVertex3f(-wd2, -wd2, wd2);
    glTexCoord2f(1, 1);
    glVertex3f(wd2, -wd2, wd2);
    glTexCoord2f(1, 0);

    glVertex3f(wd2, wd2, wd2);
    glTexCoord2f(0, 0);

    glVertex3f(-wd2, wd2, wd2);

    //left side..
    //right side..
    //back..
    //top..
    //bottom..

    glEnd();
}

int LoadTexture(const char* tex) {
    texID = SOIL_load_OGL_texture(tex, 4, 0, 0);
    if (!texID) {
        cout << "Texture not loaded!\n";

    }
    return texID;
}

但我的问题是，我怎么才能只得到整个纹理中的一个精灵呢

图为：

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纹理坐标将几何体的顶点（点）映射到纹理图像中的一个点。因此，它指定了纹理的哪一部分放置在几何体的特定部分上，并与纹理参数一起指定了纹理如何包裹几何体。
通常，纹理的左下点由纹理坐标（0，0）寻址，纹理的右上点由（1，1）寻址

纹理由8列4行的瓷砖组成。若要将纹理的单个平铺放置在四边形上，必须以相同的方式分割纹理坐标。单个磁贴的角的地址如下所示：

float tiles_U = 8.0f;
float tiles_V = 4.0f;
float index_U = .... ; // index of the column in [0, tiles_U-1];
float index_V = .... ; // index of the row in [0, tiles_V-1];

float left_U  = index_U        / tiles_U;
float right_U = (index_U+1.0f) / tiles_U;

float top_V    = (tiles_V - index_V)        / tiles_V;
float bottom_V = (tiles_V - index_V - 1.0f) / tiles_V;

float wd2 = width / 2;
glColor3f(0, 0, 1);
glBegin(GL_QUADS);

glTexCoord2f( left_U, top_V );
glVertex3f(-wd2, -wd2, wd2);

glTexCoord2f( right_U, top_V );
glVertex3f(wd2, -wd2, wd2);

glTexCoord2f( right_U, bottom_V );
glVertex3f(wd2, wd2, wd2);

glTexCoord2f( left_U, bottom_V );
glVertex3f(-wd2, wd2, wd2);

glEnd();

将其应用于代码，如下所示：

float tiles_U = 8.0f;
float tiles_V = 4.0f;
float index_U = .... ; // index of the column in [0, tiles_U-1];
float index_V = .... ; // index of the row in [0, tiles_V-1];

float left_U  = index_U        / tiles_U;
float right_U = (index_U+1.0f) / tiles_U;

float top_V    = (tiles_V - index_V)        / tiles_V;
float bottom_V = (tiles_V - index_V - 1.0f) / tiles_V;

float wd2 = width / 2;
glColor3f(0, 0, 1);
glBegin(GL_QUADS);

glTexCoord2f( left_U, top_V );
glVertex3f(-wd2, -wd2, wd2);

glTexCoord2f( right_U, top_V );
glVertex3f(wd2, -wd2, wd2);

glTexCoord2f( right_U, bottom_V );
glVertex3f(wd2, wd2, wd2);

glTexCoord2f( left_U, bottom_V );
glVertex3f(-wd2, wd2, wd2);

glEnd();

---

纹理坐标将几何体的顶点（点）映射到纹理图像中的点。因此，它指定了纹理的哪一部分放置在几何体的特定部分上，并与纹理参数一起指定了纹理如何包裹几何体。
通常，纹理的左下点由纹理坐标（0，0）寻址，纹理的右上点由（1，1）寻址

纹理由8列4行的瓷砖组成。若要将纹理的单个平铺放置在四边形上，必须以相同的方式分割纹理坐标。单个磁贴的角的地址如下所示：

float tiles_U = 8.0f;
float tiles_V = 4.0f;
float index_U = .... ; // index of the column in [0, tiles_U-1];
float index_V = .... ; // index of the row in [0, tiles_V-1];

float left_U  = index_U        / tiles_U;
float right_U = (index_U+1.0f) / tiles_U;

float top_V    = (tiles_V - index_V)        / tiles_V;
float bottom_V = (tiles_V - index_V - 1.0f) / tiles_V;

float wd2 = width / 2;
glColor3f(0, 0, 1);
glBegin(GL_QUADS);

glTexCoord2f( left_U, top_V );
glVertex3f(-wd2, -wd2, wd2);

glTexCoord2f( right_U, top_V );
glVertex3f(wd2, -wd2, wd2);

glTexCoord2f( right_U, bottom_V );
glVertex3f(wd2, wd2, wd2);

glTexCoord2f( left_U, bottom_V );
glVertex3f(-wd2, wd2, wd2);

glEnd();

将其应用于代码，如下所示：

float tiles_U = 8.0f;
float tiles_V = 4.0f;
float index_U = .... ; // index of the column in [0, tiles_U-1];
float index_V = .... ; // index of the row in [0, tiles_V-1];

float left_U  = index_U        / tiles_U;
float right_U = (index_U+1.0f) / tiles_U;

float top_V    = (tiles_V - index_V)        / tiles_V;
float bottom_V = (tiles_V - index_V - 1.0f) / tiles_V;

float wd2 = width / 2;
glColor3f(0, 0, 1);
glBegin(GL_QUADS);

glTexCoord2f( left_U, top_V );
glVertex3f(-wd2, -wd2, wd2);

glTexCoord2f( right_U, top_V );
glVertex3f(wd2, -wd2, wd2);

glTexCoord2f( right_U, bottom_V );
glVertex3f(wd2, wd2, wd2);

glTexCoord2f( left_U, bottom_V );
glVertex3f(-wd2, wd2, wd2);

glEnd();

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但是你怎么计算分数呢？我的意思是，如果我想要例如第一行的第二个tile，我怎么知道glTexCoord2f的参数呢？但是你怎么计算分数呢？我的意思是，如果我想要例如第一行中的第二个tile，我如何知道glTexCoord2f的参数？