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T-SQL SQL 2008
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T-SQL SQL 2008
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bitbucket-25253
bitbucket-25253
Posted Saturday, June 19, 2010 9:21 PM
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Comments posted to this topic are about the item
T-SQL SQL 2008
If everything seems to be going well, you have obviously overlooked something.
Ron
Please help us, help you -before posting a question please
read
Before posting a performance problem please
read
Post #940014
Stewart "Arturius" Campbell
Stewart "Arturius" Campbell
Posted Sunday, June 20, 2010 11:42 PM
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Good question.
These operators are often overlooked and are definately underused.
____________________________________________
Space, the final frontier? not any more...
All limits henceforth are self-imposed.
“libera tute vulgaris ex”
Post #940098
tommyh
tommyh
Posted Sunday, June 20, 2010 11:53 PM
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Personally i would rather see them not being used. You dont really gain anything by using them. Only makes the code harder to read. Saving a few keypresses vs loosing readability is in my mind a bad trade.
Well i would think that you wont gain any performance on it anyway. SQL still have to do the math. But havent been able to test... no SQL2008
Post #940099
honza.mf
honza.mf
Posted Monday, June 21, 2010 12:19 AM
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+= or *= are quite easy, but |= or even ^= wanted a little thinking on bitwise arithmetics.
Nice equestion.
See, understand, learn, try, use efficient
© Dr.Plch
Post #940105
vk-kirov
vk-kirov
Posted Monday, June 21, 2010 12:40 AM
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stewartc-708166 (6/20/2010)
These operators ... are definately underused.
Maybe because they are not compatible with SQL Server 2005 and below?
Post #940110
Hugo Kornelis
Hugo Kornelis
Posted Monday, June 21, 2010 12:47 AM
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Good question, but the explanation is lacking. I think that the +=, -=, etc operators are generally even better known than the bitwise operators used here (even if they were done in SQL-2005 compatible format such as SET @b = @b & 1).
The explanation that should have been there is:
& - bitwise AND. Convert operators to binary, do a bit-by-bit AND operation (result bit is 1 if both input bits are 1), then convert the result back to the result data type. In the example here, 5 AND 1 becomes:
5 = 0101
1 = 0001
---- &
0001 = 1
| - bitwise OR. Convert operators to binary, do a bit-by-bit OR operation (result bit is 1 if either input bit is 1), then convert the result back to the result data type. In the example here, 5 OR 1 becomes:
5 = 0101
1 = 0001
---- |
0101 = 5
^ - bitwise exclusive OR. Convert operators to binary, do a bit-by-bit exclusive OR operation (result bit is 1 if exactly one bit is 1), then convert the result back to the result data type. In the example here, 5 OR 1 becomes:
5 = 0101
1 = 0001
---- ^
0100 = 4
% - modulo. I assume the readers here to be familiar with this operator.
(EDIT - attempt to edit formatting of leading spaces)
Hugo Kornelis, SQL Server MVP
Visit my SQL Server blog:
http://sqlblog.com/blogs/hugo_kornelis
Post #940114
paul.knibbs
paul.knibbs
Posted Monday, June 21, 2010 2:46 AM
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Picture didn't appear on the e-mail--had to come here before it showed up. That happen to anyone else?
Post #940163
cengland0
cengland0
Posted Monday, June 21, 2010 2:55 AM
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I had to guess because I didn't know the ^= operator but I knew all the others. I was lucky that the other three were specific enough that I guessed correctly.
So why doesn't the Explanation discuss the ^= operator? What is that?
Post #940173
Toreador
Toreador
Posted Monday, June 21, 2010 3:00 AM
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I agree with Hugo - I knew the modulus answer, but had no idea what the others were. So got it correct as only one option had the right value for @e :)
Post #940177
Ninja's_RGR'us
Ninja's_RGR'us
Posted Monday, June 21, 2010 5:03 AM
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Toreador (6/21/2010)
I agree with Hugo - I knew the modulus answer, but had no idea what the others were. So got it correct as only one option had the right value for @e :)
Same here, I was instantly 100% sure about the % and low and behold, I didn't have to think any further
.
Post #940235
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