以下内容可以编码为一个函数。您需要去掉前导零以满足问题的要求。

declare @intvalue int
set @intvalue=5

declare @vsresult varchar(64)
declare @inti int
select @inti = 64, @vsresult =   
while @inti>0
  begin
    select @vsresult=convert(char(1), @intvalue % 2)+@vsresult
    select @intvalue = convert(int, (@intvalue / 2)), @inti=@inti-1
  end
select @vsresult

事实上，使用普通的旧SQL非常简单。只需使用位AND。我有点惊讶，网上没有发布一个简单的解决方案（不调用UDF）。在我的情况下，我真的想检查比特是开还是关（数据来自dotnet eNums）。

因此，这里有一个例子，它将分别给出位值和二进制字符串（大并集只是一种生成跨数据库的数字的方法：

    select t.Number
    , cast(t.Number & 64 as bit) as bit7
    , cast(t.Number & 32 as bit) as bit6
    , cast(t.Number & 16 as bit) as bit5
    , cast(t.Number & 8 as bit) as bit4
    , cast(t.Number & 4 as bit) as bit3
    , cast(t.Number & 2 as bit)  as bit2
    ,cast(t.Number & 1 as bit) as bit1

    , cast(cast(t.Number & 64 as bit) as CHAR(1)) 
    +cast( cast(t.Number & 32 as bit) as CHAR(1))
    +cast( cast(t.Number & 16 as bit)  as CHAR(1))
    +cast( cast(t.Number & 8 as bit)  as CHAR(1))
    +cast( cast(t.Number & 4 as bit)  as CHAR(1))
    +cast( cast(t.Number & 2 as bit)   as CHAR(1))
    +cast(cast(t.Number & 1 as bit)  as CHAR(1)) as binary_string
    --to explicitly answer the question, on MSSQL without using REGEXP (which would make it simple)
    ,SUBSTRING(cast(cast(t.Number & 64 as bit) as CHAR(1)) 
                    +cast( cast(t.Number & 32 as bit) as CHAR(1))
                    +cast( cast(t.Number & 16 as bit)  as CHAR(1))
                    +cast( cast(t.Number & 8 as bit)  as CHAR(1))
                    +cast( cast(t.Number & 4 as bit)  as CHAR(1))
                    +cast( cast(t.Number & 2 as bit)   as CHAR(1))
                    +cast(cast(t.Number & 1 as bit)  as CHAR(1))
                    ,
                    PATINDEX( %1% , cast(cast(t.Number & 64 as bit) as CHAR(1)) 
                                        +cast( cast(t.Number & 32 as bit) as CHAR(1))
                                        +cast( cast(t.Number & 16 as bit)  as CHAR(1))
                                        +cast( cast(t.Number & 8 as bit)  as CHAR(1))
                                        +cast( cast(t.Number & 4 as bit)  as CHAR(1))
                                        +cast( cast(t.Number & 2 as bit)   as CHAR(1))
                                        +cast(cast(t.Number & 1 as bit)  as CHAR(1)  )
                    )
,99)


from (select 1 as Number union all select 2 union all select 3 union all select 4 union all select 5 union all select 6 
    union all select 7 union all select 8 union all select 9 union all select 10) as t

产生以下结果：

num  bit7 bit6 bit5 bit4 bit3 bit2 bit1 binary_string   binary_string_trimmed 
1    0    0    0    0    0    0    1    0000001         1
2    0    0    0    0    0    1    0    0000010         10
3    0    0    0    0    0    1    1    0000011         11
4    0    0    0    1    0    0    0    0000100         100
5    0    0    0    0    1    0    1    0000101         101
6    0    0    0    0    1    1    0    0000110         110
7    0    0    0    0    1    1    1    0000111         111
8    0    0    0    1    0    0    0    0001000         1000
9    0    0    0    1    0    0    1    0001001         1001
10   0    0    0    1    0    1    0    0001010         1010

这是一个通用的基本转换器

http://dpatrickcaldwell.blogspot.com/2009/05/converting-decimal-to-hexadecimal-with.html

你可以

select reverse(dbo.ConvertToBase(5, 2))   -- 101

这里对Sean接受的回答做了一些更改，因为我发现它限制了在输出中只允许硬编码的数字。在我的日常使用中，我发现只获取最高的1位数或指定我期望返回的位数更有用。它会自动在边上填充0，这样它就可以排到8、16或任何你想要的位数。

