60行js代码实现俄罗斯方块

2024-05-06 16:17:59

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这篇文章主要介绍了60行js代码实现俄罗斯方块，可能会有些小bug，不过总体上还是比较不错的。大伙来研究下代码，给出意见。

这是我之前网上看到的,很牛逼的一位大神写的,一直膜拜中

<!doctype html><html><head></head><body>

<div id="box" style="width:252px;font:25px/25px 宋体;background:#000;color:#9f9;border:#999 20px ridge;text-shadow:2px 3px 1px #0f0;"></div>

<script>

var map=eval("["+Array(23).join("0x801,")+"0xfff]");

var tatris=[[0x6600],[0x2222,0xf00],[0xc600,0x2640],[0x6c00,0x4620],[0x4460,0x2e0,0x6220,0x740],[0x2260,0xe20,0x6440,0x4700],[0x2620,0x720,0x2320,0x2700]];

var keycom={"38":"rotate(1)","40":"down()","37":"move(2,1)","39":"move(0.5,-1)"};

var dia, pos, bak, run;

function start(){

dia=tatris[~~(Math.random()*7)];

bak=pos={fk:[],y:0,x:4,s:~~(Math.random()*4)};

rotate(0);

}

function over(){

document.onkeydown=null;

clearInterval(run);

alert("GAME OVER");

}

function update(t){

bak={fk:pos.fk.slice(0),y:pos.y,x:pos.x,s:pos.s};

if(t) return;

for(var i=0,a2=""; i<22; i++)

a2+=map[i].toString(2).slice(1,-1)+"<br/>";

for(var i=0,n; i<4; i++)

if(/([^0]+)/.test(bak.fk[i].toString(2).replace(/1/g,"/u25a1")))

a2=a2.substr(0,n=(bak.y+i+1)*15-RegExp.$_.length-4)+RegExp.$1+a2.slice(n+RegExp.$1.length);

document.getElementById("box").innerHTML=a2.replace(/1/g,"/u25a0").replace(/0/g,"/u3000");

}

function is(){

for(var i=0; i<4; i++)

if((pos.fk[i]&map[pos.y+i])!=0) return pos=bak;

}

function rotate(r){

var f=dia[pos.s=(pos.s+r)%dia.length];

for(var i=0; i<4; i++)

pos.fk[i]=(f>>(12-i*4)&15)<<pos.x;

update(is());

}

function down(){

++pos.y;

if(is()){

for(var i=0; i<4 && pos.y+i<22; i++)

if((map[pos.y+i]|=pos.fk[i])==0xfff)

map.splice(pos.y+i,1), map.unshift(0x801);

if(map[1]!=0x801) return over();

start();

}

update();

}

function move(t,k){

pos.x+=k;

for(var i=0; i<4; i++)

pos.fk[i]*=t;

update(is());

}

document.onkeydown=function(e){

eval(keycom[(e?e:event).keyCode]);

};

start();

run=setInterval("down()",400);

</script></body></html>