如何在画布上绘制一个简单的五边形

Brk*_*Brk 5 html javascript canvas polygon

我想创建一个五角大楼,我成功地创建了这样的五角大楼。

但是我的五边形是不对的,因为它在表面上不正确。

我该如何解决?我需要一个优雅的答案,而不仅仅是快速修复

更新:

我还想知道另一件事:

如何仅使用坐标绘制五边形,我的意思是五边形的 5 个坐标?

我想根据五个已知坐标(v1,v2..v5)绘制一个五边形并且没有任何循环,以在五个点之间绘制某种路径。

$(function(){
var canvas=document.getElementById("canvas");
var cxt=canvas.getContext("2d");
    // hexagon
    var numberOfSides = 5,
        size = 100,
        Xcenter = 150,
        Ycenter = 150;

    cxt.beginPath();
    cxt.moveTo (Xcenter +  size * Math.cos(0), Ycenter +  size *  Math.sin(0));          

    for (var i = 1; i <= numberOfSides;i += 1) {
      cxt.lineTo (Xcenter + size * Math.cos(i * 2 * Math.PI / numberOfSides), Ycenter + size * Math.sin(i * 2 * Math.PI / numberOfSides));
    }

    cxt.strokeStyle = "#000000";
    cxt.lineWidth = 1;
    cxt.stroke();
  });
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<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<canvas id="canvas" width=650 height=500></canvas>
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小智 5

有趣的问题,您可以移动正弦/余弦值,使底线水平对齐:

$(function(){
   var canvas=document.getElementById("canvas");
   var cxt=canvas.getContext("2d");
    // hexagon
    var numberOfSides = 5,
        size = 100,
        Xcenter = 150,
        Ycenter = 150,
        step  = 2 * Math.PI / numberOfSides,//Precalculate step value
        shift = (Math.PI / 180.0) * -18;//Quick fix ;)

    cxt.beginPath();
    //cxt.moveTo (Xcenter +  size * Math.cos(0), Ycenter +  size *  Math.sin(0));          

    for (var i = 0; i <= numberOfSides;i++) {
    	var curStep = i * step + shift;
        cxt.lineTo (Xcenter + size * Math.cos(curStep), Ycenter + size * Math.sin(curStep));
    }

    cxt.strokeStyle = "#000000";
    cxt.lineWidth = 1;
    cxt.stroke();
  });
Run Code Online (Sandbox Code Playgroud)
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<canvas id="canvas" width=650 height=500></canvas>
Run Code Online (Sandbox Code Playgroud)