如何将 char 值转换为特定的 int 值

Question

我有一个任务，我必须编写一个程序，该程序允许人们输入一个七个字母的单词并将其转换为电话号码（例如 1-800-PAINTER 到 1-800-724-6837）。我正在尝试将每个字母转换为要输出给用户的特定数字，每个字母对应于电话键盘上的数字（因此a，A，b，B或c，C等于1，即更多信息：https://en.wikipedia.org/wiki/Telephone_keypad)。

目前我已将其设置为输入单词的每个字母分别代表一、二、三、四、五、六或七的 char 变量。然后，使用 switch 和 if 语句，想法是将 char 转换为 xtwo = 2、xthree = 3 等的 int 变量。但这不起作用。有没有更好的方法来做到这一点？

代码示例（直到第一次切换，尽管大多数情况下它是这样的重复模式）：

int main()
{
    char one, two, three, four, five, six, seven;

    cout << "Enter seven letter word (1-800-***-****): " << "\n";

    cin >> one >> two >> three >> four >> five >> six >> seven;
    int xtwo = 2; int xthree = 3; int xfour = 4; int xfive = 5; int xsix = 6; int xseven = 7; int xeight = 8;
    int xnine = 9;

    switch (one)
    {
    case 1:
        if (one == 'a' || one == 'b' || one == 'c' || one == 'A' || one == 'B' || one == 'C')
        {
            one = xtwo;
        }
        break;
    case 2:
        if (one == 'd' || one == 'e' || one == 'f' || one == 'D' || one == 'E' || one == 'F')
        {
            one = xthree;
        }
        break;
    case 3:
        if (one == 'g' || one == 'h' || one == 'l' || one == 'G' || one == 'H' || one == 'L')
        {
            one = xfour;
        }
        break;
    case 4:
        if (one == 'j' || one == 'k' || one == 'l' || one == 'J' || one == 'K' || one == 'L')
        {
            one = xfive;
        }
        break;
    case 5:
        if (one == 'm' || one == 'n' || one == 'o' || one == 'M' || one == 'N' || one == 'O')
        {
            one = xsix;
        }
        break;
    case 6:
        if (one == 'p' || one == 'q' || one == 'r' || one == 's' || one == 'P' || one == 'Q' || one == 'R' || one == 'S')
        {
            one = xseven;
        }
        break;
    case 7:
        if (one == 't' || one == 'u' || one == 'v' || one == 'T' || one == 'U' || one == 'V')
        {
            one = xeight;
        }
        break;
    case 8:
        if (one == 'w' || one == 'x' || one == 'y' || one == 'z' || one == 'W' || one == 'X' || one == 'Y' || one == 'Z')
        {
            one = xnine;
        }
        break;
    }

那么，本质上，一个字母的char变量如何转化为具体的int变量呢？

Answer 1

自从我编写任何 c/c++ 代码以来已经有好几年了，我什至没有安装编译器来测试 .. 但这应该让你开始走上正轨 检查函数和语法……完全出乎我的意料。需要检查。 //

int numArray[7];
char inputStr[10]; 
cout << " give me 7 characters";
cin >> input;

/*
use a for loop to read the string letter by letter (a string in c is an 
array of characters)
convert the characters to uppercase 
fall through case statements for each group of letters
assing value to output array to do wiht as you like.
*/


for(i=0; i < 7; i++){
    inputStr[i] = toupper(inputStr[i]);
    switch(input[i]){
      case 'A':
      case 'B':
      case 'C':
        numArray[i] = 2;
        break;
      case 'D':
      case 'E':
      case 'F':
        numArray[i] = 3;
        break;

    and so on and so foth....


      }

}

Answer 2

您可以使用std::map。

例如，你可以有

std::map<char,int> char_to_dig {
  {'a',1}, {'b',1}, {'c',1},
  {'d',2}, {'e',2}, {'f',2}
};

然后

char_to_dig['a']

会给你1。

---

或者，您可以编写一个执行映射的函数。类似这样的东西：

int char_to_dig(char c) {
  static const char _c[] = "abcdefghi";
  static const int  _i[] = { 1,1,1,2,2,2,3,3,3 };
  for (unsigned i=0; i<9; ++i) {
    if (_c[i]==c) return _i[i];
  }
  return -1; // some value to signal error
}

---

或者，您可以不使用数组，而是对 chars 执行算术运算（因为它们只是小整数）。

int char_to_dig(char c) {
  c = std::toupper(c);
  if (c < 'A' || c > 'Z') return -1;
  if (c == 'Z') return 9;
  if (c > 'R') --c;
  return ((c-'A')/3)+2;
}

这会给你像这个垫子上的数字：

---

显然，有一个类似的code gold question。