如何更好地检查两个 char 变量是否在某些值集中？

Question

最近，我们的教授要求我们使用两个char变量（天）来接收用户的输入。

下面的代码可以很好地作为检查，以确保 Mo、Tu、We、Th、Fr、Sa、Su 是仅有的两个作为一对一起输入的字符。如果接收到其他任何内容作为输入，它将循环并要求用户提供有效输入。

输入应该不区分大小写，这意味着，例如，"mO"并且"tu"是可以接受的。似乎有很多重复正在发生。有没有办法清理这个？

cout << "Please enter the day of the week did you made the long distance call (Mo Tu We Th Fr Sa Su): ";
cin >> dayOne >> dayTwo;

while ((dayOne != 'M' && dayOne != 'm' || dayTwo != 'O' && dayTwo != 'o') &&
       (dayOne != 'T' && dayOne != 't' || dayTwo != 'U' && dayTwo != 'u') &&
       (dayOne != 'W' && dayOne != 'w' || dayTwo != 'e' && dayTwo != 'E') &&
       (dayOne != 'T' && dayOne != 't' || dayOne != 'H' && dayTwo != 'h') &&
       (dayOne != 'F' && dayOne != 'f' || dayTwo != 'R' && dayTwo != 'r') &&
       (dayOne != 'S' && dayOne != 's' || dayTwo != 'A' && dayTwo != 'a') &&
       (dayOne != 'S' && dayOne != 's' || dayTwo != 'U' && dayTwo != 'u'))
{
    cin.clear();
    cin.ignore(std::numeric_limits<std::streamsize>::max(), '\n');
    cout << endl << "You have entered an invalid day. Please re-enter a day in the correct format (Mo Tu We Th Fr Sa Su): ";
    cin >> dayOne >> dayTwo;
}

Answer 1

您可以编写一个折叠表达式，将 2 个字符与一个字符串进行比较：

template<typename ...Days>
bool any_of(char a, char b, Days ...days)
{
    return (... || (a == days[0] && b == days[1]));
}

然后像这样使用它：

while (! any_of(std::tolower(dayOne), std::tolower(dayTwo), "mo", "tu", "we", "th", "fr", "sa", "su"))
    // keep asking for input

这是一个演示。

这应该满足使用 2 个char输入的要求。

Answer 2

您通常先使用tolower或toupper将char变量转换为正确的大小写。我喜欢使用tolower- 它看起来稍微好一点。

dayOne = tolower(dayOne);
dayTwo = tolower(dayTwo);

while (
    (dayOne != 'm' || dayTwo != 'o') &&
    (dayOne != 't' || dayTwo != 'u') &&
    (dayOne != 'w' || dayTwo != 'e') &&
    (dayOne != 't' || dayTwo != 'h') &&
    (dayOne != 'f' || dayTwo != 'r') &&
    (dayOne != 's' || dayTwo != 'a') &&
    (dayOne != 's' || dayTwo != 'u'))
{
    ...
}

您可以通过memcmp一次比较两个字符来进一步更改它，但我不确定它会简化代码。

Answer 3

另一种可能值得一提的方法是组织您的数据，以便您可以对它使用 std 函数 ( std::find)

// Example program
#include <algorithm>
#include <string>
#include <vector>
#include <iostream>

int main()
{
    const std::vector<std::string> days = {
        "mo", "tu", "we", "th", "fr", "sa", "su"
    };

    bool found = false;

    while (found == false) {
        char dayOne, dayTwo;
        std::cout << "Please enter the first letter of the day" << std::endl;
        std::cin >> dayOne;
        std::cout << "Please enter the second letter of the day" << std::endl;
        std::cin >> dayTwo;

        std::string fullDay;
        fullDay += std::tolower(dayOne);
        fullDay += std::tolower(dayTwo);

        found = std::find(days.begin(), days.end(), fullDay) != days.end();
        std::cout << (found ? "correct day " : "invalid day, please try again ")
                  << fullDay
                  << std::endl;
    }
}

在这里运行

Answer 4

怎么样

switch (256 * tolower(dayOne) + tolower(dayTwo))
{
    case 256 * 'm' + 'o':
        // Monday
    case 256 * 't' + 'u':
        // Tuesday
}

等等？

Answer 5

不知道您是否使用/允许使用正则表达式，但我会这样解决：

bool isDayOfTheWeek(char a, char b)
{
    std::string day({a, b});
    std::regex pattern("Mo|Tu|We|Th|Fr|Sa|Su", std::regex_constants::icase);
    return std::regex_search(day, pattern);
}

然后简单地：

cout << "Please enter the day of the week did you made the long distance call (Mo Tu We Th Fr Sa Su): ";
cin >> dayOne >> dayTwo;

while (!isDayOfTheWeek(dayOne, dayTwo))
{
    cin.clear();
    cin.ignore(std::numeric_limits<std::streamsize>::max(), '\n');
    cout << endl << "You have entered an invalid day. Please re-enter a day in the correct format (Mo Tu We Th Fr Sa Su): ";
    cin >> dayOne >> dayTwo;
}