## SET ROWCOUNT and table variable

 Author Message mm_sharma82 Grasshopper Group: General Forum Members Points: 11 Visits: 23 following script doesn't go in infinite loopDECLARE @i float,@rc intset @i = 0while @i <> 1begin set @i = @i + 0.8 set @i = @i + 0.1 set @i = @i + 0.1 print @iendBut below script is going in infinite loopDECLARE @i float,@rc intset @i = 0while @i <> 1begin set @i = @i + 0.7 set @i = @i + 0.1 set @i = @i + 0.1 set @i = @i + 0.1 print @iendCan any one know the answer on this??RegardsManmohan Open Minded SSC Eights! Group: General Forum Members Points: 908 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. paul.knibbs SSCertifiable Group: General Forum Members Points: 7916 Visits: 6240 Gopinath Srirangan (8/27/2010)Hi,But another sample query as shown below gives expected result which contradicts from while loop.DECLARE @i floatset @i = .9set @i = @i + .1You'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. Steve Eckhart Ten Centuries Group: General Forum Members Points: 1428 Visits: 8664 Actually, we're doing the exact same thing here. This was the original query:`while @i <> 1begin declare @a table(a int) set @rc = @i + 0.9 set rowcount @rc insert into @a select id from sysobjects set @i = @i + 0.1end`So, there was only one floating point addition which made @i<>1 TRUE, but in the if statement @i<>1 is FALSE. Steve Eckhart paul.knibbs SSCertifiable Group: General Forum Members Points: 7916 Visits: 6240 Steve Eckhart (8/27/2010)Actually, we're doing the exact same thing here. This was the original query:`while @i <> 1begin declare @a table(a int) set @rc = @i + 0.9 set rowcount @rc insert into @a select id from sysobjects set @i = @i + 0.1end`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? Steve Eckhart Ten Centuries Group: General Forum Members Points: 1428 Visits: 8664 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 Michael Poppers UDP Broadcaster Group: General Forum Members Points: 1485 Visits: 416 mccork (8/25/2010)The answer was almost obvious from the "do not run it on production server" recommendation.<>Agreed (and frankly, I wouldn't have chosen the right answer otherwise :-))! nisan.al SSC-Enthusiastic Group: General Forum Members Points: 116 Visits: 164 I can't say i completely understand this issue.The while statement dont not stop because @i will never be equal to 1So how come the following select return 1 ?DECLARE @i floatset @i = 0set @i = @i + 0.2set @i = @i + 0.2set @i = @i + 0.2set @i = @i + 0.2set @i = @i + 0.2SELECT 1WHERE @i=1 Carlo Romagnano SSChampion Group: General Forum Members Points: 12303 Visits: 3514 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 1So how come the following select return 1 ?DECLARE @i floatset @i = 0set @i = @i + 0.2set @i = @i + 0.2set @i = @i + 0.2set @i = @i + 0.2set @i = @i + 0.2SELECT 1WHERE @i=1As said above, the increment of 0.2 (precise) is different from 0.1 (approximate) Hugo Kornelis SSC-Dedicated Group: General Forum Members Points: 34189 Visits: 13108 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 1So how come the following select return 1 ?DECLARE @i floatset @i = 0set @i = @i + 0.2set @i = @i + 0.2set @i = @i + 0.2set @i = @i + 0.2set @i = @i + 0.2SELECT 1WHERE @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 1WHERE @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 2WHERE @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/Data Platform MVP (2006-2016)Visit my SQL Server blog: http://sqlblog.com/blogs/hugo_kornelis