c++ - 无效的操作数和非法赋值强制转换

Question

该程序的目的是获取一个美元金额并打印将返回的钞票数量，以最有效地提供该金额和剩余的零钱。（即 523.33 美元 = 50 美元 x 10 / 20 美元 x1 / 1 美元 x 3/ .33 剩余）。除了我在附加图像中遇到的错误外，它应该可以工作。我已经尝试了我被教导的每一次铸造，但没有任何效果。

#include <iostream>
using namespace std;


int main()
{
    double withdrawAmount; //declare variable to store use input for desired withdraw amount

    do { //Ask for valid user input
       cout << "Please enter an amount to withdraw from yor account" << endl;
       cin >> withdrawAmount; //save user input into withdrawAmount variable
   } while (withdrawAmount < 1);

   //Print greatest # of bills that can be taken from the withdrawlAmount
     cout << "$50 bills :"<< (int) withdrawAmount / 50 << endl;    
   //Print number of $50 bills
     (int) withdrawAmount %= 50;
     cout << "$20 bills: " << (int) (withdrawAmount %= 50) / 20 << endl; 
   //Print number of $20 bills
     (int) withdrawAmount %= 20;
     cout << "$10 bills: " << (int) (withdrawAmount %= 20) / 10 << endl; 
  //Print number of $10 bills
   (int)withdrawAmount %= 10;
   cout << "$5 bills: " << (int) (withdrawAmount %= 10) / 5 << endl; 
  //Print number of $5 bills
   (int)withdrawAmount %= 5;
   cout << "$1 bills: " << (int) (withdrawAmount %= 5) / 1 << endl; 
  //Print number of $1 bills
   (int) withdrawAmount %= 1;
   cout << "Remaining: " << withdrawAmount / 1 << endl;
   return 0;
}

标签： c++casting

Answer 1

(int) withdrawAmount %= 50;应该替换为

withdrawAmount = std::fmod(withdrawAmount, 50);

其他值也一样。（不要忘记#include <cmath>）。

作为替代方案：

double input; //declare variable to store use input for desired withdraw amount

do { //Ask for valid user input
   cout << "Please enter an amount to withdraw from yor account" << endl;
   cin >> input; //save user input into withdrawAmount variable
} while (input < 1);
int withdrawAmount = input; // rounded toward 0.

std::cout << "$50 bills :"<< withdrawAmount / 50 << std::endl;
withdrawAmount %= 50;

// And so on...

// Final remaining
std::cout << "Remaining: " << input - int(input) << std::endl;