java - 无法获得输入 .11 的正确输出

Question

我在学校有一项作业，我必须显示用户输入的小于 1.00 但大于 0 的金额的正确更改。除了带有 1 的两位数以外，每个金额都有效。或第十位上的 6。例如 .11、.16、.21、.26 等。

这是我的代码

import java.util.Scanner;

public class AmountChange
{
   public static void main(String[] args)
   {
      //
      double amt; 
      int cents, quarter, dime, nickle, penny;

      Scanner keyboard = new Scanner(System.in);

      //To get the users input
      System.out.println("Change in Coins");
      System.out.println("---------------");
      System.out.println("Enter the amount less than $1.00, " +
                         "\nbut more than zero.");
      System.out.print("\nEnter amount: ");
      amt = keyboard.nextDouble();

      //Loop for incorrect input
      while ( amt < 0 || amt > 1.00 )
      {
         System.out.println("Please enter the amount less than $1.00,"  
                           + "\nbut more than zero.");
         System.out.print("\nRe-enter amount: ");
         amt = keyboard.nextDouble();
      }

      // 
      cents = (int)( amt * 100 + .1 );

      quarter = cents/25;
      cents %= 25;

      dime = cents/10;
      cents %= 10;

      nickle = cents/5;
      cents %= 5;

      penny = cents;

      // ----------------------------------------------------------
      if (quarter > 1)
      {
         System.out.print("\nYou will need " + quarter + " quarters, ");
      }
      else if (quarter == 1)
      {
         System.out.print("\nYou will need " + quarter + " quarter ,");
      }
      else 
      {
         System.out.print("\nYou will need no quarters, ");
      }

      // ----------------------------------------------------------
      if (dime > 1)
      {
         System.out.print(dime + " dimes, ");
      }
      else if (dime == 1)
      {
         System.out.print(dime + " dime, ");
      }
      else 
      {
         System.out.print("no dimes, ");
      }

      // ----------------------------------------------------------
      if (nickle > 1)
      {
         System.out.print(nickle + " nickles, ");
      }
      else if (nickle == 1)
      {
         System.out.print(nickle + " nickle, ");
      }
      else 
      {
         System.out.print("no nickles, ");
      }

      // ----------------------------------------------------------
      if (penny > 1)
      {
         System.out.print("and " + penny + " pennies.");
      }
      else if (quarter == 1)
      {
         System.out.print("and " + penny + " penny.");
      }
      else 
      {
         System.out.print("and no pennies.");
      }      
   }
}

Answer 1

啊，剪切和粘贴的乐趣:-)

if (penny > 1)
{
    System.out.print("and " + penny + " pennies.");
}
else if (quarter == 1)  // <<<<< LOOK HERE !!!
{
     System.out.print("and " + penny + " penny.");
}
else 
{
     System.out.print("and no pennies.");
}      

那应该是便士，而不是四分之一。

而且，事实上，它实际上确实适用于.26(尽管你有这样的断言)，因为quarter设置为1 ，与便士相同。事实上，它适用于任何硬币数量等于便士数量的值(.26、.52、.78) ，但只是偶然。

<小时/>

顺便说一句，您可能需要考虑的另一件事是使用以下内容重构所有重复的代码:

import java.util.Scanner;

public class Test
{
    static double getAmount(Scanner keyboard) {
        System.out.println("Enter the amount between zero and $1.00.");
        System.out.print("\nEnter amount: ");
        return keyboard.nextDouble();
    }

    static String mkeTxt (int val, String prefix, String singular, String plural) {
        if (val == 0)
            return prefix + "no " + plural;
        if (val == 1)
            return prefix + "1 " + singular;
        return prefix + val + " " + plural;
    }

    public static void main(String[] args)
    {
        double amt; 
        int cents, quarter, dime, nickle, penny;

        Scanner keyboard = new Scanner(System.in);

        System.out.println("Change in Coins");
        System.out.println("---------------");
        amt = getAmount(keyboard);
        while ( amt < 0 || amt > 1.00 )
            amt = getAmount(keyboard);

        cents = (int)( amt * 100 + .1 );
        quarter = cents/25;
        cents %= 25;
        dime = cents/10;
        cents %= 10;
        nickle = cents/5;
        cents %= 5;
        penny = cents;

        System.out.print("\nYou will need ");
        System.out.print(mkeTxt(quarter,"", "quarter", "quarters"));
        System.out.print(mkeTxt(dime,", ", "dime", "dimes"));
        System.out.print(mkeTxt(nickle,", ", "nickle", "nickles"));
        System.out.print(mkeTxt(penny," and ", "penny", "pennies"));
        System.out.println(".");
    }
}

使用函数来输出提示并接受输入使用户输入代码更容易维护，因为您只需要在一处更改交互。

真正的保护者是 mkTxt() 函数，它为您提供一个可以自动根据硬币数量进行调整的字符串。它摆脱了 main() 中大量的 if/then/else block ，在一定程度上提高了可读性。

如果您发现自己多次执行类似的操作但具有不同的值，那么肯定需要将其更改为某种描述的函数或循环。