一天有个淘宝网店客服问我OpenGL俄罗斯方块, 没错就是XXX作业代写。
俄罗斯方块分成两部分游戏逻辑和画面渲染.
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