PHYData DBA (5/1/2014)
TomThomson (5/1/2014)
gbritton1 (5/1/2014)
Before I answered this question, I readIt reads, in part:
The hash join has two inputs: the build input and probe input. The query optimizer assigns these roles so that the smaller of the two inputs is the build input.
then later:
The hash join first scans or computes the entire build input and then builds a hash table in memory.
I'm having trouble reconciling the official description with the "correct" answer.
Without looking anything up, I answered smallest (ie build) input first. When I saw the answer, I constructed a test (using SQL Server 2012) to see what happened. Given one table with 65537 rows and another with 393222 rows (the sizes of teh test tables I built - admittedly not very big, but big enough to refute a silly "correct" answer) I found that whether I wrote the join so that the small table was the first or second referenced in the select clause and whether the small table was teh first or second listed in teh from clause, the actual (not estimated but actual) execution plan took the smaller table as first (ie build) table.
So I don't just have trouble reconciling the official correct answer with the documentation, I have clear and solid experimantal evidence that the officially correct answer is just plain wrong.
Tom - See my earlier post... there is no need for the answer to be wrong ( run test using HASH join hint) to match with your tests so far. When writing a hash join using the Hint you put the smaller to the right to make it the build table. Try you experiment with the Hash Join hint and see what happens.
Your experiment proves what the Query Optimizer does, but not how it is done.
Of course I would not know about any of this if I had not learned all of this by having to work with an Execution Plan that for some reason had chosen the Larger (2 million records) set as the Build list years ago.
I suppose it depends on what the answer means by "first". If it means the "the table which ir processed first in order to build the the hash table against which rows from teh other table will be tested" then " that's first" is a pointless tautology. It's not at all clear to me that any sane person can take it as meaning that. If on the other hand it means that the table used as build table can be any of the tables involved, regardless of size, which is the only sensible interpretation of the words given that (a) "is the one that is used first used first?" is not a sensible question and (b) the statement "Size does not matter for this operator" in the explanation it seems clear both that the BOL documentation quoted by gbritton1 indicates that the "correct" answer is wrong fpr SQL 2008R2 (not a terribly interesting argument, given how often BOL gets it wrong, but in fact - see below - BOL didn't go wrong here) and that the experiment I undertook after seeing this crazy answer proved conclusively that the "correct" answer is wrong for SQL 2012. It may of course be different for SQL 2014, but there are several releases currently in full support and we can see that it is wrong for the majority of them (I ran the experiment for 2008 R2 as well before composing this reply, and this is not one of the places where BOL got it wrong).
So we have to work out what Jason (author of the referenced document) meant by "first(top)" and "second(bottom)". Jason doesn't usually get things wrong. I think he was referring to the layout of the operators in the query plan in the standard query plan display provided by SSMS - certainly that's what I thought he meant when I read it a month or two ago. There's no imaginable way that today's question could be interpreted so that "first" mant that, so I think that Steve has misunderstood what Jason was saying and produced a question and answer based on that misunderstanding.
Tom