C ++將char值設置為字符串？

Question

無論用戶選擇是否鍵入小寫字母，這都會輸出大寫字母“ S”或“ P”。 當我用代碼中的其他語句進行提示時，輸出將起作用，但是...我想在最終的cout語句中顯示STANDARD或PREMIUM。

如何更改char的值以輸出STANDARD或PREMIUM？

#include <string>
#include <iostream>

char meal;

cout << endl << "Meal type:  standard or premium (S/P)?  ";
cin >> meal;

meal = toupper(meal);
    if (meal == 'S'){
      meal = 'S';
  }

    else{
      meal = 'P';
}

我嘗試過進餐=“標准”，進餐=“高級”，它不起作用。

Answer 1

聲明額外的可變string mealTitle; ，然后執行if (meal == 'P') mealTitle = "Premium"

#include <string>
#include <cstdio>
#include <iostream>
using namespace std;
int main(void) {
        string s = "Premium";
        cout << s;
}

Answer 2

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

int main(int argc, char* argv)
{
    char meal = '\0';
    cout << "Meal type:  standard or premium (s/p)?" << endl;;
    string mealLevel = "";
    cin >> meal;
    meal = toupper(meal);
    if (meal == 'S'){
        mealLevel = "Standard";
    }

    else{
        mealLevel = "Premium";
    }
    cout << mealLevel << endl;
    return 0;
}

Answer 3

您不能將變量meal更改為字符串，因為其類型為char 。 只需使用另一個名稱不同的對象即可：

std::string meal_type;
switch (meal) {
case 'P':
    meal_type = "Premium";
    break;
case 'S':
default:
    meal_type = "Standard";
    break;
}

Answer 4

#include <string>
#include <iostream>

std::string ask() {
  while (true) {
    char c;
    std::cout << "\nMeal type:  standard or premium (S/P)?  ";
    std::cout.flush();
    if (!std::cin.get(c)) {
      return ""; // error value
    }
    switch (c) {
    case 'S':
    case 's':
      return "standard";
    case 'P':
    case 'p':
      return "premium";
    }
  }
}
int main() {
  std::string result = ask();
  if (!result.empty()) {
    std::cout << "\nYou asked for " << result << '\n';
  } else {
    std::cout << "\nYou didn't answer.\n";
  }
  return 0;
}