Question

这是我第一次在这里发帖，所以如果代码太长，我会非常乐意编辑它。

我正在制作收银员，这是一种改变的功能。

奇怪的是：如果我为1.15购买1.14，我会得到一分钱。但是，如果我为.15购买.14，我就没有回头了。

另外，如果我为.20购买.15，我会买回镍。但是，如果我以.25购买.20，我会得到4便士。我需要帮助理解为什么我得到四便士而不是0便士和1镍就像我应该的那样。我相信在某个地方划分某些事情是错误的。

#include <iostream>
#include <cmath>

using namespace std;

int main() {

  double buyAmount = 0.0;
  double paidAmount = 0.0;
  double diffAmount = 0.0;
  double smallChange = 0.0;
  int dollars = 0;
  int quarters = 0;
  int dimes = 0;
  int nickels = 0;
  int pennies = 0;

  cout << "Enter purchase amount: ";
  cin >> buyAmount;
  cout << endl;

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

  diffAmount = paidAmount - buyAmount;

  if (diffAmount < 0 ) {
     cout << "Error: Make sure paid amount exceeds purchase amount. ";
     cout << endl;
  }

  dollars = floor(diffAmount);
  smallChange = (diffAmount - dollars) * 100  

  quarters = (smallChange / 25);
  smallChange = smallChange - (quarters * 25);
  dimes = (smallChange / 10);
  smallChange = smallChange - (dimes * 10);
  nickels = floor(smallChange / 5);
  smallChange = smallChange - (nickels * 5);
  pennies = floor(smallChange / 1); 

  cout << "Total Change: $" << diffAmount;
  cout << endl << endl;
  cout << "dollars " << dollars << endl;
  cout << "quarters " << quarters << endl;
  cout << "dimes " << dimes << endl;
  cout << "nickels " << nickels << endl;
  cout << "pennies " << pennies << endl;
}

Answer 1

浮点运算并不准确。有时你得到的错误很少。例如0.25 / 0.25可导致1或0.99999999999999 如果您使用地板，除非您添加额外的“epsilon”，否则无法使用地板。补偿计算错误，例如

quarters = floor(smallChange / 0.25);

可能会成为

const double epsilon = 1E-5;
quarters = floor(smallChange / 0.25 + epsilon);

管理货币仍然是一个更好的主意：

int diffAmount = ...; // (pennies or cents).
int dollars = diffAmount / 100;
int smallChange = diffAmount % 100;
int quarters = smallChange / 25;
smallChange -= quarters * 25;
int dimes = smallChange / 10;
smallChange -= dimes * 10;
...

您还应该对数据进行驱动：创建包含不同硬币及其值的数据结构，然后编写代码以循环该数据结构。这样可以轻松切换到另一种货币（大多数不会有四分之一硬币）或修复错误。

Answer 2

你不应该使用浮点，因为它们不准确。

相反，将美元和美分分别读入整数，然后完全转换为美分整数除法将使这项工作完美无缺如果您只是假设用户输入正确，您可以这样读：

int purchaseDollars = 0;
int purchaseCents = 0;
char decimalPoint;  // Can be anything except a number in this snippet.
std::cin >> purchaseDollars >> decimalPoint >> purchaseCents;
int buyAmount = purchaseCents + 100 * purchaseDollars; // Convert to cents
// Same for paid amount

int diffAmount = paidAmount - buyAmount;
int dollars = diffAmount / 100;   // Integer division makes this work as you want
diffAmount -= dollars;            // Now diffAmount is just the cents
/// and so on...

收银员功能不适用于特定情况

2 个答案: