Float Storage And Usage

Jun 7, 2007

Greetings.

I need to convert some columns of type numeric(12, 0) to hold floating point
information scale and precision I can't determine in advance (customer data
can vary wildly) so I wanted to use the datatype that offers the maximum scale
and precision..

I'm targeting sql server 2005 systems (not 2000).

It seems my choices are real and float, and the docs seem to indicate that
float offers with widest ranges.

I'm trying out using the 'float' for the new data type as the default
precision is said to be 53.. Does this mean the total number of digits is up
to 53?

I don't know if there is anything else I need to take into account since these
two columns are part of a primary key, and I supposed, therefore, are indexed.

thanks
Jeff Kish

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Storage Usage For Enterprise Applications

Sep 4, 2006

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I can't take full credit for this. I want to share this with Jeff Moden who did the important research for this calculation here.

All I did was just adapting some old code according to the mantissa finding Jeff made and optimized it a little


Some test codeDECLARE@SomeNumber FLOAT,
@BinFloat BINARY(8)

SELECT@SomeNumber = -185.6125,
@BinFloat = CAST(@SomeNumber AS BINARY(8))

SELECT@SomeNumber AS [Original],
CAST(@SomeNumber AS BINARY(8)) AS [Binary],
dbo.fnBinaryFloat2Float(CAST(@SomeNumber AS BINARY(8))) AS [Converted],
@SomeNumber - dbo.fnBinaryFloat2Float(CAST(@SomeNumber AS BINARY(8))) AS [Error]

And here is the code for the function.CREATE FUNCTION dbo.fnBinaryFloat2Float
(
@BinaryFloat BINARY(8)
)
RETURNS FLOAT
AS
BEGIN
DECLARE@Part TINYINT,
@PartValue TINYINT,
@Mask TINYINT,
@Mantissa FLOAT,
@Exponent SMALLINT,
@Bit TINYINT,
@Ln2 FLOAT,
@BigValue BIGINT

SELECT@Part = 1,
@Mantissa = 1,
@Bit = 1,
@Ln2 = LOG(2),
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WHILE @Part <= 8
BEGIN
SELECT@Part = @Part + 1,
@PartValue = CAST(SUBSTRING(@BinaryFloat, @Part, 1) AS TINYINT),
@Mask =CASE WHEN @Part = 2 THEN 8 ELSE 128 END

WHILE @Mask > 0
BEGIN
IF @PartValue & @Mask > 0
SET @Mantissa = @Mantissa + EXP(-@Bit * @Ln2)

SELECT@Bit = @Bit + 1,
@Mask = @Mask / 2
END
END

RETURNSIGN(@BigValue) * @Mantissa * POWER(CAST(2 AS FLOAT), @Exponent - 1023)
END
Thanks again Jeff!


Peter Larsson
Helsingborg, Sweden

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