Javascript 这是一个无限循环,但我一生都不知道为什么
因此,我正在制作一个小的网络应用程序,以更好地使用画布,但我被卡住了。我想要一个旋转的n边多边形(画线已经可以了)。游戏循环通过一个网格数组循环(网格上的每个点都包含一个point()对象的子类),并对每个点调用tick()方法。在碰到ShapePoint()对象(鼠标中键放置在画布上)之前,一切都正常。ShapePoint的tick()方法在某种程度上是一个无限循环。如果你在里面放一个console.log(“hi”),它会给你2000条“hi”消息,所以它(理论上)可以工作。有趣的是,即使它在stoke()中循环,也没有发生任何事情Javascript 这是一个无限循环,但我一生都不知道为什么,javascript,jquery,loops,canvas,infinite,Javascript,Jquery,Loops,Canvas,Infinite,因此,我正在制作一个小的网络应用程序,以更好地使用画布,但我被卡住了。我想要一个旋转的n边多边形(画线已经可以了)。游戏循环通过一个网格数组循环(网格上的每个点都包含一个point()对象的子类),并对每个点调用tick()方法。在碰到ShapePoint()对象(鼠标中键放置在画布上)之前,一切都正常。ShapePoint的tick()方法在某种程度上是一个无限循环。如果你在里面放一个console.log(“hi”),它会给你2000条“hi”消息,所以它(理论上)可以工作。有趣的是,即使它
//################################################################
// THIS IS THE PROBLEM CLASS.
// So pretty much, when the game loop calls the tick() funciton
// of ANY ShapePoint object, everything hangs. The game is still
// looping through the ENTIRE tick() function (put console.log()
// functions in and you'll see what I mean) continually, but the
// effects it is supposed to display aren't shown.
//
//################################################################
function ShapePoint(x, y, sides) {
//position variable
this.positionOnCanvas = [x, y];
//number of sides
this.N = sides;
//current step
this.step = 0;
//the array to store all the angle data
this.thetas = new Array(this.N);
//the array to store all the vertex data
this.vertList = new Array(this.N);
//function to increase all the angels by an even amount
this.stepPoints = function(s) {
//for every side
for (i=0; i<this.N; i++) {
//multiply the current 'i' value by ((360/number of sides) + current step). This serves to create points at even intervals all the way around a circle, and have it increase by s every loop
this.thetas[i] = i*((360/this.N) + s);
//get the x value with 40*cos(angle for this point). Same for y, only with sin. Round both to 2 decimal places
this.vertList[i] = [Math.round((40*(Math.cos(this.thetas[i])))*100)/100, Math.round((40*(Math.sin(this.thetas[i])))*100)/100];
//if the current angle is between 90 and 180...
if (this.thetas[i]>=90 && this.thetas[i]<=180) {
//invert the x value
this.vertList[i][0] *= -1;
//else if the angle is between 180 and 270...
} else if (this.thetas[i]>=180 && this.thetas[i]<=270) {
//invert both the x and the y values
this.vertList[i][0] *= -1;
this.vertList[i][1] *= -1;
//else if the angle is between 270 and 360...
} else if (this.thetas[i]>=270 && this.thetas[i]<=360) {
//invert the y value
this.vertList[i][1] *= -1;
}
//nothing needed for 0-90 because both are positive
}
}
this.tick = function() { //<<<<<<<<THIS IS THE PROBLEM FUNCTION!
//setup all the points forward
this.stepPoints(this.step);
//for every side in this polyogn...
for (i=0; i<this.N; i++) {
//shorten the location of the positions
var posX = this.vertList[i][0] + this.positionOnCanvas[0];
var posY = this.vertList[i][1] + this.positionOnCanvas[1];
//begin drawing
ctx.beginPath();
//move to the x and y location of the current point
ctx.moveTo(posX, posY);
//if point is not the last in the array...
if (i < this.N-1) {
//draw a line to the next point in the array
ctx.lineTo(this.vertList[i+1][0] + this.positionOnCanvas[0], this.vertList[i+1][1] + this.positionOnCanvas[1]);
//else...
} else {
//draw a line to the first point in the array
ctx.lineTo(this.vertList[0][0] + this.positionOnCanvas[0], this.vertList[0][1] + this.positionOnCanvas[1]);
}
//draw a line
ctx.strokeStyle = "#000000";
ctx.lineWidth = 0.5;
//end
ctx.stroke();
//draw the vertex
ctx.fillStyle = "orange";
ctx.fillRect(posX-2, posY-2, 4, 4);
}
//draw the origin of the polygon
ctx.fillStyle = "lightPurple";
ctx.fillRect(this.positionOnCanvas[0]-2, this.positionOnCanvas[1]-2, 4, 4);
//if the step is greater than 360, set it to 0
this.step = this.step % 36; //(thanks Neikos!)
}
}
ShapePoint.prototype = new Point();
//################################################################
//这是问题课。
//差不多,当游戏循环调用tick()函数时
//在任何ShapePoint对象中,所有对象都挂起。比赛还在继续
//循环遍历整个tick()函数(put console.log())
//函数,你会明白我的意思)不断地,但是
//它应该显示的效果没有显示。
//
//################################################################
函数形状点(x、y、边){
//位置变量
this.positionOnCanvas=[x,y];
//边数
这个。N=边;
//当前步骤
这一步=0;
//用于存储所有角度数据的数组
this.thetas=新数组(this.N);
//用于存储所有顶点数据的数组
this.vertList=新数组(this.N);
//函数将所有天使的数量增加到偶数
this.stepPoints=函数{
//四面八方
因为(i=0;i=90&&this.thetas[i]=180&&this.thetas[i]=270&&this.thetas[i]在他/她的上述评论中是正确的:您在步骤点
和勾选
中都使用了一个全局i
变量,因此这些方法相互干扰
有些语言中,方法中使用的变量是隐式局部变量,除非另有声明,但JavaScript不是这些语言中的一种。在JavaScript中,您需要使用var
关键字声明局部变量,否则它们是隐式全局变量。不是直接答案,但您可以执行this.step=this.step%360
将其限制在一定范围内。这只是猜测,但两个循环中的i
的声明是同一个变量吗?@user2310289否,不同的方法,因此它们不是interfere@Lampitosgames但是您未能将i
声明为局部变量。在没有显式var
声明的情况下,它是全局变量。@Lampitosgames-wh他当时说的是…只是把var i=0
改为i=0
谢谢,直到现在才意识到这一点!也要归功于用户,对不起,伙计XD