C＃中的按位操作 [英] Bit wise operations in C#

问题描述

大家好，



看来我之前做的二进制混音是对的（但错了）但我需要
将二进制数作为位字段，下面的代码粗略地说明了我希望二进制数字中的4位数值

Hi all,

It appears that the binary mucking I did earlier was right (but wrong) it but I
need to go over the binary number as bit field the below code is a rough lash up of how I want the 4 bit values from the Binary number

char[] Input;

          for (int i = 0; i < 4; i++)
          {
              if (Input[i] > 'Z')
                  Input[i] -= 0x20;  // Convert lower case to upper case

              if ((Input[i] >= '0') && (Input[i] <= '9')) // First check to see if the number is 0 to 9
              {
                  Input[i] -= '0';  //If so, then subtract ascii ‘0’ from it

              }

              else if ((Input[i] >= 'A') && (Input[i] <= 'F'))// is this the hex codes A to F?
              {
                  Input[i] -= 0x37;  // Subtract 55 (0x37) so ascii A becomes 10
              }
             else
                  Input[i] = 0;  // Else invalid so substitute 0
          }

如果您和&使用0x08，结果不为0，然后设置第3（4）位



如果你&使用0x04并且结果不为0然后设置位2（3）



如果你&使用0x02并且结果不为0然后设置位1（2）



如果你&使用0x01并且结果不为0然后设置位0（1）



我想做旧的C路线（a&& amp; 2）等男人我得回家了！



格伦

（如果这个问题达到顶峰，我会改写并明天更清洁）





它有，所以改为时间我有一个0到15之间的字符，上面的代码（有点）将它转换为尝试获得0到15的价值。我相信应该有一个ConvertTo来做这个但是....

格伦

If you & with 0x08 and the result is not 0 then bit 3(4) is set

If you & with 0x04 and the result is not 0 then bit 2(3) is set

If you & with 0x02 and the result is not 0 then bit 1(2) is set

If you & with 0x01 and the result is not 0 then bit 0(1) is set

I want to do the old C route of doing it bit wise (a&&2) etc. Man I have to go home!

Glenn
(If the question peaks interest I will rephrase and make it cleaner tomorrow)


Umm it has, so rephrase time I have a character between 0 to 15 the code above (sort of) converted it to try to get 0 to 15 value out. I believe there should be a ConvertTo to do this but....
Glenn

推荐答案

在这段代码中，没有类似的东西位操作本身。请参阅运营商列表：

http： //msdn.microsoft.com/en-us/library/6a71f45d%28v=vs.110%29.aspx [ ^ ]。



你会的需要二进制移位''>>''，''<<''，二进制AND，OR，XOR和NOT（''&''，''|''，''''''，' '〜''）。另请参阅：

http ：//blog.typps.com/2007/10/bitwise-operators-in-c-or-xor-and-not.html [ ^ ]。

-SA

In this code, there is nothing with resembles bit operations per se. Refer to the operator list:
http://msdn.microsoft.com/en-us/library/6a71f45d%28v=vs.110%29.aspx[^].

You will need binary shift ''>>'', ''<<'', binary AND, OR, XOR and NOT (''&'', ''|'', ''^'', ''~''). See also:
http://blog.typps.com/2007/10/bitwise-operators-in-c-or-xor-and-not.html[^].

—SA

你可以参考检查位标志你可以在C＃中执行此操作：

On the off hand chance you are refering to checking bit flags you could do this in C#:

public enum BitFlag : byte
        {
            First=1,
            Second=2,
            Third=4,
            Fourth=8
        }

        public static string CheckFlag()
        {
            BitFlag bFlag = BitFlag.First | BitFlag.Third;
            StringBuilder result = new StringBuilder();

            if ((bFlag & BitFlag.First) == BitFlag.First)
                result.AppendLine("First");
            if ((bFlag & BitFlag.Second) == BitFlag.Second)
                result.AppendLine("Second");
            if ((bFlag & BitFlag.Third) == BitFlag.Third)
                result.AppendLine("Third");
            if ((bFlag & BitFlag.Fourth) == BitFlag.Fourth)
                result.AppendLine("Fourth");
            return result.ToString();
        }

你做的是像

What you do is something like

int i = int.TryParse(s, System.Globalization.NumberStyles.AllowHexSpecifier, null, out i)
      ? i
      : 0;

这与位运算符无关。

Andi

This has nothing to do with bit operators.
Andi

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