一天有个小朋友问我OpenGL俄罗斯方块怎么写。
俄罗斯方块分成两部分游戏逻辑和画面渲染.
1. 游戏逻辑
一个简单的俄罗斯方块的逻辑部分需要考虑的情况如下:
1. 方块的表示(坐标, 旋转, 上下左右移动)
2. 格子的状态记录, 移动中的方块和边界的碰撞检测和已固定的方块的碰撞检测
3. 行满检测与消除
具体的面向对象实现如下:
class Game{
public:
int tiles[20][10] = {0};
void new_tile() {
tile << 5, 0, random()%7, random()%4;
}
void write_tile() {
if(!running || !started) return ;
for(int i = 0 ; i < 4; i++) {
tiles[tile[1]+tileMap[tile[2]][tile[3]][i][1]][tile[0]+tileMap[tile[2]][tile[3]][i][0]] = tile[2] + 1;
}
}
void wipe_tile() {
if(!running || !started) return ;
for(int i = 0 ; i < 4; i++) {
tiles[tile[1]+tileMap[tile[2]][tile[3]][i][1]][tile[0]+tileMap[tile[2]][tile[3]][i][0]] = 0;
}
}
void rotation() {
if(!running || !started) return ;
tile[3] = (tile[3]+1)%4;
for(int i = 0 ; i < 4; i++) {
std::pair<int,int> pos= {
tile[0]+tileMap[tile[2]][tile[3]][i][0],
tile[1]+tileMap[tile[2]][tile[3]][i][1],
};
if(pos.first < 0 || pos.first >= 10 || pos.second >= 20 || tiles[pos.second][pos.first]) {
tile[3] = (tile[3]+3)%4;
break;
}
}
}
bool move(int x, int y, bool passive = false) {
if(!running || !started) return true;
tile[0] += x;
tile[1] += y;
for(int i = 0 ; i < 4; i++) {
std::pair<int,int> pos= {
tile[0]+tileMap[tile[2]][tile[3]][i][0],
tile[1]+tileMap[tile[2]][tile[3]][i][1],
};
if(pos.first < 0 || pos.first >= 10 || pos.second >= 20 || tiles[pos.second][pos.first]) {
tile[0] -= x; tile[1] -= y;
if(passive) {
write_tile();
if(tile[1] != 0) new_tile();
else {started = false; write_end(8);}
}
return false;
}
}
check_row();
return true;
}
void update() {
if(!running || !started) return ;
if(frame_cnt++ == 24) {
wipe_tile();
move(0,1,true);
write_tile();
frame_cnt = 0;
}
}
void restart() {
memset(tiles, 0, sizeof(int)*200);
new_tile();
started = true;
running = true;
}
void resume_or_pause() {
running = !running;
}
void write_end(int i = 8) {
memset(tiles[std::max(i-1,0)], 0, 4*10*7);
tiles[i+1][0] = tiles[i+3][0] = 1;
tiles[i+0][0] = tiles[i+0][1] = tiles[i+0][2] = 1;
tiles[i+2][0] = tiles[i+2][1] = tiles[i+2][2] = 1;
tiles[i+4][0] = tiles[i+4][1] = tiles[i+4][2] = 1;
tiles[i+0][3] = tiles[i+1][3] = tiles[i+2][3] = tiles[i+3][3] = tiles[i+4][3] = 2;
tiles[i+0][6] = tiles[i+1][6] = tiles[i+2][6] = tiles[i+3][6] = tiles[i+4][6] = 2;
tiles[i+1][4] = tiles[i+2][4] = tiles[i+2][5] = tiles[i+3][5] = 2;
tiles[i+0][7] = tiles[i+1][7] = tiles[i+2][7] = tiles[i+3][7] = tiles[i+4][7] = 3;