Create function f_DecimalToBinaryString
    (
        @Dec int,
        @MaxLength int = null
    )
Returns varchar(max)
as Begin

    Declare @BinStr varchar(max) =   ;

    -- Perform the translation from Dec to Bin
    While @Dec > 0 Begin

        Set @BinStr = Convert(char(1), @Dec % 2) + @BinStr;
        Set @Dec = Convert(int, @Dec /2);

    End;

    -- Either pad or trim the output to match the number of digits specified.
    If (@MaxLength is not null) Begin
        If @MaxLength <= Len(@BinStr) Begin -- Trim down
            Set @BinStr = SubString(@BinStr, Len(@BinStr) - (@MaxLength - 1), @MaxLength);
        End Else Begin -- Pad up
            Set @BinStr = Replicate( 0 , @MaxLength - Len(@BinStr)) + @BinStr;
        End;
    End;

    Return @BinStr;

End;

SELECT   r.RN
FROM     dbo.rangeAB(0,30,1,0) AS r
ORDER BY r.RN;

当我尝试降序排序时，我会在执行计划中进行排序。

输入有限对手。RangeAB包括一个名为Op-Op RN s有限Op位置编号的列。我所说的“有限对立”是指，0是30的对立，1是29的对立，等等。与传统的对立数字不同（-1是1的对立）。有限的对立面按降序返回。

SELECT   r.RN, r.OP
FROM     dbo.rangeAB(0,30,1,0) AS r
ORDER BY r.RN;

返回：

RN    OP
----- -------
0     30
1     29
2     28
3     27
....
27    3
28    2
29    1
30    0

我可以使用Op。我可以利用RN的有限相反来按降序获取数字，同时仍然利用虚拟索引来避免排序。这两个查询返回相同的东西，但在比较执行计划时，根据SSMS，删除排序将查询成本降低了50倍。

功能

CREATE FUNCTION dbo.NumberToBinary(@input INT)
RETURNS TABLE WITH SCHEMABINDING AS RETURN
/* Created By Alan Burstein 20191112, Requires RangeAB (code below) */
SELECT BIN = (
  SELECT       @input/f.Np2%2
  FROM        dbo.rangeAB(0,30,1,0)   AS r
  CROSS APPLY (VALUES(POWER(2,r.Op))) AS f(NP2)
  WHERE       (@input = 0 AND f.Np2 = 1) OR @input >= f.Np2
  ORDER BY     ROW_NUMBER() OVER (ORDER BY (SELECT NULL))
  FOR XML PATH(  ));

范围AB

CREATE FUNCTION dbo.rangeAB
(
  @low  bigint, 
  @high bigint, 
  @gap  bigint,
  @row1 bit
)
/****************************************************************************************
[Purpose]:
 Creates up to 531,441,000,000 sequentia1 integers numbers beginning with @low and ending 
 with @high. Used to replace iterative methods such as loops, cursors and recursive CTEs 
 to solve SQL problems. Based on Itzik Ben-Gan s getnums function with some tweeks and 
 enhancements and added functionality. The logic for getting rn to begin at 0 or 1 is 
 based comes from Jeff Moden s fnTally function. 

 The name range because it s similar to clojure s range function. The name "rangeAB" as 
 used because "range" is a reserved SQL keyword.

[Author]: Alan Burstein

[Compatibility]: 
 SQL Server 2008+ and Azure SQL Database

[Syntax]:
 SELECT r.RN, r.OP, r.N1, r.N2
 FROM dbo.rangeAB(@low,@high,@gap,@row1) AS r;

[Parameters]:
 @low  = a bigint that represents the lowest value for n1.
 @high = a bigint that represents the highest value for n1.
 @gap  = a bigint that represents how much n1 and n2 will increase each row; @gap also
         represents the difference between n1 and n2.
 @row1 = a bit that represents the first value of rn. When @row = 0 then rn begins
         at 0, when @row = 1 then rn will begin at 1.

[Returns]:
 Inline Table Valued Function returns:
 rn = bigint; a row number that works just like T-SQL ROW_NUMBER() except that it can 
      start at 0 or 1 which is dictated by @row1.
 op = bigint; returns the "opposite number that relates to rn. When rn begins with 0 and
      ends with 10 then 10 is the opposite of 0, 9 the opposite of 1, etc. When rn begins
      with 1 and ends with 5 then 1 is the opposite of 5, 2 the opposite of 4, etc...
 n1 = bigint; a sequential number starting at the value of @low and incrimentingby the
      value of @gap until it is less than or equal to the value of @high.
 n2 = bigint; a sequential number starting at the value of @low+@gap and  incrimenting 
      by the value of @gap.

