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Round

Pradyothana Shastry

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September 16, 2010 at 10:15 pm

#238405

Comments posted to this topic are about the item Round

- Pradyothana http://www.msqlserver.com

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tommyh SSCertifiable Points: 6302 More actions · Answer 1

Nice question. Though the explanation isnt entirely accurate. Its not rounded, its truncated.

If the question had been, lets say

declare @r decimal(5,2)

set @r= round(5/3.2,1,3)

select @r

Then we would get 1.5 wheras a round would have given us 1.6 (change last parameter in Round to 0... or remove it all together).

SQLRNNR SSC Guru Points: 281334 More actions · Answer 2

thanks for the question

Jason...AKA CirqueDeSQLeil
_______________________________________________
I have given a name to my pain...MCM SQL Server, MVP
SQL RNNR
Posting Performance Based Questions - Gail Shaw[/url]
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honza.mf SSCertifiable Points: 5519 More actions · Answer 3

tommyh (9/16/2010)
Nice question. Though the explanation isnt entirely accurate. Its not rounded, its truncated.
If the question had been, lets say
declare @r decimal(5,2)
set @r= round(5/3.2,1,3)
select @r
Then we would get 1.5 wheras a round would have given us 1.6 (change last parameter in Round to 0... or remove it all together).

Right, the 3rd parameter of round is the most important one. It's value 3 is not typical but greater than zero.

See, understand, learn, try, use efficient
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Koen Verbeeck SSC Guru Points: 259085 More actions · Answer 4

Good question, but a real easy one since all the rounding questions of Hugo Kornelis. 🙂

It would've been more a challenge if you used 3.2 or 3.0, as someone already suggested in this thread.

Need an answer? No, you need a question
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MCSE Business Intelligence - Microsoft Data Platform MVP

cengland0 SSCertifiable Points: 6102 More actions · Answer 5

I got it right but hesitated to click the submit button. It was too easy and I was looking for the tricky part of the question. I've been bitten by so many QOTDs with an obscure typo or something else tricky.

Daniel Bowlin SSC-Dedicated Points: 34566 More actions · Answer 6

Good question. Round is a function with a few gotchas that it is important to know about and understand. Thanks.

Dave62 SSCertifiable Points: 7459 More actions · Answer 7

cengland0 (9/17/2010)
... I've been bitten by so many QOTDs with an obscure typo or something else tricky.

I like composing questions that are on the tricky side. Must be because of all my bad experiences with certification exams.

If it's of any consolation, the editors here do have a limit to what they will tolerate for trick questions. I submitted one a few months ago that never got published.

I get it. The QOTDs are more about the learning experience than the certification experience.

I'm working on one now that's a little more tame... 🙂

hrvoje.piasevoli Ten Centuries Points: 1377 More actions · Answer 8

The hardest part of the question was figuring out what is the result of 5/3.1

IMHO the point of the question would make more sense if it had been changed to 5/3.0 as the result of the function would actually be affected by the third parameter of the ROUND function unlike in this question.

To illustrate:

select

Example = 'Your example'

,input_data = '5/3.1'

,Truncation = ROUND(1.612903, 1, 1) -- 1.600000

,Rounding = ROUND(1.612903, 1, 0) -- 1.600000

union all

select

'Better example'

,'5/3.0'

,ROUND(1.666666,1,1) -- 1.600000

,ROUND(1.666666,1,0) -- 1.700000

Example input_data Truncation Rounding

-------------- ---------- ------------ ----------

Your example 5/3.1 1.600000 1.600000

Better example 5/3.0 1.600000 1.700000

Regards,

Hrvoje Piasevoli

john.arnott SSChampion Points: 11882 More actions · Answer 9

honza.mf (9/17/2010)
tommyh (9/16/2010)
Nice question. Though the explanation isnt entirely accurate. Its not rounded, its truncated.
If the question had been, lets say
declare @r decimal(5,2)
set @r= round(5/3.2,1,3)
select @r
Then we would get 1.5 wheras a round would have given us 1.6 (change last parameter in Round to 0... or remove it all together).
Right, the 3rd parameter of round is the most important one. It's value 3 is not typical but greater than zero.

Just to be picky, it's not because the third parameter is greater than zero, but because it is not equal to zero. Here's the entry in BOL for SQL 2005:

Syntax
ROUND ( numeric_expression , length [ ,function ] )
Arguments
numeric_expression
Is an expression of the exact numeric or approximate numeric data type category, except for the bit data type.
length
Is the precision to which numeric_expression is to be rounded. length must be tinyint, smallint, or int. When length is a positive number, numeric_expression is rounded to the number of decimal positions specified by length. When length is a negative number, numeric_expression is rounded on the left side of the decimal point, as specified by length.
function
Is the type of operation to perform. function must be tinyint, smallint, or int. When function is omitted or has a value of 0 (default), numeric_expression is rounded. When a value other than 0 is specified, numeric_expression is truncated.

niki-659689 Valued Member Points: 57 More actions · Answer 10

niki-659689

Valued Member

Points: 57

September 20, 2010 at 6:42 am

#1223966

Very Nicely Explained!!

Thanks.

terrykzncs Ten Centuries Points: 1324 More actions · Answer 11

Thanks for the question.

I hardly use math functions at work, so its sometimes a shock that these simple functions can be so tricky..... A misunderstood (or missing) parameter would mean that the calculations would be inaccurate:w00t: