

Grasshopper
Group: General Forum Members
Last Login: Tuesday, June 25, 2013 10:32 AM
Points: 11,
Visits: 23


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




SSCEnthusiastic
Group: General Forum Members
Last Login: Monday, August 18, 2014 4:04 AM
Points: 176,
Visits: 567


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.




SSCommitted
Group: General Forum Members
Last Login: Tuesday, June 21, 2016 2:43 AM
Points: 1,805,
Visits: 6,103


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.




Ten Centuries
Group: General Forum Members
Last Login: Monday, March 23, 2015 11:04 AM
Points: 1,050,
Visits: 8,664


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




SSCommitted
Group: General Forum Members
Last Login: Tuesday, June 21, 2016 2:43 AM
Points: 1,805,
Visits: 6,103


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?




Ten Centuries
Group: General Forum Members
Last Login: Monday, March 23, 2015 11:04 AM
Points: 1,050,
Visits: 8,664


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




Right there with Babe
Group: General Forum Members
Last Login: Thursday, December 31, 2015 11:30 AM
Points: 736,
Visits: 414


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 )!




Valued Member
Group: General Forum Members
Last Login: Monday, May 30, 2016 6:01 AM
Points: 72,
Visits: 164


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




Hall of Fame
Group: General Forum Members
Last Login: Today @ 12:50 AM
Points: 3,206,
Visits: 3,052


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)




SSCertifiable
Group: General Forum Members
Last Login: Today @ 12:31 AM
Points: 7,646,
Visits: 10,615


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=1As 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



