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Home » Article Discussions » Article Discussions by Author » Discuss content posted by Carlo Romagnano » SET ROWCOUNT and table variable
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mm_sharma82
Posted Friday, August 27, 2010 4:13 AM
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following script doesn't go in infinite loop
DECLARE @i float,@rc int
set @i = 0

while @i <> 1
begin
set @i = @i + 0.8
set @i = @i + 0.1
set @i = @i + 0.1
print @i
end

But below script is going in infinite loop
DECLARE @i float,@rc int
set @i = 0

while @i <> 1
begin
set @i = @i + 0.7
set @i = @i + 0.1
set @i = @i + 0.1
set @i = @i + 0.1

print @i
end


Can any one know the answer on this??

Regards
Manmohan
Post #976264
Open Minded
Posted Friday, August 27, 2010 4:33 AM
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Please try to see the trailing decimal digits using
 print convert(decimal(38,36),@i)



I think the 1st example behaved well because it was dealing with an even number (8), while the 2nd does not because it was with odd numbers (7).

Because float comparison does not behave uniformly in all cases, I think this is why float comparison should not be used to test for equality, this is "equal", = and "not equal", <>.

But float may be used for less than or greater than, if accuracy/significance is required then the ideal less actual result may be used re: "while (ideal - actual) > 0.0001". The latter assumes the operations will converge close to the ideal point, otherwise there must be additional codes to check if divergence is happening in order to put a stop operation.
Post #976273
paul.knibbs
Posted Friday, August 27, 2010 4:37 AM
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Gopinath Srirangan (8/27/2010)
Hi,
But another sample query as shown below gives expected result which contradicts from while loop.

DECLARE @i float
set @i = .9
set @i = @i + .1


You're only adding one float there, so the cumulative error isn't that great. In the original WHILE loop there were ten separate additions of 0.1, which caused more of an issue.
Post #976275
Steve Eckhart
Posted Friday, August 27, 2010 7:55 AM
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Actually, we're doing the exact same thing here. This was the original query:

while @i <> 1
begin 
 declare @a table(a int)
 set @rc = @i + 0.9
 set rowcount @rc
 insert into @a select id from sysobjects
 set @i = @i + 0.1
end

So, there was only one floating point addition which made @i<>1 TRUE, but in the if statement @i<>1 is FALSE.


Steve Eckhart
Post #976410
paul.knibbs
Posted Friday, August 27, 2010 8:03 AM
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Steve Eckhart (8/27/2010)
Actually, we're doing the exact same thing here. This was the original query:

while @i <> 1
begin 
 declare @a table(a int)
 set @rc = @i + 0.9
 set rowcount @rc
 insert into @a select id from sysobjects
 set @i = @i + 0.1
end

So, there was only one floating point addition which made @i<>1 TRUE, but in the if statement @i<>1 is FALSE.


I don't see that. RC is set to @i + 0.9, then later 0.1 is added to @i. I'm not seeing the point where the value of @rc is copied into @i, which it would have to be for your statement to be correct. Maybe I'm just misreading the code?
Post #976419
Steve Eckhart
Posted Friday, August 27, 2010 8:11 AM
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Thanks, Paul. That was the part that I missed that caused the misunderstanding on my part. I got in my head that we were starting @i at 0.9, but we're not. We're starting @i at 0 and incrementing by 0.1 which causes us to "miss" 1.0.


Steve Eckhart
Post #976426
Michael Poppers
Posted Friday, August 27, 2010 10:18 AM
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mccork (8/25/2010)
The answer was almost obvious from the "do not run it on production server" recommendation.
<<snip>>

Agreed (and frankly, I wouldn't have chosen the right answer otherwise )!
Post #976557
nisan.al
Posted Sunday, August 29, 2010 11:25 PM
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I can't say i completely understand this issue.
The while statement dont not stop because @i will never be equal to 1
So how come the following select return 1 ?

DECLARE @i float

set @i = 0

set @i = @i + 0.2
set @i = @i + 0.2
set @i = @i + 0.2
set @i = @i + 0.2
set @i = @i + 0.2

SELECT 1
WHERE @i=1
Post #977163
Carlo Romagnano
Posted Monday, August 30, 2010 12:53 AM


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nisan.al (8/29/2010)
I can't say i completely understand this issue.
The while statement dont not stop because @i will never be equal to 1
So how come the following select return 1 ?

DECLARE @i float

set @i = 0

set @i = @i + 0.2
set @i = @i + 0.2
set @i = @i + 0.2
set @i = @i + 0.2
set @i = @i + 0.2

SELECT 1
WHERE @i=1

As said above, the increment of 0.2 (precise) is different from 0.1 (approximate)
Post #977174
Hugo Kornelis
Posted Monday, August 30, 2010 1:46 AM


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Carlo Romagnano (8/30/2010)
nisan.al (8/29/2010)
I can't say i completely understand this issue.
The while statement dont not stop because @i will never be equal to 1
So how come the following select return 1 ?

DECLARE @i float

set @i = 0

set @i = @i + 0.2
set @i = @i + 0.2
set @i = @i + 0.2
set @i = @i + 0.2
set @i = @i + 0.2

SELECT 1
WHERE @i=1

As said above, the increment of 0.2 (precise) is different from 0.1 (approximate)

No, that is not the reason. Neither 0.1, nor 0.2 can be represented precisely in binary floating point.
The binary equivalent of 0.2 is 0.000110001100011000110......
The binary equivalent of 0.1 is 0.000011000110001100011......

The real reason is that, with all rounding errors, you only see the effect if it stacks sufficiently. Because the calculation is done with some extra bits, most rounding errors remain invisible, especially becuase in random operations the rounding tend to even out each other. Repeating the same operation means repeating the same rounding error - and that will become visible when repeated often enough.

DECLARE @i float;

set @i = 0;

set @i = @i + 0.2;
set @i = @i + 0.2;
set @i = @i + 0.2;
set @i = @i + 0.2;
set @i = @i + 0.2;

SELECT 1
WHERE @i=1;

set @i = @i + 0.2;
set @i = @i + 0.2;
set @i = @i + 0.2;
set @i = @i + 0.2;
set @i = @i + 0.2;

SELECT 2
WHERE @i=2;

The first SELECT returns a value (as indicated by nisan.al), because the sum of the five rounding errors is so small that it has not yet reached the significant bits for the comparison. The second SELECT does not return a value, so obviously the sum of ten rounding errors does bring it into the significant bits.

A good way to visualize this, is to check the corresponding rounding errors when using decimal notation. Calculate 1/with one decimal place (0.3). Add it three times (0.9). Compare it to the integer 1 (0.9 rounds to 1, so conclusion is: 1 = the sum of three (1/3)).
Now add the same rounded 1/3 fifteen times more (5.4). Compare to the integer 6 (5.4 rounds to 5; not equal, so now the conclusion is that 6 <> the sum of eighteen (1/3)).

The flloating point calculations obviously use lots more siginificant digits, but so do the comparisons; the problem remains the same.

Hugo Kornelis, SQL Server MVP
Visit my SQL Server blog: http://sqlblog.com/blogs/hugo_kornelis
Post #977186
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