Question

我试图重新创建井字游戏，以再次使用Processing。我记得在将函数放入keyPressed（）ifs之前，该程序运行良好，每次用户在控制台中输入数字时，都会在控制台中输入文本行。现在，每次我按下按键时，都不会显示任何内容，甚至不会显示我分配的错误消息，以防用户决定在数字之外键入任何内容。现在我不确定哪里出了问题。由于我并非完全擅长编写Processing，因此我不确定如何解决此错误。任何帮助，将不胜感激。谢谢。

代码：

String pos = "";
boolean isx = true;
boolean onepressed = false;
boolean twopressed = false;
boolean threepressed = false;
boolean fourpressed = false;
boolean fivepressed = false;
boolean sixpressed = false;
boolean sevenpressed = false;
boolean eightpressed = false;
boolean ninepressed = false;
void setup() {
  size(1000,1000);
  frameRate(60);
  background(255,255,255);
  smooth();
  noStroke();
}
void keyPressedUpdate() {
  if (isx == true) {
    print("An x has been placed in the ");
    print(pos);
    print(" of the board");
    print("Currently x's turn");
    isx = false;
  } else if (isx == false) {
    print("An o has been placed in the ");
    print(pos);
    print(" of the board");
    print("Currently o's turn");
    isx = true;
  }
}
void keyPressed() {
  if (key == '1' && onepressed == false) {
    pos = "top-left corner";
    keyPressedUpdate();
    onepressed = true;
  } else if (key == '2' && twopressed == false) {
    pos = "top-center square";
    keyPressedUpdate();
    twopressed = true;
  } else if (key == '3' && threepressed == false) {
    pos = "top-right corner";
    keyPressedUpdate();
    threepressed = true;
  } else if (key == '4' && fourpressed == false) {
    pos = "center-left square";
    keyPressedUpdate();
    fourpressed = true;
  } else if (key == '5' && fivepressed == false) {
    pos = "center square";
    keyPressedUpdate();
    fivepressed = true;
  } else if (key == '6' && sixpressed == false) {
    pos = "center-right square";
    keyPressedUpdate();
    sixpressed = true;
  } else if (key == '7' && sevenpressed == false) {
    pos = "bottom-left corner";
    keyPressedUpdate();
    sevenpressed = true;
  } else if (key == '8' && eightpressed == false) {
    pos = "center-bottom square";
    keyPressedUpdate();
    eightpressed = true;
  } else if (key == '9' && ninepressed == false) {
    pos = "bottom-right corner";
    keyPressedUpdate();
    ninepressed = true;
  } else {
    print("Please enter a number, not a letter or symbol.");
  }
}

Answer 1

如果您正在使用keyPressed()之类的事件函数，则即使没有实际绘制任何内容，也需要具有draw()函数。

换句话说，添加一个draw()函数：

void draw(){}

现在，您将在控制台中看到打印语句。

keyPressed（）函数未响应（正在处理3.0）

1 个答案: