c++ - 在Money更改程序中正确计算美元-C++

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请看下面的代码

#include <QtCore/QCoreApplication>
#include <iostream>

int main(int argc, char *argv[])
{
    using namespace std;

    double purchaseAmount;
    double paidAmount;
    float balance;

     int change, quarters, dimes, nickels, pennies, tenDollar, fiveDollar; // declare variables

    cout << "Enter Total purchased amount" << endl;
    cin >> purchaseAmount;

    cout << "Enter Total paid amount" << endl;
    cin >> paidAmount;

    balance = paidAmount - purchaseAmount ;

     tenDollar = balance / 10; // calculate the number of Ten Dollars
     change =   tenDollar  % 10  ; // calculate the change needed
     change = balance * 100;
     quarters = change / 25; // calculate the number of quarters
     change = change % 25; // calculate remaining change needed
     dimes = change / 10; // calculate the number of dimes
     change = change % 10; // calculate remaining change needed
     nickels = change / 5; // calculate the number of nickels
     pennies = change % 5; // calculate pennies

     cout << "\nQuarters: " << quarters << endl; // display # of quarters
     cout << " Dimes: " << dimes << endl; // display # of dimes
     cout << " Nickels: " << nickels << endl; // display # of nickels
     cout <<" Pennies: " << pennies << endl; // display # of pennies
     cout <<" Ten dollar: " << tenDollar << endl; // display # of Ten dollar
     //cout <<" Five dollar: " << fiveDollar << endl; // display # of Ten dollar

     return (0);

 }

我在这里尝试做的事，以十美元，四分之一硬币，硬币，镍和几美分计算剩下的零钱。例如，当我以这种方式运行程序时

Enter Total purchased amount
9.75
Enter Total paid amount
20

Quarters: 4
 Dimes: 0
 Nickels: 0
 Pennies: 0
 Ten dollar: 1

哪有错话虽这么说，上面的输出是错误的。而是应该

Enter Total purchased amount
9.75
Enter Total paid amount
20

Quarters: 1
 Dimes: 0
 Nickels: 0
 Pennies: 0
 Ten dollar: 1

那我在做什么错呢？

谢谢

最佳答案

正如反复提到的那样，错误是在痛苦的重复代码中。考虑以下：

int currencyCount(int& pennies, int penniesInDenomination) {
  const int count = penniesInBase / penniesInDenomination;
  pennies = pennies % penniesInDenomination;

  return count;
}

它可以用于每种面额-重复且一行。通过使函数获得两个值来进行此工作：新余额和该面额的计数。这种排序通过引用余额来作弊，并且作为调用此显然独立功能的“副作用”，它会根据返回的面额数量来减少余额。显然，您需要对此进行记录。

...
const int numberOfQuarters = currencyCount(balance, 25);
const int numberOfDimes    = currencyCount(balance, 10);
...

您还可以将货币信息（例如，货币名称和代表的美分数量）放入矢量中，并在其上循环执行相同的操作：

typedef std::pair<std::string, int> Currency; 
typedef std::vector<Currency> Currencies;
typedef Currencies::const_iterator CIter;

...

for(CIter iter = currencies.begin(); iter != currencies.end(); ++iter) {
  const int quantity = currencyCount(balance, iter->second);
  std::cout << iter->first << ": " << quantity << std::endl;
}

这样，您可以避免重复的代码及其所涉及的错误。

关于c++ - 在Money更改程序中正确计算美元-C++，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/12441488/

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Answer 1

正如反复提到的那样，错误是在痛苦的重复代码中。考虑以下：

int currencyCount(int& pennies, int penniesInDenomination) {
  const int count = penniesInBase / penniesInDenomination;
  pennies = pennies % penniesInDenomination;

  return count;
}

它可以用于每种面额-重复且一行。通过使函数获得两个值来进行此工作：新余额和该面额的计数。这种排序通过引用余额来作弊，并且作为调用此显然独立功能的“副作用”，它会根据返回的面额数量来减少余额。显然，您需要对此进行记录。

...
const int numberOfQuarters = currencyCount(balance, 25);
const int numberOfDimes    = currencyCount(balance, 10);
...

您还可以将货币信息（例如，货币名称和代表的美分数量）放入矢量中，并在其上循环执行相同的操作：

typedef std::pair<std::string, int> Currency; 
typedef std::vector<Currency> Currencies;
typedef Currencies::const_iterator CIter;

...

for(CIter iter = currencies.begin(); iter != currencies.end(); ++iter) {
  const int quantity = currencyCount(balance, iter->second);
  std::cout << iter->first << ": " << quantity << std::endl;
}

这样，您可以避免重复的代码及其所涉及的错误。