一个16进制转10进制函数运行异常

一个16进制转10进制函数运行错误？
CREATE FUNCTION dbo.f_hex_dec(@s varchar(32))
RETURNS bigint
AS
BEGIN
  DECLARE @i bigint,@result bigint
  SELECT @i=0,@result=0,@s=RTRIM(LTRIM(UPPER(REVERSE(@s))))
  WHILE @i<LEN(@s)
  BEGIN
  IF SUBSTRING(@s,@i+1,1) not between '0' and '9' and SUBSTRING(@s,@i+1,1) not between 'A' and 'F'
  BEGIN
  SELECT @result=0
  break
  END
  SELECT @result=@result+(CHARINDEX(SUBSTRING(@s,@i+1,1),'0123456789ABCDEF')-1)*POWER(16,@i),@i=@i+1
  END
  RETURN @result
END
GO

调用函数运行：select oid from lists where oid=dbo.f_hex_dec('7a7a003500bc0001') and state=0 and tid=13出现错误为：类型 int 发生算术溢出错误，值 = 4294967296.000000。
为什么呢？RETURNS为bigint，@i bigint,@result bigint都为bigint，这些来存储16个16进制数应该没问题啊？请知道者帮忙修改下代码？谢谢了

[解决办法]
修改后

SQL code

create FUNCTION dbo.f_hex_dec(@s varchar(max))  RETURNS bigint  AS  BEGIN      DECLARE @i BIGINT ,          @result BIGINT              DECLARE @data BIGINT  --添加了这个    SET @data =16        SELECT  @i = 0 ,              @result = 0 ,              @s = RTRIM(LTRIM(UPPER(REVERSE(@s))))      WHILE @i < DATALENGTH(@s)           BEGIN              IF SUBSTRING(@s, @i + 1, 1) NOT BETWEEN '0' AND '9'                  AND SUBSTRING(@s, @i + 1, 1) NOT BETWEEN 'A' AND 'F'                   BEGIN                      SELECT  @result = 0                      BREAK                  END              SELECT  @result = @result + ( CHARINDEX(SUBSTRING(@s, @i + 1, 1),                                                      '0123456789ABCDEF') - 1 )                      * POWER(@data, @i) ,                      @i = @i + 1          END      RETURN @result  END  SELECT dbo.f_hex_dec('7a7a003500bc0001')--8825366647431495681