As Kimberly blogged about earlier this year, SQLskills has an ongoing initiative to blog about basic topics, which we’re calling SQL101. We’re all blogging about things that we often see done incorrectly, technologies used the wrong way, or where there are many misunderstandings that lead to serious problems. If you want to find all of our SQLskills SQL101 blog posts, check out

An interesting situation was discussed online recently which prompted me to write this post. A fellow MVP was seeing periodic corruption messages in his error log, but DBCC CHECKDB on all databases didn’t find any corruptions. A subsequent restart of the instance caused the problem to go away.

My diagnosis? Memory corruption. Something had corrupted a page in memory – maybe it was bad memory chips or a memory scribbler (something that writes into SQL Server’s buffer pool, like a poorly-written extended stored procedure), or maybe a SQL Server bug. Whatever it was, restarting the instance wiped the buffer pool clean, removing the corrupt page.

So why didn’t DBCC CHECKDB encounter the corrupt page?

The answer is to do with DBCC CHECKDB‘s use of database snapshots (and all other DBCC CHECK* commands). It creates a database snapshot and then runs the consistency-checking algorithms on the database snapshot. The database snapshot is a transactionally-consistent, unchanging view of the database, which is what DBCC CHECKDB requires.

More info on DBCC CHECKDB’s use of snapshots, and potential problems can be found at:

A database snapshot is a separate database as far as the buffer pool is concerned, with its own database ID. A page in the buffer pool is owned by exactly one database ID, and cannot be shared by any other databases. So when DBCC CHECKDB reads a page in the context of the database snapshot, that page must be read from the source database on disk; it cannot use the page from the source database if it’s already in memory, as that page has the wrong database ID.

This means that DBCC CHECKDB reads the entire source database from disk when it uses a database snapshot. This is not a bad thing.

This also means that if there’s a page in the source database that’s corrupt in memory but not corrupt on disk, DBCC CHECKDB will not encounter it if it uses a database snapshot (the default).

If you suspect that a database has some corruption in memory, the only way to have DBCC CHECKDB use the in-memory pages, is to use the WITH TABLOCK option, which skips using a database snapshot and instead uses locks to quiesce changes in the database.

Hope this helps clear up any confusion!