如何在C ++中获取局部变量的值？

问题描述

我一直在努力从选择结构中获取局部变量。我通常不熟悉编程，因此在计算其值在if函数内的项目的总购买额方面需要帮助，并且我在while函数中遇到麻烦，因为我打算使用它来无限地接受进餐代码和数量，直到用户输入0，但这是行不通的。如何使我的无限while函数起作用，如何从选择结构中正确获取值？任何帮助，提示和建议都将受到赞赏！

更新：我能够通过函数内部的变量解决问题。当前的问题是我需要两次输入0，因为第一个0被忽略了，我也不知道为什么。

int main()
{
    char mealCode;
    bool noOrder = false;
    int quantity = -1,subTotalA = 0,subTotalB = 0,subTotalC = 0,subTotalD = 0,subTotalE = 0,total = 0;

while (quantity != 0)
{
    cin >> mealCode >> quantity;
    if (quantity == 0)
        break;
    else
    {
        if (mealCode == 'A')
        {
            subTotalA += 45 * quantity;
            cout << "Subtotal is " << subTotalA << endl;
        }
        else if (mealCode == 'B')
        {
            subTotalB += 50 * quantity;
            cout << "Subtotal is " << subTotalB << endl;
        }
        else if (mealCode == 'C')
        {
            subTotalC += 55 * quantity;
            cout << "Subtotal is " << subTotalC << endl;
        }
        else if (mealCode == 'D')
        {
            subTotalD += 60 * quantity;
            cout << "Subtotal is " << subTotalD << endl;
        }
        else if (mealCode == 'E')
        {
            subTotalE += 75 * quantity;
            cout << "Subtotal is " << subTotalE << endl;
        }
    }
    total = subTotalA + subTotalB + subTotalC + subTotalD + subTotalE;
    cout << "Total Purchase is" << total;
}

}

解决方法

如果您希望data _null_; monyyyy = 'January 2013'; date = input(substr(strip(monyyyy),1,12),YYMMN.);; put date=YYMMN.; run; 循环在数量等于0之前是无限的，则需要这样写。您不需要两个相同的while。

另外，如果要用这么多的常量进行比较，最好使用cin控制结构。

例如，您可能想要执行以下操作：

switch

,

不确定要做什么，但有几点评论

• 如果要跟踪小计和总计，则需要将变量初始化为0，然后将每个产品的新小计求和
• 如果要使用多种产品，则需要将操作包括在while循环中（当用户不断添加数量的餐食！= 0时，您将继续计算小计）

类似

    char mealCode;
    int subTotalA=0,subTotalB=0,total,quantity;

   
    while (cin >> mealCode >> quantity)
    {
        if (quantity <= 0) { break; }

        if (mealCode == 'A')
        {
            subTotalA += 45 * quantity;
            cout << "Subtotal is " << subTotalA << endl;
        }
        if (mealCode == 'B')
        {
            subTotalB += 50 * quantity;
            cout << "Subtotal is " << subTotalB << endl;
        }
      
        total = subTotalA + subTotalB ;
        cout << "Total Purchase is Php" << total;
    }

,

我不喜欢if和switch解决方案，所以让我们来尝试一下并消除重复。

#include <iostream> //cin. cout
#include <cstring> //strchr
#include <numeric> //std::accumulate
#include <iterator> //std::begin and std::end
using namespace std;

int main()
{
    char mealCode;
    int quantity;

    // use tables to replace the different code for different meal codes
    // list valid meal codes
    const char mealCodes[] = "ABCDE";
    // list cost of each meal code.
    const int mealCosts[] = {45,50,55,60,75};

    // Sucks when you ass an item to one table and not the other
    // so lets give ourselves a reminder,hmmm?
    // Refuse to compile if the costs and codes don't line up
    // note the -1 to ignore the null terminator in the string
    static_assert(std::size(mealCodes) - 1 == std::size(mealCosts),"Meal codes and meal costs aren't the same size");

    // when you have a bunch of sequentially numbered variables that's nature
    // telling you you want an array. This one is forced to be the same size
    // as the number of meal codes. They're all 0,so no magic is required.
   int subTotal[std::size(mealCodes)] = {0};

    while (cin >> mealCode >> quantity && quantity != 0) // loop until bad input
                                                         // or quantity 0
    {
        // verify meal code
        const char * found = strchr(mealCodes,mealCode);
        if (found) // found a valid meal code
        {
            //get index of mealCode in mealCodes
            size_t index = found - mealCodes;
            // Look up the cost in the table and accumulate
            subTotal[index] += mealCosts[index] * quantity;
            // not 100% sure about the += but not summing multiple 
            // runs of the same meal code run seems wrong

            cout << "Subtotal is " << subTotal[index] << endl;
        }
    }
    // sum up all of the subtotals
    int total = std::accumulate(std::begin(subTotal),std::end(subTotal),0);
    cout << "Total Purchase is Php" << total;
}

优点：添加新的进餐代码是代码中的两个变化：将代码添加到mealCodes并将成本添加到mealCosts。

没有多余的注释，该程序就很短，而且随着进餐次数的增加，它并不会变得太大。

#include <iostream> 
#include <cstring> 
#include <numeric> 
#include <iterator> 
using namespace std;

int main()
{
    char mealCode;
    int quantity;

    const char mealCodes[] = "ABCDEFGHIJK";
    const int mealCosts[] = {45,75,13,62,88,42,10,99 };
    static_assert(std::size(mealCodes) - 1 == std::size(mealCosts),"Meal codes and meal costs aren't the same size");
   int subTotal[std::size(mealCodes)] = {0};

    while (cin >> mealCode >> quantity && quantity != 0) 
    {
        const char * found = strchr(mealCodes,mealCode);
        if (found)
        {
            size_t index = found - mealCodes;
            subTotal[index] += mealCosts[index] * quantity;
            cout << "Subtotal is " << subTotal[index] << endl;
        }
    }
    int total = std::accumulate(std::begin(subTotal),0);
    cout << "Total Purchase is Php" << total;
}