將很多if轉化爲更小的東西

Question

我需要檢查一個char數組並將每個char轉換爲一個int數組;

我設法做到了這一點，但有很多代碼。 無論如何，我可以寫更少的代碼？

for (int i = 0; i < 8; i++) 
     { 
      if (oddF[i] == '0') 
      { 
       oddFlValue[i] = 1; 
       continue; 
      } 
      else if (oddF[i] == '1') 
      { 
       oddFValue[i] = 0; 
       continue; 
      } 
      else if (oddFiscal[i] == '8') 
      { 
       oddFiscalValue[i] = 19; 
       continue; 
      } 
      else if (oddF[i] == '9') 
      { 
       oddFValue[i] = 21; 
       continue; 
      } 
      else if ((oddF[i] == 'a') || (oddF[i] == 'A')) 
      { 
       oddFValue[i] = 1; 
       continue; 
      } 
      else if ((oddF[i] == 'b') || (oddF[i] == 'B')) 
      { 
       oddFValue[i] = 0; 
       continue; 
      } 

這只是一個示例，因爲我需要檢查每個數字和字母類型。 謝謝！

Answer 1

我認爲最簡單的方法是使用Dictionary<char, int>並獲得它的價值。事情是這樣的：

Dictionary<char, int> MyCombinations = new Dictionary<char, int>{ 
    {'0', 1}, 
    {'5', 15}, 
    /*etc etc etc*/ 
}; 

，然後用它喜歡：

for (int i = 0; i < 8; i++) 
{ 
    oddFValue[i] = MyCombinations[oddFiscal[i]]; 
} 

Answer 2

class Program 
{ 
    static void Main(string[] args) 
    { 
     char[] oddF   = new char[8]{'0','1','8','A','b','9','a','B'}; 
     int[] oddFValue  = new int[8]; 

     Dictionary<char, int> table = new Dictionary<char, int>(){ 
      { '0', 1 }, 
      { '1', 0 }, 
      { '8', 19 }, 
      { '9', 21 }, 
      { 'a', 1 }, 
      { 'A', 1 }, 
      { 'b', 0 }, 
      { 'B', 0 } 
     }; 

     for (int i = 0; i < 8; i++) 
     { 
      oddFValue[i] = table[oddF[i]]; // Lookup which function should be called, and call it. 
     } 
    } 
} 

Answer 3

如何：

public int? ConvertCharToInt(char c) 
{ 
    switch (c) 
    { 
     case '0': return 1; 
     case '1': return 0; 
     case '8': return 19; 
     case '9': return 21; 
     case 'a': 
     case 'A': return 1; 
     case 'b': 
     case 'B': return 0; 
     // ... 
     default: return null; 
    } 
} 

然後：

for (int i = 0; i < 8; i++) 
{ 
    int? number = ConvertCharToInt(oddF[i]); 
    if (number.HasValue) 
     oddFValue[i] = number.Value; 
    else 
     // Character cannot be converted to int value 
}