tiles[i+0][8] = tiles[i+4][8] = tiles[i+1][9] = tiles[i+2][9] = tiles[i+3][9] = 3;
}
Eigen::Vector4i tile; // (x, y, type, rotation)
private:
// 一个二维数组表示所有可能出现的方块和方向。
static int tileMap[7][4][4][2];
void check_row() {
for(int i = 19, ii = 19; i >= 0; i--) {
int tile_cnt = 0;
for(int j = 0; j < 10; j++) tile_cnt += tiles[i][j] > 0;
for(int j = 0; j < 10; j++) tiles[ii][j] = tiles[i][j] ;
if(tile_cnt != 10) ii--;
}
}
int frame_cnt = 0;
bool running = false, started = false;
} game;
int Game::tileMap[7][4][4][2] = {
{ 0, 0, -1, 0, 1, 0, -1, -1, // "L"
0, 1, 0, 0, 0, -1, 1, -1,
1, 1, -1, 0, 0, 0, 1, 0,
-1, 1, 0, 1, 0, 0, 0, -1},
{ 0, 0, -1, -1, -1, 0, 0, -1, //"O"
0, 0, -1, -1, -1, 0, 0, -1,
0, 0, -1, -1, -1, 0, 0, -1,
0, 0, -1, -1, -1, 0, 0, -1},
{ 0, 0, 1, 0, -1, 0, -2, 0, //"I"
0, 0, 0, -2, 0, -1, 0, 1,
0, 0, 1, 0, -1, 0, -2, 0,
0, 0, 0, -2, 0, -1, 0, 1},
{ 0, 0, 1, 0, -1, -1, 0, -1, //"S"
0, 0, 0, 1, 1, 0, 1, -1,
0, 0, 1, 0, -1, -1, 0, -1,
0, 0, 0, 1, 1, 0, 1, -1},
{ 0, 0, -1, 0, 1, 0, 1, -1, //"J"
0, 0, 0, 1, 0, -1, 1, 1,
0, 0, 1, 0, -1, 0, -1, 1,
0, 0, 0, 1, 0, -1, -1, -1},
{ 0, 0, -1, 0, 0, -1, 1, -1, //"Z"
0, -1, 0, 0, 1, 0, 1, 1,
0, 0, -1, 0, 0, -1, 1, -1,
0, -1, 0, 0, 1, 0, 1, 1},
{ 0, 0, 1, 0, -1, 0, 0, -1, //"T"
0, 0, 0, 1, 0, -1, 1, 0,
0, 0, 0, 1, -1, 0, 1, 0,
-1, 0, 0, 1, 0, -1, 0, 0}
};
2. 渲染实现
对于渲染的要求,只使用一组着色器实现,即通过Uniform传所有格子的状态,具体如下:
// vertex shader
#version 330 core
in vec2 position;
out vec2 pos;
void main()
{
pos = vec2(position.x*1.14,position.y*-1.09);
gl_Position = vec4(position, 0.0, 1.0);
} ;
// fragment shader
#version 330 core
out vec4 outColor;
in vec2 pos;
uniform int tile[200];
uniform sampler2D tileTex;
void main()
{
vec2 border = smoothstep(-0.1, 0.0, -abs(sin(3.1415926*pos*vec2(5.0,10.0))));
if(max(abs(pos.x),abs(pos.y))>1.002) {outColor.xyz = texture(tileTex,vec2(pos.x/2.28 + 0.5, 360.0/393.0*(pos.y/2.18 + 0.5 )) ).xyz;return;}int tile_id = int(floor(10.0*pos.y+10.0)/*xiconxi.github.io*/*10)+int(floor(5.0*pos.x+5.0));
tile_id = tile[tile_id<200&&tile_id>=0?tile_id:0];
vec3 tile_color = texture(tileTex,vec2( (tile_id-1+mod(abs(pos.x*5),1.0))/7.0,360.0/393.0+33.0/393.0*mod(abs(pos.y*10.0), 1.0))).xyz;
tile_color = tile_id == 0 ? vec3(0.55)*length(tile_color): tile_color;
outColor.xyz = mix(tile_color ,vec3(0.0),max(border.x,border.y));
} ;
总体渲染效果如下:
具体代码在Github::Tetris