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c++ - 在 C++ 中将美元转换为美分

问题描述

我正在学习 C++，我正在尝试使用一个函数将美元转换为美分，该函数具有一个静态变量，每次调用都会累积总数。不幸的是，我的函数似乎造成了上溢或下溢的情况。这个函数的任何指针都会有很大的帮助。这是代码。

#include <iostream>
#include <iomanip>
using namespace std;

void normalizeMoney(float& dollars, int cents = 150);
// This function takes cents as an integer and converts it to dollars
// and cents. The default value for cents is 150 which is converted
// to 1.50 and stored in dollars

int main()
{
   int cents;
   float dollars;

   cout << setprecision(2) << fixed << showpoint;

   cents = 95;
   cout << "\n We will now add 95 cents to our dollar total\n";

   normalizeMoney(dollars, cents);//    Fill in the code to call normalizeMoney to add 95 cents

   cout << "Converting cents to dollars resulted in " << dollars << " dollars\n";



   cout << "\n We will now add 193 cents to our dollar total\n";

   normalizeMoney(dollars, 193);// Fill in the code to call normalizeMoney to add 193 cents

   cout << "Converting cents to dollars resulted in " << dollars << " dollars\n";

   cout << "\n We will now add the default value to our dollar total\n";

   normalizeMoney(dollars);// Fill in the code to call normalizeMoney to add the default value of cents

   cout << "Converting cents to dollars resulted in " << dollars << " dollars\n";

   return 0;
}

//*******************************************************************************
//  normalizeMoney
//
//  task:     This function is given a value in cents. It will convert cents
//            to dollars and cents which is stored in a local variable called
//            total which is sent back to the calling function through the
//            parameter dollars. It will keep a running total of all the money
//            processed in a local static variable called sum.
//
//  data in:  cents which is an integer
//  data out: dollars (which alters the corresponding actual parameter)
//
//*********************************************************************************

void normalizeMoney(float& dollars, int cents)
{
    float total = 0;

    // Fill in the definition of sum as a static local variable
    static float sum = 0.0;

    // Fill in the code to convert cents to dollars
    if (cents >= 100) {
        cents -= 100;
        dollars += 1;
        total = total + dollars;
        sum = static_cast <float> (sum + dollars + (cents / 100));

    }
    else {
        total += (cents / 100);
        static_cast <float> (sum += (cents / 100));
    }
    cout << "We have added another $" << dollars << "   to our total" << endl;
    cout << "Our total so far is    $" << sum << endl;

    cout << "The value of our local variable total is $" << total << endl;
}

我得到的输出是：

We will now add 95 cents to our dollar total
We have added another $-107374176.00    to our total
Our total so far is     $0.00
The value of our local variable total is $0.00
Converting cents to dollars resulted in -107374176.00 dollars

 We will now add 193 cents to our dollar total
We have added another $-107374176.00    to our total
Our total so far is     $-107374176.00
The value of our local variable total is $-107374176.00
Converting cents to dollars resulted in -107374176.00 dollars

 We will now add the default value to our dollar total
We have added another $-107374176.00    to our total
Our total so far is     $-214748352.00
The value of our local variable total is $-107374176.00
Converting cents to dollars resulted in -107374176.00 dollars
Press any key to continue . . .

如果有人能告诉我我在哪里搞砸了，我将不胜感激。

标签： c++c++11

解决方案

就您最初的问题而言，dollars从未初始化。因此，当时内存中的任何价值都将是您的函数添加到的起始美元价值。但是这个问题不是修复你的主要问题，而是因为你没有将新计算的总和分配给你dollars的函数，这将是根据函数描述的预期行为。

要在评论中回答您的其他问题，要将您的美分转换为美元，您只需计算cents / 100.0f，注意您除以浮点数100.0f而不是整数100，这样您的结果本身就会变成 afloat而不是 a int。

笔记：

虽然从表面上看，这是一项学校作业，但仍然值得一提的是：

  1. 不要将金额存储在浮点值中。
  2. 函数签名/行为是疯狂的。

目前，您想要实现的更多是“标准化这些美分并将它们添加到这个值”。如果您想编写一个将您的美分转换为美元的函数，将其写为

float normalizeMoney(const int cents = 150);

然后用作

dollars += normalizeMoney(95);

忘记完全没有根据的静态变量。

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