利用 char 确定输出不起作用

Posted 2023-02-25

【中文标题】利用 char 确定输出不起作用

【英文标题】：Utilizing char to determine output isnt working

【发布时间】：2015-10-06 14:45:27

【问题描述】：

我正在尝试这样做，因此当用户输入 r 或 p 时，将执行不同的部分，但由于某种原因，它无法识别单个字符输入。它只是将我带到第一个 if 语句，说我没有正确的值，然后不管我输入什么都为 r 执行。为什么这对我不起作用？我用字母本身和相应字母的 ascii 数字尝试了它，每次它都做同样的事情。它甚至没有输出说无效服务代码重试的最终语句。我的程序看起来像这样。提前感谢您的宝贵时间！

#include <iostream>
#include <iomanip>
#include <string>

using namespace std;

int main()

    char servicetype;
    int account;
    double minutes;
    double initialCharge;
    double overCharge;
    double day;
    double night;
    double dayrate;
    double nightrate;
    double balance;
    string service;

    cout << "Please enter your account number: ";
    cin >> account;
    cout << endl;

    cout << "Please enter your service code: ";
    cin >> servicetype;
    cout << endl;

    if (servicetype != 'r' || 'R' || 'p' || 'P')
    
        cout << "Invalid service code entered. Please enter a valid service code of P for premium service or R for regular service: ";
        cin >> servicetype;
        cout << endl;
    

    else if (servicetype == 'r' || 'R')
    
        service = "regular";

        initialCharge = 10.00;
        overCharge = .2;

        cout << "Please enter the number of minutes the service was used: ";
        cin >> minutes;
        cout << endl;

        balance = initialCharge + (minutes * overCharge);

        cout << "Account number " << account << " with the " << service << " service which was utilized for " << minutes << " minutes and therefore a balance is due of $" << fixed << setprecision(2) << balance << endl;
    
    else if (servicetype == 'p' || 'P')
    
        service = "premium";

        initialCharge = 25.00;
        cout << "Please enter the number of minutes which were used between the hours of 6:00 am and 6:00 pm: ";
        cin >> day;
        cout << endl;

        cout << "Please eneter the number of minutes which were used between the hours of 6:00 pm and 6:00 am: ";
        cin >> night;
        cout << endl;

        if (day < 75)
            dayrate = 0;
        else
            dayrate = (day - 75) * .1;

        if (night < 100)
            nightrate = 0;
        else
            nightrate = (night - 100) * .05;

            balance = initialCharge + nightrate + dayrate;

            minutes = day + night;

            cout << "Account number " << account << " with the " << service << " service which was utilized for " << minutes << " minutes and therefore a balance is due of $" << fixed << setprecision(2) << balance << endl;
    
    else
        cout << "An invalid service code was entered, please try again." << endl;


    system("pause");

    return 0;

【问题讨论】：

我知道我可以使用 switch 代替 if 并且它可能会起作用，但我现在有点想知道为什么它不能使用 if 语句

|| 是一个 operator ，这意味着它接受输入并给出输出。 || 运算符接受两个输入，并输出true 或false。您已将'r' 和'R' 的输入提供给您的第一个||。结果将是true，因为至少有一个输入不是假的。

【参考方案1】：

您需要将serviceType 与每个字母进行比较。事实上，你基本上是在说：

“如果 serviceType 不是 'r' 或如果 'R' ... etc”总是评估为true，因为'R'（它被转换为它的等效数字）评估为true。这类似于说：

if ('R')

您可以通过将serviceType 与每个字符进行比较来修复它：

if (serviceType != 'r' && serviceType != 'R' && serviceType != 'p' && serviceType != 'P')

您可以使用tolower（或toupper）稍微简化它：

if (tolower(serviceType) != 'r' && tolower(serviceType) != 'p')

【讨论】：

应该是tolower((unsigned char)serviceType)，或者tolower(serviceType, std::locale())。 Reference