[Dependencies]:
N/A

[Developer Notes]:

 1. The lowest and highest possible numbers returned are whatever is allowable by a 
    bigint. The function, however, returns no more than 531,441,000,000 rows (8100^3). 
 2. @gap does not affect rn, rn will begin at @row1 and increase by 1 until the last row
    unless its used in a query where a filter is applied to rn.
 3. @gap must be greater than 0 or the function will not return any rows.
 4. Keep in mind that when @row1 is 0 then the highest row-number will be the number of
    rows returned minus 1
 5. If you only need is a sequential set beginning at 0 or 1 then, for best performance
    use the RN column. Use N1 and/or N2 when you need to begin your sequence at any 
    number other than 0 or 1 or if you need a gap between your sequence of numbers. 
 6. Although @gap is a bigint it must be a positive integer or the function will
    not return any rows.
 7. The function will not return any rows when one of the following conditions are true:
      * any of the input parameters are NULL
      * @high is less than @low 
      * @gap is not greater than 0
    To force the function to return all NULLs instead of not returning anything you can
    add the following code to the end of the query:

      UNION ALL 
      SELECT NULL, NULL, NULL, NULL
      WHERE NOT (@high&@low&@gap&@row1 IS NOT NULL AND @high >= @low AND @gap > 0)

    This code was excluded as it adds a ~5% performance penalty.
 8. There is no performance penalty for sorting by rn ASC; there is a large performance 
    penalty for sorting in descending order WHEN @row1 = 1; WHEN @row1 = 0
    If you need a descending sort the use op in place of rn then sort by rn ASC. 

Best Practices:
--===== 1. Using RN (rownumber)
 -- (1.1) The best way to get the numbers 1,2,3...@high (e.g. 1 to 5):
 SELECT RN FROM dbo.rangeAB(1,5,1,1);
 -- (1.2) The best way to get the numbers 0,1,2...@high-1 (e.g. 0 to 5):
 SELECT RN FROM dbo.rangeAB(0,5,1,0);

--===== 2. Using OP for descending sorts without a performance penalty
 -- (2.1) The best way to get the numbers 5,4,3...@high (e.g. 5 to 1):
 SELECT op FROM dbo.rangeAB(1,5,1,1) ORDER BY rn ASC;
 -- (2.2) The best way to get the numbers 0,1,2...@high-1 (e.g. 5 to 0):
 SELECT op FROM dbo.rangeAB(1,6,1,0) ORDER BY rn ASC;

--===== 3. Using N1
 -- (3.1) To begin with numbers other than 0 or 1 use N1 (e.g. -3 to 3):
 SELECT N1 FROM dbo.rangeAB(-3,3,1,1);
 -- (3.2) ROW_NUMBER() is built in. If you want a ROW_NUMBER() include RN:
 SELECT RN, N1 FROM dbo.rangeAB(-3,3,1,1);
 -- (3.3) If you wanted a ROW_NUMBER() that started at 0 you would do this:
 SELECT RN, N1 FROM dbo.rangeAB(-3,3,1,0);

--===== 4. Using N2 and @gap
 -- (4.1) To get 0,10,20,30...100, set @low to 0, @high to 100 and @gap to 10:
 SELECT N1 FROM dbo.rangeAB(0,100,10,1);
 -- (4.2) Note that N2=N1+@gap; this allows you to create a sequence of ranges.
 --       For example, to get (0,10),(10,20),(20,30).... (90,100):
 SELECT N1, N2 FROM dbo.rangeAB(0,90,10,1);
 -- (4.3) Remember that a rownumber is included and it can begin at 0 or 1:
 SELECT RN, N1, N2 FROM dbo.rangeAB(0,90,10,1);

[Examples]:
--===== 1. Generating Sample data (using rangeAB to create "dummy rows")
 -- The query below will generate 10,000 ids and random numbers between 50,000 and 500,000
 SELECT
   someId    = r.rn,
   someNumer = ABS(CHECKSUM(NEWID())%450000)+50001 
 FROM rangeAB(1,10000,1,1) r;

--===== 2. Create a series of dates; rn is 0 to include the first date in the series
 DECLARE @startdate DATE =  20180101 , @enddate DATE =  20180131 ;

 SELECT r.rn, calDate = DATEADD(dd, r.rn, @startdate)
 FROM dbo.rangeAB(1, DATEDIFF(dd,@startdate,@enddate),1,0) r;
 GO

--===== 3. Splitting (tokenizing) a string with fixed sized items
 -- given a delimited string of identifiers that are always 7 characters long
 DECLARE @string VARCHAR(1000) =  A601225,B435223,G008081,R678567 ;

 SELECT
   itemNumber = r.rn, -- item s ordinal position 
   itemIndex  = r.n1, -- item s position in the string (it s CHARINDEX value)
   item       = SUBSTRING(@string, r.n1, 7) -- item (token)
 FROM dbo.rangeAB(1, LEN(@string), 8,1)  r;
 GO

--===== 4. Splitting (tokenizing) a string with random delimiters
 DECLARE @string VARCHAR(1000) =  ABC123,999F,XX,9994443335 ;

 SELECT
   itemNumber = ROW_NUMBER() OVER (ORDER BY r.rn), -- item s ordinal position 
   itemIndex  = r.n1+1, -- item s position in the string (it s CHARINDEX value)
   item       = SUBSTRING
               (
                 @string,
                 r.n1+1,
                 ISNULL(NULLIF(CHARINDEX( , ,@string,r.n1+1),0)-r.n1-1, 8000)
               ) -- item (token)
 FROM dbo.rangeAB(0,DATALENGTH(@string),1,1) r
 WHERE SUBSTRING(@string,r.n1,1) =  ,  OR r.n1 = 0;
 -- logic borrowed from: http://www.sqlservercentral.com/articles/Tally+Table/72993/

--===== 5. Grouping by a weekly intervals
 -- 5.1. how to create a series of start/end dates between @startDate & @endDate
 DECLARE @startDate DATE =  1/1/2015 , @endDate DATE =  2/1/2015 ;
 SELECT 
   WeekNbr   = r.RN,
   WeekStart = DATEADD(DAY,r.N1,@StartDate), 
   WeekEnd   = DATEADD(DAY,r.N2-1,@StartDate)
 FROM dbo.rangeAB(0,datediff(DAY,@StartDate,@EndDate),7,1) r;
 GO

 -- 5.2. LEFT JOIN to the weekly interval table
 BEGIN
  DECLARE @startDate datetime =  1/1/2015 , @endDate datetime =  2/1/2015 ;
  -- sample data 
  DECLARE @loans TABLE (loID INT, lockDate DATE);
  INSERT @loans SELECT r.rn, DATEADD(dd, ABS(CHECKSUM(NEWID())%32), @startDate)
  FROM dbo.rangeAB(1,50,1,1) r;

  -- solution 
  SELECT 
    WeekNbr   = r.RN,
    WeekStart = dt.WeekStart, 
    WeekEnd   = dt.WeekEnd,
    total     = COUNT(l.lockDate)
  FROM dbo.rangeAB(0,datediff(DAY,@StartDate,@EndDate),7,1) r
  CROSS APPLY (VALUES (
    CAST(DATEADD(DAY,r.N1,@StartDate) AS DATE), 
    CAST(DATEADD(DAY,r.N2-1,@StartDate) AS DATE))) dt(WeekStart,WeekEnd)
  LEFT JOIN @loans l ON l.lockDate BETWEEN  dt.WeekStart AND dt.WeekEnd
  GROUP BY r.RN, dt.WeekStart, dt.WeekEnd ;
 END;

--===== 6. Identify the first vowel and last vowel in a along with their positions
 DECLARE @string VARCHAR(200) =  This string has vowels ;

 SELECT TOP(1) position = r.rn, letter = SUBSTRING(@string,r.rn,1)
 FROM dbo.rangeAB(1,LEN(@string),1,1) r
 WHERE SUBSTRING(@string,r.rn,1) LIKE  %[aeiou]% 
 ORDER BY r.rn;

 -- To avoid a sort in the execution plan we ll use op instead of rn
 SELECT TOP(1) position = r.op, letter = SUBSTRING(@string,r.op,1)
 FROM dbo.rangeAB(1,LEN(@string),1,1) r
 WHERE SUBSTRING(@string,r.rn,1) LIKE  %[aeiou]% 
 ORDER BY r.rn;

---------------------------------------------------------------------------------------
[Revision History]:
 Rev 00 - 20140518 - Initial Development - Alan Burstein
 Rev 01 - 20151029 - Added 65 rows to make L1=465; 465^3=100.5M. Updated comment section
                   - Alan Burstein
 Rev 02 - 20180613 - Complete re-design including opposite number column (op)
 Rev 03 - 20180920 - Added additional CROSS JOIN to L2 for 530B rows max - Alan Burstein
****************************************************************************************/
RETURNS TABLE WITH SCHEMABINDING AS RETURN
WITH L1(N) AS 
(
  SELECT 1
  FROM (VALUES
   (0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),
   (0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),
   (0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),
   (0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),
   (0),(0)) T(N) -- 90 values 
),
L2(N)  AS (SELECT 1 FROM L1 a CROSS JOIN L1 b CROSS JOIN L1 c),
iTally AS (SELECT rn = ROW_NUMBER() OVER (ORDER BY (SELECT 1)) FROM L2 a CROSS JOIN L2 b)
SELECT
  r.RN,
  r.OP,
  r.N1,
  r.N2
FROM
(
  SELECT
    RN = 0,
    OP = (@high-@low)/@gap,
    N1 = @low,
    N2 = @gap+@low
  WHERE @row1 = 0
  UNION ALL -- ISNULL required in the TOP statement below for error handling purposes
  SELECT TOP (ABS((ISNULL(@high,0)-ISNULL(@low,0))/ISNULL(@gap,0)+ISNULL(@row1,1)))
    RN = i.rn,
    OP = (@high-@low)/@gap+(2*@row1)-i.rn,
    N1 = (i.rn-@row1)*@gap+@low,
    N2 = (i.rn-(@row1-1))*@gap+@low
  FROM iTally AS i
  ORDER BY i.rn
) AS r
WHERE @high&@low&@gap&@row1 IS NOT NULL AND @high >= @low AND @gap > 0;
GO

我相信这种方法简化了其他人提出的许多其他想法。它使用逐位算术和FOR XML技巧以及CTE来生成二进制数字。

DECLARE @my_int INT = 5

;WITH CTE_Binary AS
(
    SELECT 1 AS seq, 1 AS val
    UNION ALL
    SELECT seq + 1 AS seq, power(2, seq)
    FROM CTE_Binary
    WHERE
        seq < 8
)
SELECT
(
    SELECT
        CAST(CASE WHEN B2.seq IS NOT NULL THEN 1 ELSE 0 END AS CHAR(1))
    FROM
        CTE_Binary B1
    LEFT OUTER JOIN CTE_Binary B2 ON
        B2.seq = B1.seq AND
        @my_int & B2.val = B2.val
    ORDER BY
        B1.seq DESC
    FOR XML PATH(  )
) AS val

You can use a recursive CTE table to do this. In this example code, it is set for 16 bits, but you can do any length by changing 16-> your choice. Also, the data you want to convert is table DecimalTable

WITH DecimalTable AS (SELECT 10 decimal_num UNION SELECT 20),
     DtoB AS (SELECT decimal_num
                     ,1 n
                     ,CAST(CAST(decimal_num%2 AS bit) AS VARCHAR(16)) binary_num
               FROM DecimalTable 
               UNION ALL
               SELECT decimal_num
                     ,n*2 n
                     ,CAST(CONCAT(CAST(decimal_num&n as bit), binary_num)
                           AS VARCHAR(16)) binary_num
                FROM DtoB
                WHERE n<POWER(2,16))

    SELECT decimal_num, binary_num
    FROM DtoB

此函数是一个通用转换器，允许将整数转换为任何基本编号系统的字符串描述，如二进制、八进制、十六进制等。

-- specify a string and numbering system Base value, for example 16 for hexadecimal
CREATE FUNCTION udf_IntToBaseXStr(@baseVal BIGINT,
                                    @baseX   BIGINT)
returns VARCHAR(63)
AS
    BEGIN
        --bigint : -2^63 (-9,223,372,036,854,775,808) to 2^63-1 (9,223,372,036,854,775,807)  
        --         or 63 ones (1111111,11111111,11111111,11111111,11111111,11111111,11111111,11111111) in binary
        DECLARE @val BIGINT -- value from all
        DECLARE @cv BIGINT -- value from a single char
        DECLARE @baseStr VARCHAR(63)
        SET @baseStr =   ;
        -- assumes a numbering method of 0123456789ABCDEF..... 
        SET @val = @baseVal
        WHILE ( @val > 0 )
            BEGIN
                SET @cv = @val % @basex -- calculate the right most char s value
                SET @baseStr = -- add it to (any existing) string
                CASE
                    WHEN @cv < 10 THEN Char(Ascii( 0 ) + @cv)
                    ELSE Char(Ascii( A ) + ( @cv - 10 ))
                END
                + @baseStr
                SET @val = ( @val - @cv ) / @basex
            END
        RETURN @baseStr
    END
GO

