Oop 如何在处理过程中单独控制以编程方式创建的对象的实例?

Oop 如何在处理过程中单独控制以编程方式创建的对象的实例?,oop,object,processing,Oop,Object,Processing,在处理过程中,我试图创建一个显示x数量形状的可视化。由于数量可能相当大且不确定,因此我希望以编程方式(使用循环)创建它们。这看起来像这样: First firstDot1; float offset; float radius = 0; float endRad = 100; float speed = 0.1; String[] lines,colors; void setup(){ size(800,600); smooth(); background(255);

在处理过程中,我试图创建一个显示x数量形状的可视化。由于数量可能相当大且不确定,因此我希望以编程方式(使用循环)创建它们。这看起来像这样:

First firstDot1;
float offset;
float radius = 0;
float endRad = 100;
float speed = 0.1;
String[] lines,colors;

void setup(){
    size(800,600);
    smooth();
    background(255); 
    firstDot1 = new First(5);
}

void draw(){
    background(255);
    for(int z=0; z<36; z++){
        offset = z * 10;
        firstDot1.display();
        firstDot1.start(offset);
    }
}    

class First{
    float angle;
    int id;
    float eRad = 5;
    float xpos, ypos, rad, i;
    Boolean start = true;

    First(float tempAngle){
        angle = tempAngle;
    } 

    void display(){
        noStroke();
        fill(247,147,30); 
        ellipseMode(CENTER);
        ellipse(xpos, ypos, eRad, eRad);
    }

    void start(float offset){
        if(i<endRad){
            i = i+speed/2;
            xpos = width/2 + cos(radians(-angle+offset))*(radius+i);
            ypos = height/2 + sin(radians(-angle+offset))*(radius+i);
        }
        else{
            turn(offset);
        }
    }

    void turn(float offset){
        angle = angle-speed/50;
        xpos = width/2 + cos(radians(-angle+offset))*(endRad);
        ypos = height/2 + sin(radians(-angle+offset))*(endRad);  
    }
}
First firstDot1;
浮动偏移量;
浮动半径=0;
浮点数endRad=100;
浮动速度=0.1;
字符串[]行、颜色;
无效设置(){
规模(800600);
光滑的();
背景(255);
firstDot1=新的第一个(5);
}
作废提款(){
背景(255);

对于(int z=0;z,您需要创建第一个对象的数组:

First[] dots = new First[36];//declare and initialize an array of First object with size/length 36
然后需要初始化数组中的对象:

for(int i = 0 ; i < dots.length; i++) dots[i] = new First(5);

另外,请查看示例>主题>GUI>按钮

非常感谢。我一直在玩数组的概念,但没有弄明白。您的答案很好,感谢您的努力!不用担心。使用单独的真正基本的草图来测试新事物(如数组或原语、对象数组)可能会更容易,然后编写更复杂的草图。很高兴答案有帮助:)
First[] dots;
float offset;
float radius = 0;
float endRad = 100;
float speed = 0.1;
String[] lines,colors;

void setup(){
    size(800,600);
    smooth();
    background(255); 
    dots = new First[36];
    for(int i = 0 ; i < dots.length; i++) dots[i] = new First(5);
}

void draw(){
    background(255);
    for(int z=0; z<36; z++){
        offset = z * 10;
        dots[z].update(mouseX,mouseY,mousePressed);
        dots[z].display();
        dots[z].start(offset);
    }
}    

class First{
    float angle;
    int id;
    float eRad = 5;
    float xpos, ypos, rad, i;
    Boolean start = true;

    color up = color(247,147,30);
    color over = color(0);
    color down = color(217,117,0);
    boolean isOver,isDown;

    First(float tempAngle){
        angle = tempAngle;
    } 
    void update(int mx,int my,boolean pressed){
      isOver = (dist(mx,my,xpos,ypos) < eRad);
      isDown = pressed;
    }

    void display(){
        noStroke();
        fill(up);//default
        if(isOver) fill(over);
        if(isOver && isDown) fill(down);
        ellipseMode(CENTER);
        ellipse(xpos, ypos, eRad, eRad);
    }

    void start(float offset){
        if(i<endRad){
            i = i+speed/2;
            xpos = width/2 + cos(radians(-angle+offset))*(radius+i);
            ypos = height/2 + sin(radians(-angle+offset))*(radius+i);
        }
        else{
            turn(offset);
        }
    }

    void turn(float offset){
        angle = angle-speed/50;
        xpos = width/2 + cos(radians(-angle+offset))*(endRad);
        ypos = height/2 + sin(radians(-angle+offset))*(endRad);  
    }
}