Question

所以这是我制作一堆桌子的代码。（我还是初学者）

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循环所以然后我不需要做所有这些序列垃圾，有人可以为我做一个有效的while循环代码并教我如何做它，我已经坚持了很久。

Answer 1

您可以使用内部类来存储坐标。我不确定你是否想要专注于多维数组。

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;
        }
    }

}

Answer 2

为什么不这样做，而不是为每个坐标设置单独的数组：

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

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

现在你只需要迭代这些使用循环。请记住，隐含的假设似乎是x和y都具有相同的坐标数，因此您应该能够同时迭代它们两个以获得相应的x和{ {1}}的

要求你为你编写代码可能不会起作用，因为我们并没有真正给出代码，除非你表现出很重要的工作。

Answer 3

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 [] []，其中x为大小[4] [4]，y为[4] [4]。

然后你可以创建一个while循环来迭代数组。即使是for循环也足够好。

类似这样的事情



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

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

Answer 4

这是一种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);
        }
    }
}

Answer 5

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循环也是很棒的方式，但在那个时候我设法通过一个while循环找到了如何做到这一点，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.
    }

使用带有g.drawPolygon代码的While循环

5 个答案: