使用C＃計算二進制數中連續的數

Question

我想讓程序遍歷從00000000到11111111的每個可能的二進制數，並計算出連續的“運行次數”。 例如）00000001和11100000都計為一次的運行00001010和11101110都計為兩次的運行

問題是，它忽略了AND遮罩部分，我不知道為什么。

{
    static void Main(string[] args)
    {
        //Start
        int stuff = BitRunner8();

        //Display
        Console.Write(stuff);
        Console.ReadKey();
    }
    public static int BitRunner8()
    {
        int uniRunOrNot = 0;
        int uniRunCount = 0;
        int uniRunTotal = 0;

        //iterating from numbers 0 to 255
        for (int x = 0; x < 255; x++)
        {
            //I use 128 as my AND mask because 128 is 10000000 in binary
            for ( int uniMask = 128; uniMask != 0; uniMask >>= 1)
            {

                //This is the if statement that doesn't return true ever
                if ((x & uniMask) != 0)

                {
                    //If the and mask is true, and ther were no previous ones before it, add to the the uniRunCount
                    if (uniRunOrNot == 0)
                    {
                        //Total count of the runs
                        uniRunCount++;
                    }
                    // Making it so that if two consective ones are in a row, the 'if' statement right above would return false,
                    //so that it wouldn't add to the uniRunCount
                    uniRunOrNot++;
                }
                else
                {
                    //add the total number of runs to uniRunTotal, and then reset both uniRunOrNot, and uniRunCount
                    uniRunTotal += uniRunCount;
                    uniRunOrNot = uniRunCount = 0;
                }
            }
        }
        //Divide the final amount by 256 total numbers
        uniRunTotal /= 256;
        return uniRunCount;
    }
}

Answer 1

問題是您的代碼將忽略包含最低有效位的運行。 您的代碼僅在發現零位時更新uniRunTotal 。 當最低有效位不為零時， uniRunCount永遠不會添加到總數中。

在循環之后添加代碼以添加uniRunCount來解決此問題。

您還可以通過應用哨兵策略來解決此問題：計數另一端的位數，並使用9位而不是8位，因為位數9始終為零：

for (int uniMask = 1; uniMask <= 256; uniMask <<= 1)