匹配數字中的一系列位，然後將匹配轉換爲零？

Question

我的任務是搜索數字的二進制表示並替換數字的另一個二進制表示的匹配模式。如果我得到一個匹配，我將第一個整數的匹配位轉換爲零，然後繼續。 例如，數字469將是111010101，我必須將其與5（101）匹配。這是我迄今爲止編寫的程序。不按預期工作。

using System; 

namespace Conductors 
{ 
    class Program 
    { 
     static void Main(string[] args) 
     { 
      //this is the number I'm searching for a match in 
      int binaryTicket = 469; 
      //This is the pattern I'm trying to match (101) 
      int binaryPerforator = 5; 

      string binaryTicket01 = Convert.ToString(binaryTicket, 2); 

      bool match = true; 
      //in a 32 bit integer, position 29 is the last one I would 
      //search in, since I'm searching for the next 3 
      for (int pos = 0; pos < 29; pos++) 
      { 
       for (int j = 0; j <= 3; j++) 
       { 
        var posInBinaryTicket = pos + j; 
        var posInPerforator = j; 

        int bitInBinaryTicket = (binaryTicket & (1 << posInBinaryTicket)) >> posInBinaryTicket; 
        int bitInPerforator = (binaryPerforator & (1 << posInPerforator)) >> posInPerforator; 

        if (bitInBinaryTicket != bitInPerforator) 
        { 
         match = false; 
         break; 
        } 
        else 
        { 
         //what would be the proper bitwise operator here? 
         bitInBinaryTicket = 0; 
        } 
       } 

       Console.WriteLine(binaryTicket01); 
      } 
     } 
    } 
} 

Answer 1

幾件事情：

1. 使用uint這一點。在處理二進制數字時使事情變得更加簡單。
2. 你並沒有真正設置任何東西 - 你只是存儲信息，這就是爲什麼你經常打印出相同的數字。
3. 你應該循環x次，其中x =二進制字符串的長度（不僅僅是29）。有沒有必要爲內環

static void Main(string[] args) 
{ 
    //this is the number I'm searching for a match in 
    uint binaryTicket = 469; 
    //This is the pattern I'm trying to match (101) 
    uint binaryPerforator = 5; 

    var numBinaryDigits = Math.Ceiling(Math.Log(binaryTicket, 2)); 
    for (var i = 0; i < numBinaryDigits; i++) 
    { 
     var perforatorShifted = binaryPerforator << i; 

     //We need to mask off the result (otherwise we fail for checking 101 -> 111) 
     //The mask will put 1s in each place the perforator is checking. 
     var perforDigits = (int)Math.Ceiling(Math.Log(perforatorShifted, 2)); 
     uint mask = (uint)Math.Pow(2, perforDigits) - 1; 

     Console.WriteLine("Ticket:\t" + GetBinary(binaryTicket)); 
     Console.WriteLine("Perfor:\t" + GetBinary(perforatorShifted)); 
     Console.WriteLine("Mask :\t" + GetBinary(mask)); 

     if ((binaryTicket & mask) == perforatorShifted) 
     { 
      Console.WriteLine("Match."); 
      //Imagine we have the case: 

      //Ticket: 
      //111010101 
      //Perforator: 
      //000000101 

      //Is a match. What binary operation can we do to 0-out the final 101? 
      //We need to AND it with 
      //111111010 

      //To get that value, we need to invert the perforatorShifted 
      //000000101 
      //XOR 
      //111111111 
      //EQUALS 
      //111111010 

      //Which would yield: 
      //111010101 
      //AND 
      //111110000 
      //Equals 
      //111010000 

      var flipped = perforatorShifted^((uint)0xFFFFFFFF); 
      binaryTicket = binaryTicket & flipped; 
     } 
    } 

    string binaryTicket01 = Convert.ToString(binaryTicket, 2); 
    Console.WriteLine(binaryTicket01); 
} 

static string GetBinary(uint v) 
{ 
    return Convert.ToString(v, 2).PadLeft(32, '0'); 
} 

請仔細閱讀了上面的代碼 - 如果有什麼你不明白，給我留下了評論，我可以通過它陪你一起跑。