What is the Output?

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What is the Output?

Shiva N

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July 9, 2014 at 11:57 pm

#293269

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Shiva N
Database Consultant
July 10, 2014 at 3:38 am

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Hardy21 SSCrazy Eights Points: 9708 More actions · Answer 1

Hardy21

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July 9, 2014 at 11:58 pm

#1728266

Nice one, thanks for sharing

Thanks

Koen Verbeeck SSC Guru Points: 259105 More actions · Answer 2

Nice question, thanks.

Need an answer? No, you need a question
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Iulian -207023 SSCertifiable Points: 7639 More actions · Answer 3

Iulian -207023

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July 10, 2014 at 5:26 am

#1728329

thanks for this

Raghavendra Mudugal SSChampion Points: 10658 More actions · Answer 4

Thank you for the post, very good one.

ww; Raghu
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The first and the hardest SQL statement I have wrote- "select * from customers" - and I was happy and felt smart.

Gazareth One Orange Chip Points: 27737 More actions · Answer 5

Nice Q, thanks. My logic on data type promotion was right, but my arithmetic was lacking 🙂

Ed Wagner SSC Guru Points: 287026 More actions · Answer 6

And here I didn't count the declare as line 1. DOH!

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rhythmk SSCertifiable Points: 7271 More actions · Answer 7

Thanks for question Shiva.

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stephen.long.1 SSCrazy Points: 2582 More actions · Answer 8

stephen.long.1

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July 10, 2014 at 7:34 am

#1728375

Good question, Shiva. Thanks!

TomThomson SSC Guru Points: 104773 More actions · Answer 9

Good question, and teh correct answer is right (unlike quite a few QotDs). But I have a niggle about the explanation.

Clearly the conversion of @intB to bigint in line 4 is not done explicitly as implied by the explanation: there is no cast or conversion operator that applies to @intB. In fact in line 4, @intB is automatically converted to bigint (implicit conversion caused because bigint comes before int in type precedence order) so there certainly is an automatic promotion from int to bigint.

What's happening in line 7 is that the first calculation is @intB+1 because in the absence of brackets expressions are evaluated from left to right except when this is overridden by operator precedence and here there are no brackets and both operators are + so they have the same precedence, and there is no bigint element in that subexpression to cause implicit conversion from int, so the calculation is attempted as an int calculation which fails because the result would not be a valid int.

Not really a technical error, just bad/misleading phrasing.

Tom

Umer Akram SSC-Insane Points: 22209 More actions · Answer 10

Umer Akram

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July 11, 2014 at 4:07 am

#1728661

Nice question, Thanks for sharing

Hany Helmy SSChampion Points: 13921 More actions · Answer 11

Good back to basics, thanx a lot 🙂

Thanks & Best Regards,
Hany Helmy
SQL Server Database Consultant

Ashok Teotia SSCrazy Points: 2407 More actions · Answer 12

Ashok Teotia

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August 20, 2014 at 7:40 am

#1739577

Nice Q? However I did not get it right.

mailtonareshchowdary SSC Enthusiast Points: 160 More actions · Answer 13

Nice question.. Good to know that following is not failing.. Order does matter

DECLARE @intA INT = 2147483647, @intB INT = -2147483648, @intC INT --1

SELECT @intC = @intB + 1 + @intA --6