如果您需要保证最小长度，下一个函数会包装上面的函数，在前面加上一个ZEROES，迫使返回的字符串达到您想要的最小长度。它不会截断到指定的长度。

-- specify a string and numbering system Base value, for example, 16 for hexadecimal
-- prepends LEADING ZEROS to force length of returned string to be AT LEAST minLength chars
CREATE FUNCTION udf_IntToBaseXStr_MinLength(@baseVal   BIGINT,
                                            @baseX     BIGINT,
                                            @minLength INT)
returns VARCHAR(63)
AS
    BEGIN
        DECLARE @baseStr VARCHAR(63)
        SET @baseStr = dbo.udf_IntToBaseXStr(@baseVal, @baseX)
        IF Len(@baseStr) < @minLength
            SET @baseStr = Replicate( 0 , @minLength - Len(@baseStr))
                                         + @baseStr
        RETURN @baseStr
    END
GO

udf_IntToBaseXStr用法：

;with CTE as 
(
    SELECT BaseX = 2, AKA =  binary  
    UNION SELECT 8,  octal  
    UNION SELECT 10,  decimal  
    UNION SELECT 15,  pentadecimal  
    UNION SELECT 16,  hexadecimal  
)
SELECT BaseX, AKA, Result = dbo.udf_IntToBaseXStr(328239523, BaseX) FROM CTE

udf_IntToBaseXStr结果：

udf_IntToBaseXStr_MinLength用法：

;with CTE as 
(
    SELECT BaseX = 2, AKA =  binary  
    UNION SELECT 8,  octal  
    UNION SELECT 10,  decimal  
    UNION SELECT 15,  pentadecimal  
    UNION SELECT 16,  hexadecimal  
)
SELECT BaseX, AKA, Result = dbo.udf_IntToBaseXStr_MinLength(328239523, BaseX, 24) FROM CTE

udf_IntToBaseXStr_MinLength结果：

想要轻松吗？做一些逐位计算来映射出每个二进制数字。

CREATE FUNCTION dbo.BinaryRep (@val INT)
RETURNS VARCHAR(32)
WITH EXECUTE AS CALLER
AS

BEGIN
    DECLARE @ret VARCHAR(32)
    DECLARE @cnt INT = 30; -- 30 to 0 inclusive in loop
    
    -- handle negative (we re using signed magnitude because that s simple)
    SET @ret = IIF(@val < 0,  1 ,  0 );
    SET @val = ABS(@val); -- totally cheating here.
    
    -- bitwise masking madness, one digit at a time.   
    WHILE @cnt > -1
    BEGIN
        SET @ret = CONCAT(@ret, IIF(@val & POWER(2, @cnt) = 0, 0, 1));
        SET @cnt = @cnt - 1;
    END;
    
    RETURN @ret;
END

唯一的转折点正是康斯坦丁注意到的：你觉得你的底片怎么样？

此版本价格低廉，使用有符号幅度，其中第一位为1，表示负值，没有其他更改。-123和123仅因其高位不同。

select dbo.BinaryRep(123) as plus, dbo.BinaryRep(-123) as minus

plus                             minus
-------------------------------- --------------------------------
00000000000000000000000001111011 10000000000000000000000001111011

请注意，SQLServerINT支持2^-31到2³¹，因此我们需要循环31次（包括30到0），而不是32次。

为什么不简单地。。。

declare @num int = 75

select
@num [Dec]
, convert (varchar(1), @num / 128 % 2)
+ convert (varchar(1), @num / 64 % 2)
+ convert (varchar(1), @num / 32 % 2)
+ convert (varchar(1), @num / 16 % 2)
+ convert (varchar(1), @num / 8 % 2)
+ convert (varchar(1), @num / 4 % 2)
+ convert (varchar(1), @num / 2 % 2)
+ convert (varchar(1), @num % 2) as [Bin]