Question

（语言正在处理）嘿，我有一项我不太懂的任务。这是另一个赋值的第2部分，我必须创建一个名为drawButton的函数，它使用数组作为我们正在绘制的按钮的参数。总共有3个具有预定值（x，y，大小，颜色）的按钮这里是第2部分;

编写另一个名为的函数定义 isPositionWithinButton。功能应该取三个参数：一个点（x，y）和一个按钮阵列。在测试的函数中实现代码这点是否在范围内按钮并使该函数返回true或false 取决于结果。调用该函数从代码中的适当位置（例如使用 mouseX和mouseY作为点参数）和打印button1，-2或-3基于传递参数和返回的值。

所以，我现在所处的位置是我已经完成了这个功能，它可以说你输入的坐标是否在按钮内，但是我不确定最后一部分是什么，我必须做。我的猜测是，如果坐标在按钮坐标内，我必须以某种方式使程序绘制按钮？无论如何，我不知道该怎么做，所以非常感谢帮助！

这是我到目前为止的地方：

int[] button1;
int[] button2;
int[] button3;
boolean within;


void setup(){
  size(800,800);
  int[] button1 = {75,250,200,200,150,160,170};
  int[] button2 = {315,250,200,200,150,160,170};
  int[] button3 = {550,250,200,200,150,160,170};


  isPositionWithinButton(100,280,button1);


}


void drawButton(int[] buttonArray) {
  fill(buttonArray[4],buttonArray[5],buttonArray[6]);
  rect(buttonArray[0],buttonArray[1],buttonArray[2],buttonArray[3]);
}

void isPositionWithinButton(int x, int y, int[] buttonArray){
  if (x>buttonArray[0] && x<buttonArray[0]+buttonArray[2] && y>buttonArray[1] && y<buttonArray[1]+buttonArray[2]){
    within=true;
  }
  else{
    within=false;
  }
    if (within==true){
    println("Within.");

  }
  else{
    println("Not within.");
  }

}

Answer 1

我调整了你的代码来完成这些事情。阅读评论以了解更改。我没有使用循环来保持事物尽可能接近你的代码。希望有所帮助。

//read below..
int[] button1 = {75,250,200,200,150,160,170};
int[] button2 = {315,250,200,200,150,160,170};
int[] button3 = {550,250,200,200,150,160,170};


void setup(){
  size(800,800);

  // by repeating the 'int[]' part here you were
  // shadowing, redeclaring, the global vars
  // leaving those unitialized.
  // you could just delete the 'int[]' part, but then the direct
  // assining won't work. You would need to do:
  //button1[0] = 75;
  //button1[1] = 250 and so on...
  // so I just moved the decaration to global scope
  //int[] button1 = {75,250,200,200,150,160,170};
  //int[] button2 = {315,250,200,200,150,160,170};
  //int[] button3 = {550,250,200,200,150,160,170};

  //calling your draw function
  drawButton(button1);
  drawButton(button2);
  drawButton(button3);



}


// needed to call draw so skecth keeps running and you can test mouse coordinates
void draw(){

  //now that the function has a return type boolean 
  // you can use it like this

  if(isPositionWithinButton(mouseX, mouseY, button1)){
    println("hello, this is button 1");
  }

  if(isPositionWithinButton(mouseX, mouseY, button2)){
    println("hello, this is button 2");
  }

    if(isPositionWithinButton(mouseX, mouseY, button3)){
    println("hello, this is button 3");
  }
}


void drawButton(int[] buttonArray) {
  fill(buttonArray[4],buttonArray[5],buttonArray[6]);
  rect(buttonArray[0],buttonArray[1],buttonArray[2],buttonArray[3]);
}


// "make the function return true or false" 
// A function that returns something is not void
// you use the returned type instead, here boolean
// So you test and return either true or false

boolean isPositionWithinButton(int x, int y, int[] buttonArray){
  if (x>buttonArray[0] && x<buttonArray[0]+buttonArray[2] && y>buttonArray[1] && y<buttonArray[1]+buttonArray[2]){
    return true;
  }
  else{
    return false;
  }

  // all this body above could be repalced by this line 
  //return x>buttonArray[0] && x<buttonArray[0]+buttonArray[2] && y>buttonArray[1] && y<buttonArray[1]+buttonArray[2];

  // It says like... return true if following condition… else return false

}

坚持作业 - 对于做什么感到困惑

1 个答案: