一个很有用的自定义函数(判断自然数是否包含2的指定次幂)

/*
           Name :    Fun_WheIncluded
           Function :   判断选定的数字是否在给定的整数中
           可以知道任何一个自然数都可以拆分成若干个2的幂的和，如：
                1 = 2^0
                2 = 2^1
                3 = 2^0 + 2^1
                4 = 2^2
                5 = 2^0 + 2^2
                6 = 2^1 + 2^4
                7 = 2^0 + 2^1 + 2^2
                8 = 2^3
                9 = 2^0 + 2^3
                10 = 2^1 + 2^3
                11 = 2^0 + 2^1 + 2^3
                12 = 2^2 + 2^3
                13 = 2^0 + 2^2 + 2^3
                14 = 2^1 + 2^2 + 2^3
                15 = 2^0 + 2^1 + 2^2 + 2^3
                16 = 2^4
                17 = 2^0 + 2^4
     
    将任意一个数解析为2的幂的和的方法——递归
    规律：
                如给定 14
                  ∵ 2^3 < 14 < 2^4
                  ∴ 14中必有8——2^3
                  14 - 8 = 6
                  ∵ 2^2 < 6 < 2^3
                  ∴ 6中必有4——2^2
                  6 - 4 = 2
       
                  ∵ 2 = 2
                  ∴ 14 = 2^3 + 2^2 + 2^1

            Parameters :  @TotalNum
           Type:   INT
           @SpecifiedNum  
           Type:   INT
            Steps :    
            Author :   Waxdoll Cheung
            Date :    2005-03-21
*/

CREATE FUNCTION
 dbo.Fun_WheIncluded
 (
  @TotalNum INT,
  @SpecifiedNum INT
 )
RETURNS BIT AS 
BEGIN

 DECLARE @varRet BIT

 DECLARE @varLoop INT

 SET @varLoop = 0

 WHILE (@TotalNum >= CAST(POWER(2, @VarLoop) AS INT))
  SET @varLoop = @varLoop + 1

 SET @TotalNum = @TotalNum - CAST(POWER(2, @varLoop - 1) AS INT)

 IF (@varLoop = @SpecifiedNum + 1)
  SET @varRet = 1
 ELSE
 BEGIN
  IF (@TotalNum >= 1)
   RETURN dbo.Fun_WheIncluded(@TotalNum, @SpecifiedNum)
  ELSE
   SET @varRet = 0
 END

 RETURN @varRet
END