| #13 Line regular (700 bytes)
|
shortest snake game.html |
Compressed 500 bytes (of course the two documents can still be compressed with eval) |
index.500bytes.html |
It was very popular before 20 lines of code address (with BUG) (900bytes) |
##hj7jay/article/details/51011269
|
One-dimensional array700char
##The snake body at (0,0) position It is represented by 0, (0,1) is represented by 1, (1,0) is represented by 10, and so on
Because it is 13 line js,
Line 4 is a statement
Line 5 is more difficult To understand, you can split the ?: operator into 4 lines of if statements; you can refer to the following two-dimensional array version
Line 90|x Both ~~x and x>>0 can remove the mantissa of x(number)
It should be easy to read
<!DOCTYPE html>
<canvas id="1" width="400" height="400"></canvas>
<script>
let dir=1,food=3,snk=[1,0],ctx=document.getElementById("1").getContext("2d")
document.onkeydown=e=>{dir = snk[0]-snk[1]==-(tmp = [-1,-10,1,10][e.keyCode-37]||dir)?dir:tmp }
setInterval(()=>{
snk.unshift(Head = snk[0] + dir)
if(Head!=food) snk.pop()
else while(snk.includes(food=0|Math.random()*10*10)) ;
if(snk.indexOf(Head,1)!=-1||(dir==1&&Head%10==0)||(dir==-1&&Head%10==9)||Head<0||Head>=100)
return document.write(0&snk.shift()) //死亡记录蛇长
for(let i=0; i<100; i++){
ctx.fillStyle = '#0'+(food==i)*9910+snk.includes(i)*1990
ctx.fillRect(i%10*40,(i-i%10)*4, 40,40)
}
},100)
</script>
Color Effect
<!DOCTYPE html>
<canvas id="1" width="400" height="400"></canvas>
<script>
let dir=1,food=3,snk=[1,0],n_=0, ctx=document.getElementById("1").getContext("2d")
document.onkeydown=e=>{ dir =snk[0]-snk[1]==-(tmp = [-1,-10,1,10][e.keyCode-37]||dir)?dir:tmp }
setInterval(()=>{
snk.unshift(Head = snk[0] + dir)
if(Head!=food) snk.pop()
else while(snk.includes(food=0|Math.random()*10*10)) ;
if(snk.indexOf(Head,1)!=-1||(dir==1&&Head%10==0)||(dir==-1&&Head%10==9)||Head<0||Head>=100)
return alert("died"+ ++n_+"times") //死亡记录死亡次数
for(let i=0 ; i<100; i++){
ctx.fillStyle = '#0'+~~((food===i)*13000*Math.random())+~~(snk.includes(i)*3000*Math.random())
ctx.fillRect(i%10*40,(i-i%10)*4, 40,40)
}
},120)
</script>
Instructions
If you want to write it yourself:You should pay attention to two points
The snake's tail should disappear before the snake's head,The snake's head should be generated before the food,
-
Snake cannot walkCurrentlyThe opposite direction
can be used A snake of length 4 is tested
##Coding style
- Omit unnecessary tags
https://google.github.io/styleguide/htmlcssguide.html#Optional_Tags
- if() return only uses one line
https://google.github.io/styleguide/cppguide.html
- Except for keyboard response where using the
3 operator can save 3 lines, there is no need to use in other places
Two-dimensional array
The above are all implemented using one-dimensional arrays, and the following are written using two-dimensional arrays; to simplify It can also be simplified to within 17 lines 900char (just shorter than the 20-line one), but there is no need for
with comments 1100char<!DOCTYPE html>
<canvas id="1" width="400" height="400" style="border: 1px solid "></canvas>
<script>
ctx = document.getElementById("1").getContext("2d") //CanvasRenderingContext2D inferface
let Len = 10, dir = 2, dirNow ; //dirNow 后面解释
food = [3, 0]; Snake = [[0, 0], [1, 0]] //食物的坐标,蛇身的坐标用Snake数组记录
Map = {'0,0':'#52a', '1,0':'#52a'} //用来记录绘图颜色的 地图
dirMat = [[-1, 0], [0, -1], [1, 0], [0, 1]] //方向矩阵
pairEq = ((p1, p2) => p1[0] == p2[0] && p1[1] == p2[1]) //检测 两数对 是否相等的函数
document.onkeydown = e =>{
if (37 <= e.keyCode == e.keyCode < 41 && dirNow != ( (e.keyCode - 35) % 4) ) //确定是 方向键 并且 保证方向与当前运动方向相反
dir = e.keyCode -37
}
!function () {
Head = Snake[Snake.length-1].map((x, i) => x + dirMat[dirNow=dir][i]); //得到头部接下来的移动位置
if (!pairEq(Head, food))
Map[Snake.shift()]='#fff' //必须先删尾巴,才能加入头部,吃没吃到食物是唯一判断标准
if (Snake.some(x=>pairEq(x,Head)) || !Head.every(x => 0<=x == x < Len)) //判断蛇头是否撞到蛇身或墙壁
return document.write("Game Over") //这样调用document.write会把页面全部清空
Snake.push(Head); //可以加入头部
while (Snake.some(x => pairEq(x, food))) //加入新头后, 生成食物更方便
food = [~~(Math.random() * Len), ~~(Math.random() * Len)]; //因为js没有整形的概念, ~~ 现在相当于向原点舍去
Map[Head] = '#52a' ; Map[food] = '#ad5'
for( k in Map){
ctx.fillStyle= Map[k]
ctx.fillRect(parseInt(k[0])*40,parseInt(k[2])*40,40,40) //e.g. k="1,3",也因此地图大小限制为10
}
setTimeout(arguments.callee, 100); //100ms后调用此函数一次
}()
</script>
Colorful(Mainly more beautiful )1100char<!DOCTYPE html>
<canvas id="1" width="400" height="400" style="border: 1px solid " ></canvas>
<script>
ctx = document.getElementById("1").getContext("2d")
let Len = 10, dir = 2, dirNow ;
food = [3, 0]; Snake = [[0, 0], [1, 0]]
Map = {'0,0':'fff', '1,0':'fff'}
dirMat = [[-1, 0], [0, -1], [1, 0], [0, 1]]
pairEq = ((p1, p2) => p1[0] == p2[0] && p1[1] == p2[1])
document.onkeydown = e =>{
if (37 <= e.keyCode == e.keyCode < 41 && dirNow != ( (e.keyCode - 35) % 4) )
dir = e.keyCode -37
}
!function () {
Head = Snake[Snake.length-1].map((x, i) => x + dirMat[dirNow=dir][i]);
if (!pairEq(Head, food)) Map[Snake.shift()]='0'
if (Snake.some(x=>pairEq(x,Head)) || !Head.every(x => 0<=x == x < Len))
return document.write("Game Over")
Snake.push(Head);
while (Snake.some(x => pairEq(x, food)))
food = [~~(Math.random() * Len), ~~(Math.random() * Len)];
Map[Head] = Map[food] = 'fff'
for( k in Map){
ctx.fillStyle='#'+(0xfff-~~(parseInt(Map[k],16)*Math.random())).toString(16)
ctx.fillRect(parseInt(k[0])*40,parseInt(k[2])*40,46,43)
}
setTimeout(arguments.callee, 100);
}()
</script> The above is the entire content of this article. I hope it will be helpful to everyone’s study. Please pay attention to more related content. PHP Chinese website!
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