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 @varRetEND

原文转自：http://www.ltesting.net