Question

既然我可以成功地遍历我的循环可能的有效浇头，我无法在我的程序中添加0.75附加费。我希望你能告诉我为什么会这样。该计划完全在底部。关注我的两个部分是

public class TestPizza
{
  public static void main(String args[])
  {
    int x;
    String top[] = {"Mushrooms", "Onions", ""};
    Pizza one = new Pizza();
    Pizza two = new Pizza();
    Pizza three = new Pizza();

one.setSize(12);
one.addTopping(top);
one.showValues();

  }
}

并且

// setPrice() assigns a price to the pie
public void addTopping(String programToppings[])
{
  for(int x = 0; x < 3; x++)
  {
    toppings[x] = programToppings[x];
  }
  for(int x = 0; x < 3; x++)
  {
    toppings[x] = toppings[x].toLowerCase();
  }
  for(int x = 0; x < 3; x++)
  {
    for(int xx = 0; xx < 6; xx++)
    {
      if(toppings[x].equals(validToppings[xx]))
      {price += 0.75;}
    }
  }
}

我不明白为什么.equals方法不起作用并将奇数的变化分配给最终价格......

// This custom class is used to create Pie objects
// It stores the data about the Pie in four variables:
// size, price, type and baked
// It lets the program that creates the Pie object set these values using four methods:
// setSize, setPrice, setType and bake

public class Pizza
{

  // Declare four variables that can store the values for each pie
  // Each Pie object will have their own, separate copy of these variables 12.
    private int size;
    private double price;
    private boolean baked;
    private int x;
    private int xx;
    private String validToppings[] = new String[6];
    private String toppings[] = new String[3];


    // The "constructor" method is called when a new pie
    // object is first created. We use it to set "default" values.
    // Our typical pie is 10 inches, costs $8 and is not baked yet
    // We don't yet know what the pie filling will be
    Pizza()
    {
      size = 8;
      price = 10.0;
      baked = false;
      String validToppings[] = {"mushrooms", "pepperonis", "onions", "mushroom", "pepperoni", "onion"};
      String toppings[] = new String[3];
    }

    // showValues() is a void method that displays the values of the
    // current Pie
    public void showValues()
    {
      System.out.println("Pie Size: " + size);
      for(int x = 0; x < 3; x++) {System.out.println("With " + toppings[x]);};
      System.out.println("Price of Pie: $" + price);
    }

    // getSize() returns the size of the pie
    public int getSize()
    {
      return size;
    }
    // getPrice() returns the price of the pie
    public double getPrice()
    {
      return price;
    }
    // baked() returns whether or not the pie is baked
    public boolean getBaked()
    {
      return baked;
    }
    // setSize() assigns a size to the pie
    public void setSize(int thisSize)
    {
      size = thisSize;
      switch(size)
      {
        case 8: price = 10.00; break;
        case 12: price = 14.00; break;
        case 16: price = 18.00; break;
        default: System.out.println("Error in Pizza class: Attempt to set invalid Pizza size."); break;
      }
    }

    // setPrice() assigns a price to the pie
    public void addTopping(String programToppings[])
    {
      for(int x = 0; x < 3; x++)
      {
        toppings[x] = programToppings[x];
      }
      for(int x = 0; x < 3; x++)
      {
        toppings[x] = toppings[x].toLowerCase();
      }
      for(int x = 0; x < 3; x++)
      {
        for(int xx = 0; xx < 6; xx++)
        {
          if(toppings[x].equals(validToppings[xx]))
          {price += 0.75;}
        }
      }
    }

  public void bake()
  {
    baked = true;
  };

}

Answer 1

Pizza中的validTopics实例数组永远不会被填充。替换

String validToppings[] = {"mushrooms", "pepperonis", "onions", "mushroom", "pepperoni", "onion"};

在Pizza的构造函数中：

this.validToppings[] = {"mushrooms", "pepperonis", "onions", "mushroom", "pepperoni", "onion"};

此外，您的代码中还有许多奇怪的内容：

addTopping关于顶部迭代的效率相当低
addTopping在for循环=＆gt;中使用（IMO）幻数。 3
您在Pizza两次
初始化和验证不同浇头的方式令人困惑

Answer 2

您的validToppings数组为空。您正在将每个String比较为null。

您以这种方式声明Pizza成员：;

private String validToppings[] = new String[6];
private String toppings[] = new String[3];

并且您尝试以这种方式在构造函数中指定值

String validToppings[] = {"mushrooms", "pepperonis", "onions", "mushroom", "pepperoni", "onion"};
String toppings[] = new String[3];

但是构造函数上面的代码所做的一切都是创建一个新的局部变量，在完成构造函数的执行后会忘记它。

修复方法是为成员分配值。例如，一个可能的正确解决方案如下：

声明这样的成员： String[] validToppings;
并在构造函数中指定一个值，如下所示： validToppings = {"mushroooms", ... };

从设计的角度来看（你可能是初学者，但仍然值得一提）你应该使用枚举（可以节省你比较字符串，制作拼写错误，......）。

Answer 3

 Pizza()
{
  size = 8;
  price = 10.0;
  baked = false;
  String validToppings[] = {"mushrooms", "pepperonis", "onions", "mushroom", "pepperoni", "onion"};
  String toppings[] = new String[3];
}

在这里，

String validToppings[] = {"mushrooms", "pepperonis", "onions", "mushroom", "pepperoni", "onion"};

是构造函数的本地，无法连接到全局'validToppings []'。看看吧。

无法正确比较两个字符串

3 个答案: