p5.j​​s中嵌套for循环的重复方形/角螺旋？

Question

我正在尝试用有角的螺旋线制作一个网格。螺旋本身由 for 循环中的单行组成。当我沿一个轴（x 或 y）复制和移动（平移）螺旋的原点时，它会起作用。但是同时沿 (x AND y) 移动，为了使它成为一个网格，不分解螺旋是行不通的。

如果有人能帮助我解决我的编码难题，我将不胜感激。顺便说一句，我对任何提示都非常开放，并有助于提高我的代码编写技巧。里面肯定有很多冗余和冗长的表达...... 到目前为止，这是我的代码：

function drawSpiral() {
  let count = 8;
  let stepX = 8;
  let stepY = 8;
  let tileSize = 100;
  let pixelX = tileSize;
  let pixelY = tileSize;
  for (let j = 0; j < 5; j++) {
    let x1 = 0;
    let y1 = 0;
    let x2 = 0;
    let y2 = 0;
    let x3 = 0;
    let y3 = 0;
    let x4 = 0;
    let y4 = 0;
    for (let i = 0; i < count; i++) {
      x1 += stepX;
      x2 -= stepX;
      x3 -= stepX;
      x4 += stepX;
      y1 += stepY;
      y2 += stepY;
      y3 -= stepY;
      y4 -= stepY;
      push();
      translate(pixelX, pixelY);
      line(x1, y1, x2 - stepX, y2)
      line(x2 - stepX, y2, x3 - stepX, y3 - stepY);
      line(x3 - stepX, y3 - stepY, x4 + stepX, y4 - stepY);
      line(x4 + stepX, y4 - stepY, x1 + stepX, y1 + stepY);
      pop();
      
    }
    pixelX += tileSize * 2; //shifting either along x-axis
  }
}

多美啊？是的，你猜对了——我是编码行业的新手；）

Answer 1

如果您尝试制作一个螺旋网格，看起来您只需要在当前拥有for (let j = 0; j < 5; j++) { 的地方使用一对 for 循环。几乎任何时候你想创建一个网格，你都需要一对嵌套的 for 循环。

function setup() {
  createCanvas(800, 800);
}

function draw() {
  background(100);
  drawSpiral();
}

function drawSpiral() {
  let count = 8;
  let stepX = 8;
  let stepY = 8;
  let tileSize = 100;
  let pixelX = tileSize;
  let pixelY = tileSize;
  // Make a 5x5 grid of spirals
  for (let row = 0; row < 5; row++) {
    for (let col = 0; col < 5; col++) {
      let x1 = 0;
      let y1 = 0;
      let x2 = 0;
      let y2 = 0;
      let x3 = 0;
      let y3 = 0;
      let x4 = 0;
      let y4 = 0;
      for (let i = 0; i < count; i++) {
        x1 += stepX;
        x2 -= stepX;
        x3 -= stepX;
        x4 += stepX;
        y1 += stepY;
        y2 += stepY;
        y3 -= stepY;
        y4 -= stepY;
        push();
        translate(pixelX, pixelY);
        line(x1, y1, x2 - stepX, y2)
        line(x2 - stepX, y2, x3 - stepX, y3 - stepY);
        line(x3 - stepX, y3 - stepY, x4 + stepX, y4 - stepY);
        line(x4 + stepX, y4 - stepY, x1 + stepX, y1 + stepY);
        pop();

      }
      // Sift right for each col
      pixelX += tileSize * 2;
    }
    // Shift down for each row
    pixelY += tileSize * 2;
    // And reset the horizontal position at the end of each row
    pixelX = tileSize;
  }
}

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