0

所以这是我并排制作一堆表格的代码。（我还是个初学者）

import java.applet.Applet;
import java.awt.*;

public class Test extends Applet{

public void init() {
setSize(500, 225);
}

public void paint (Graphics g){

//Desk #1
int [ ] x8 = {430, 430, 351, 351};
int [ ] y8 = {200, 185, 185,200};
g.drawPolygon(x8, y8, 4);

//Desk #2
int [ ] x9 = {351, 351, 272, 272};
int [ ] y9 = {200, 185, 185, 200};
g.drawPolygon(x9, y9, 4);

//Desk #3
int [ ] x10 = {272, 272, 193, 193};
int [ ] y10 = {185, 200, 200, 185};
g.drawPolygon(x10, y10, 4);

//Desk #4
int [ ] x11 = {193, 193, 114, 114};
int [ ] y11 = {185, 200, 200, 185};
g.drawPolygon(x11, y11, 4);

//Desk #5
int [ ] x12 = {114, 114, 35, 35};
int [ ] y12 = {185, 200, 200, 185};
g.drawPolygon(x12, y12, 4);

}
}

我想要做的只是做一个while循环，所以我不需要做所有这些序列垃圾，有人可以为我制作一个有效的while循环代码并教我他们是如何做到的，我'长期以来一直坚持这一点。

4

5 回答 5

1

您可以使用内部类来存储坐标。我不确定您是否想专注于多维数组。

public class Test extends Applet {

    Poly desk1 = new Poly(new int[] {430, 430, 351, 351}, new int[] {200, 185, 185,200});
    Poly desk2 = new Poly(new int[] {351, 351, 272, 272}, new int[] {200, 185, 185, 200});
    Poly desk3 = new Poly(new int[] {272, 272, 193, 193}, new int[] {185, 200, 200, 185});
    Poly desk4 = new Poly(new int[] {193, 193, 114, 114}, new int[] {185, 200, 200, 185});
    Poly desk5 = new Poly(new int[] {114, 114, 35, 35}, new int[] {185, 200, 200, 185});

    Poly[] desks = new Poly[] {desk1, desk2, desk3, desk4, desk5};

    public void init() {
        setSize(500, 225);
    }

    public void paint (Graphics g) {
        for (int i = 0; i < desks.length; i++) {
            g.drawPolygon(desks[i].xs, desks[i].ys, 4);
        }
    }

    private static class Poly {
        // public fields are sometimes frowned upon,
        // but for a private class and a simple example
        public int[] xs;
        public int[] ys;

        public Poly(int[] xs, int[] ys) {
            this.xs = xs;
            this.ys = ys;
        }
    }

}
于 2013-04-25T21:53:04.250 回答
0

与其为每个坐标设置单独的数组，不如使用这样的东西：

int[][] x = {
    {123, 534, 643},
    {123, 543, 152},
    ...
    {543, 125, 163}
};

int[][] y = {
    {123, 534, 643},
    {123, 543, 152},
    ...
    {543, 125, 163}
};

现在您只需要使用循环遍历这些。请记住，这里的隐含假设似乎是 x 和 y 具有相同数量的坐标，因此您应该能够同时迭代它们以获得相应x的 ' 和y'。

要求 SO 为您编写代码可能行不通，因为除非您展示了自己的重要工作，否则我们不会真正在此处提供代码。

于 2013-04-25T21:47:43.407 回答
0

我不会给你答案，但这可能会有所帮助。创建一个二维数组 int[][]，x 为 [4][4]，y 为 [4][4]。

然后你可以创建一个while循环来遍历数组。即使是 for 循环也足够好。

像这样的东西

for(int i = 0 ; i < 4; i++) {
  g.drawPolygon(x[i], y[i]);
}

于 2013-04-25T21:48:03.983 回答
0

这是使用简单数学计算坐标的替代解决方案。

public class Test extends Applet {
    public void init() {
        setSize(500, 225);
    }

    public void paint(Graphics g) {
        int deskCount = 5;
        int[] y = { 200, 185, 185, 200 };
        for (int i = 0; i < deskCount; i++) {
            int[] x = { 430 - i * 79, 430 - i * 79, 351 - i * 79, 351 - i * 79 };
            g.drawPolygon(x, y, 4);
        }
    }
}
于 2013-04-25T21:57:07.823 回答
0

感谢大家的工作，它确实有所帮助，所以谢谢你和最后一个人，for循环也很棒，但在那段时间我设法找到了如何用while循环自己做，如下所示！:)

    //Desk #1-4 #Using While Loop
    int c=0;
    int [ ] x1 = {65, 65, 153, 153};
    int [ ] y1 = {20, 35, 35, 20};
    g.drawPolygon(x1, y1, 4);


    while (c <= 3) {
        g.drawPolygon(x1, y1, 4);

        x1[0] += 88;
        x1[1] += 88;
        x1[2] += 88;
        x1[3] += 88;

        c += 1;  // c = c + 1; - Same thing.
    }
于 2013-05-18T01:34:32.820 回答